MCQ 11 Mark
Which of the following is the example of a reversible reaction?
- A
$\text{Pb(NO}_3)_2(\text{aq})+2\text{NaI(aq)}\xrightarrow{\ \ \ \ \ \ \ \ }\text{PbI}_2(\text{s})+2\text{NaNO}_3(\text{aq})$
- B
$\text{2Na(s)}+2\text{H}_2\text{O}(\text{l})\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{2NaOH}(\text{aq})+\text{H}_2(\text{g})$
- C
$\text{AgNO}_3(\text{aq})+\text{HCl}(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ }\text{AgCl}(\text{s})+\text{HNo}_3(\text{aq})$
- ✓
$\text{KNO}_3(\text{aq})+\text{NaCl(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{KCl(aq)}+\text{NaNO}_3\text{(aq)}$
AnswerCorrect option: D. $\text{KNO}_3(\text{aq})+\text{NaCl(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{KCl(aq)}+\text{NaNO}_3\text{(aq)}$
View full question & answer→MCQ 21 Mark
What is $\ce{pH}$ of resulting solution when equal volume when equal of $0.1M ,\ce{NaOH}$ and $0.01M, \ce{HCl}$ are mixed? $[\log 4.5 = 0.65]$
- A
$7$
- B
$1.04$
- ✓
$12.65$
- D
$2.0$
AnswerCorrect option: C. $12.65$
Number of mole of $[\text{OH}^-]=\frac{0.1}{2}=0.05$ on mixing
$[\text{H}^+]=\frac{0.01}{2}=0.005$
No. of mole of $\ce{[OH}^-] $ left $= 0.05 - 0.005 = 0.045$
$\text{pOH} = -\log 4.5 \times 10-2$
$= -0.65 + 2.00 = 1.35$
$\ce{pH } = 14 - 135 = 12.65$
View full question & answer→MCQ 31 Mark
The $\ce{pH}$ of neutral water at $25^\circ C$ is $7.0.$ As the temperature increases, ionisation of water increases, however, the concentration of $\ce{H}^+$ ions and $\ce{OH}^-$ ions are equal. What will be the $\ce{pH}$ of pure water at $60^\circ C\ ?$
- A
Equal to $7.0$
- B
Greater than $7.0$
- ✓
Less than $7.0$
- D
AnswerCorrect option: C. Less than $7.0$
As $K_w$ increases $[\text{H}^+][\text{OH}^-] > 10^{-14}$
As $[\text{H}^+]=[\text{OH}^-]$
or $[\text{H}^+]^2=10^{-14}$
or $[\text{H}^+]>10^{-7}\text{M}$
$\ce{pH } < 7$
View full question & answer→MCQ 41 Mark
$2\text{NO}_2(\text{g})\rightleftharpoons\text{N}_2\text{O}_4(\text{g})+60.0\text{KJ,}$ the increase in temperature :
AnswerCorrect option: B. Favour the decomposition of $\ce{N_2O_4}$.
It will favour backward reaction because process is exothermic.
View full question & answer→MCQ 51 Mark
Which is not a buffer solution :
- A
$ \mathrm{NH}_4 \mathrm{Cl}+\mathrm{NH}_4 \mathrm{OH} $
- B
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa} $
- ✓
$ \mathrm{CH}_3 \mathrm{COONH}_4$
- D
$ \mathrm{NH}_4 \mathrm{NO}_3$
AnswerCorrect option: C. $ \mathrm{CH}_3 \mathrm{COONH}_4$
A buffer solution either is a mixture of a weak acid and its salt with strong base or a mixture of a weak base and its salt with strong acid.
Hence, clearly $ \mathrm{CH}_3 \mathrm{COONH}_4$
is not a buffer solution.
View full question & answer→MCQ 61 Mark
What will be the correct order of vapour pressure of water, acetone and ether at $30^\circ C$. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?
- A
Water $ < $ ether $ < $ acetone.
- ✓
Water $ < $ acetone $ < $ ether.
- C
Ether $ < $ acetone $ < $ water.
- D
Acetone $ < $ ether $ < $ water.
AnswerCorrect option: B. Water $ < $ acetone $ < $ ether.
Greater the boiling point, less is the vapour pressure.
Hence, the correct order of vapour pressures will be:
water $ < $ acetone $ < $ ether.
View full question & answer→MCQ 71 Mark
The solubility of $\ce{CO_2}$ in water increases with :
- A
- B
Reduction of gas pressure.
- ✓
Increase in gas pressure.
- D
AnswerCorrect option: C. Increase in gas pressure.
When carbon dioxide reacts with water it forms carbonic acid. Increasing the pressure of carbon dioxide makes the reaction feasible in the forward direction and hence solubility of $\ce{CO_2}$ increases.
View full question & answer→MCQ 81 Mark
$\ce{H_2O}$ is a:
- A
- B
- ✓
Both proton donor and proton acceptor.
- D
Neither proton donor, nor proton acceptor.
AnswerCorrect option: C. Both proton donor and proton acceptor.
A compound such as water $\ce{(H_2O)}$ has many interesting properties. Water molecules can accept a proton to act as Bronsted $-$ Lowry bases in certain circumstances. The following is an example of $\text{HCl}$ dissolving in water :
$\text{HCl}+\text{H}_2\text{O}(\ell)\rightarrow\text{H}_3\text{O}^+_\text{(aq)}+\text{C}1^-_\text{(aq)}$
The another possibility is water can act like a Bronsted $-$ Lowry acid by donating a proton. Water donates a proton to a proton $-$ accepting amide ion in the presence of ammonia, resulting in the following product:
$\text{H}_2\text{O}\ell{(ℓ)}+\text{NH}^-_2\text{(aq)}\rightarrow\text{OH}^-_{\text{(aq)}}+\text{NH}_3\text{(aq)}$
View full question & answer→MCQ 91 Mark
$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \leftrightharpoons 2 \mathrm{SO}_3(\mathrm{~g})+\mathrm{QkJ}$ In the above reaction, how can the yield of product be increased without increasing the pressure?
- A
By increasing temperature.
- ✓
By decreasing temperature.
- C
By increasing the volume of the reaction vessel.
- D
By the addition of the catalyst.
AnswerCorrect option: B. By decreasing temperature.
Since the reaction is exothermic in the forward direction, a decrease in temperature favours the forward reaction.
View full question & answer→MCQ 101 Mark
On increasing the pressure, in which direction will the gas phase reaction proceed to re $-$ establish equilibrium, is predicted by applying the Le Chatelier’s principle. Consider the reaction.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$
Which of the following is correct, if the total pressure at which the equilibrium is established, is increased without changing the temperature?
AnswerCorrect option: A. $K$ will remain same.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$
According to Le Chatelier’s principle, at constant temperature, the equilibrium composition will change but $K$ will remain same.
View full question & answer→MCQ 111 Mark
The solubility product of a sparingly soluble salt $AB$ at room temperature is $1.21 \times 10^{-6},$ its molar solubility is :
- A
$1.21 \times 100M.$
- B
$1.1 \times 10^{-4}M.$
- ✓
$1.1 \times 10^{-3}M.$
- D
AnswerCorrect option: C. $1.1 \times 10^{-3}M.$
$\text{S}=\sqrt{\text{K}_{\text{sp}}}$
$=\sqrt{1.21\times10^{-6}}$
$=1.1\times10^{-3}\text{M.}$
View full question & answer→MCQ 121 Mark
The solubility of $\text{AgI}$ in $\text{NaI}$ solutions is less than that in pure water because :
Answer$ \mathrm{AgI} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{I}^{-}$
$ \mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{I}^{-}$
Sodium iodide is a strong electrolyte and is completely dissociated.
This increases the iodide ion concentration in solution and suppresses the ionization of $\text{AgI}.$
View full question & answer→MCQ 131 Mark
Addition of water to this solution will not change $[\mathrm{H}_3 \mathrm{O}^{+}]$.
AnswerAddition of water to $X$ solution does not change the $\ce{pH}$ of the solution, which means the concentration of the species present changes in such a way that the $\ce{pH}$ remains the same. This is possible for acid/base buffer.
$\text{pH}=\text{pk}_\text{a}+\log\frac{[\text{acid}]}{[\text{salt}]}$
Since the volume changes the same for both salt and acid, the ratio $\frac{\text{[salt]}}{\text{[acid]}}$remains the same and hence $\ce{pH}$ also remains the same. Same in the case of basic buffer solution.
View full question & answer→MCQ 141 Mark
The concentration of hydrogen ion in a sample of soft drink is $3.8 \times 10^{-3}M$. What is its $\ce{pH}\ ?$
- A
$4.32$
- B
$5.12$
- C
$3.31$
- ✓
$2.42$
AnswerCorrect option: D. $2.42$
View full question & answer→MCQ 151 Mark
Buffer solution can be obtained by mixing aqueous solution of $ .........$
- A
Sodium acetate and excess of $\text{HCl}.$
- ✓
Sodium acetate and acetic acid.
- C
Sodium chloride and $\text{HCl}.$
- D
Acetic acid and excess of $\text{NaOH}.$
AnswerCorrect option: B. Sodium acetate and acetic acid.
A mixture of acetic acid $($a weak acid$)$ and sodium acetate $($its salt with strong base sodium hydroxide$)$ acts as acidic buffer.
Option $A, C$ and $D$ does not contain weak acid/ weak base and its salt thus it does not form buffer solution.
View full question & answer→MCQ 161 Mark
Which of the following is not a general characteristic of equilibria involving physical processes?
- A
Equilibrium is possible only in a closed system at a given temperature.
- B
All measurable properties of the system remain constant.
- ✓
All the physical processes stop at equilibrium.
- D
The opposing processes occur at the same rate and there is dynamic but stable condition.
AnswerCorrect option: C. All the physical processes stop at equilibrium.
All the physical processes like melting of ice and freezing of water, etc. do not stop at equilibrium.
View full question & answer→MCQ 171 Mark
What will be $\ce{pH}$ of $0.01M , \ce{CH_3COOH}\ ? (K_a = 1.80 \times 10^{-5})$
Answer$\text{pH}=-\log(\text{H}^+)$
$=-\log\sqrt{\text{K}_1\times\text{C}}$
$=-\log\sqrt{1.80\times10^{-5}\times10^{-2}}$
$=-\log\sqrt{18\times10^{-8}}$
$=-\log4.24=0.62$
$\text{pH}=-\log4.24-\log10^{-4}$
$=0.62+4.00=3.38\simeq3.4$
View full question & answer→MCQ 181 Mark
At a temperature under high pressure $\mathrm{K}_{\mathrm{w}}\left(\mathrm{H}_2 \mathrm{O}\right)=10^{10}$, a solution of $\ce{pH}\ 5.4$ is said to be :
View full question & answer→MCQ 191 Mark
Solubility of calcium phosphate $($molecular mass $,M)$ in water is $\text{Wg}$ per $100mL$ at $25^\circ C$. Its solubility product at $25^\circ C$ will be approximately.
- A
$10^7\Big(\frac{\text{W}}{\text{M}}\Big)^3$
- ✓
$10^7\Big(\frac{\text{W}}{\text{M}}\Big)^5$
- C
$10^5\Big(\frac{\text{W}}{\text{M}}\Big)^5$
- D
$10^3\Big(\frac{\text{W}}{\text{M}}\Big)^5$
AnswerCorrect option: B. $10^7\Big(\frac{\text{W}}{\text{M}}\Big)^5$
$\text{S}=\frac{10\text{W}}{\text{M}}\text{moL/ L}$
$\text{K}_{\text{sp}}$ of $\text{Ca}_3(\text{PO}_4)_2=108\text{S}^5$
$=108\Big(\frac{10\text{W}}{\text{M}}\Big)^5$
$=10^7\Big(\frac{\text{W}}{\text{M}}\Big)^5\ ($approximately.$)$
View full question & answer→MCQ 201 Mark
At $363K,$ pure water has$[\mathrm{H}_2 \mathrm{O}^{+}]=10^{-6} \mathrm{M}$. The value of $\mathrm{K}_{\mathrm{w}}$ at this temperature will be :
- A
$10^{-6} $
- ✓
$10^{-12} $
- C
$ 10^{-13} $
- D
$ 10^{-14}$
AnswerCorrect option: B. $10^{-12} $
$[\text{H}_3\text{O}^+]=[\text{OH}^-]=10^{-6}\text{M}$
$\text{K}_{\text{w}}=[\text{H}_3\text{O}^+][\text{OH}^-]$
$=10^{-6}\times10^{-6}=10^{-12}\text{M}.$
View full question & answer→MCQ 211 Mark
A 0.2 molar solution of formic acid is 3.2% ionised. Its ionisation constant is:
- A
$9.6 \times 10^{-3} $
- ✓
$ 2.1 \times 10^{-4} $
- C
$ 1.25 \times 10^{-6} $
- D
$ 4.8 \times 10^{-5} $
AnswerCorrect option: B. $ 2.1 \times 10^{-4} $
$ 2.1 \times 10^{-4} $
Explanation:
$\text{K}=\text{C}\alpha^2=0.2\times(0.032)^2$
$=2.1\times10^{-4}$
View full question & answer→MCQ 221 Mark
For the reaction, $\text{CO(g)}+\text{ClO}_2\text{(g)}\rightleftharpoons\text{COCl}_2(\text{g}),$ the $\frac{\text{K}_{\text{p}}}{\text{K}_\text{C}}$ is equal to :
- A
$\sqrt{\text{RT}}$
- B
$\text{RT}$
- ✓
$\frac{1}{\text{R.T}}$
- D
$1.0$
AnswerCorrect option: C. $\frac{1}{\text{R.T}}$
$\text{K}_{\text{p}}=\text{K}_{\text{C}}(\text{RT})^{\Delta\text{n}}$
$\Rightarrow\text{K}_{\text{p}}=\text{K}_{\text{C}}(\text{RT})^{1-2}$
$\Rightarrow\frac{\text{K}_{\text{p}}}{\text{K}_{\text{C}}}=\frac{1}{\text{RT}}$
View full question & answer→MCQ 231 Mark
$100\ mL$ of a solution contains $0.1M\ \ce{NH_4OH}$ and $0.1M\ \ce{NH_4Cl}.$ The $\ce{pH}$ of the solution will not change on adding :
- A
$20\ mL$ of $0.1M\ \ce{NH_4OH}$ solution.
- B
$20\ mL$ of $0.1M\ \ce{NH_4Cl}$ solution.
- C
$10\ mL$ of $0.1M\ \ce{NaOH}$ solution.
- ✓
$10\ mL$ of distilled water.
AnswerCorrect option: D. $10\ mL$ of distilled water.
$\ce{pH}$ of buffer does not change on adding distilled water as there are no ions present in it.
View full question & answer→MCQ 241 Mark
Le Chatelier's principle is applicable to :
- A
Only homogeneous chemical reversible reaction.
- B
Only heterogeneous chemical reversible reaction.
- C
Only physical equilibria.
- ✓
All systems chemical or physical, in equilibrium.
AnswerCorrect option: D. All systems chemical or physical, in equilibrium.
Le Chatelier’s principle states that When some variables like temperature, pressure, concentration, etc are changed in a system at equilibrium then the system will go in that direction where it can get released from the change.
Le chatelier's principle is only applicable for the compounds at equilibrium.
So, Le Chatelier’s principle is applicable to all systems chemical or physical at equilibrium.
View full question & answer→MCQ 251 Mark
Which one of the following informations can be obtained on the basis of Le $-$ Chatelier principle?
- A
Dissociation constant of a weak acid.
- B
Entropy change in a reaction.
- C
Equilibrium constant of a chemical reaction.
- ✓
View full question & answer→MCQ 261 Mark
The addition of $\ce{HCl}$ will not suppress the ionisation of :
View full question & answer→MCQ 271 Mark
Which of the following is not affected by change in pressure?
- ✓
$ \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NO}(\mathrm{g}) $
- B
$ \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) $
- C
$ \mathrm{PCl}_5(\mathrm{~g}) \Leftrightarrow \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2$
- D
$ 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{SO}_3(\mathrm{~g})$
AnswerCorrect option: A. $ \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NO}(\mathrm{g}) $
For the reaction, N$ \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NO}(\mathrm{g}) $, the value of $\triangle n$ is $2 − (1 + 1) = 0.$ When there is no change in the number of moles of reactants and products, there is no effect on the equilibrium when pressure is changed.
In all other option, the number of mole of reactant and products are different, Hence, if pressure is changed the equilibrium will be affected.
View full question & answer→MCQ 281 Mark
It accepts a proton. It is called as :
AnswerA Bronsted acid can accept a proton $\ce{(H^+)}.$ For example a basic salt, such as $\mathrm{Na}^{\oplus} \mathrm{H}^{\ominus}$, can take up a $\mathrm{H}^{+}$from $\mathrm{H}_2 \mathrm{O}$ to form $\mathrm{OH}^{\ominus}$ ions.
$\mathrm{F}^{\ominus}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HF}(\mathrm{aq})+\mathrm{OH}^{\oplus}(\mathrm{aq})$
View full question & answer→MCQ 291 Mark
$1M\ \ce{ NaCl}$ and $1M\ \ce{ HCl }$ are present in an aqueous solution . The solution is :
- ✓
Not a buffer solution with $\ce{pH } < 7$
- B
Not a buffer solution with $\ce{pH } > 7$
- C
A buffer solution with $\ce{pH } < 7$
- D
A buffer solution with $\ce{pH } > 7$
AnswerCorrect option: A. Not a buffer solution with $\ce{pH } < 7$
Buffer can accept and donate protons at the same time and $\ce{HC_1}$ is an acid.
So, it has $\ce{pH } < 7.$
So, this is not a buffer and the solution will be acidic.
View full question & answer→MCQ 301 Mark
Which buffer solution comprising of the following has its $\ce{pH }$ value greater than $7\ ?$
- A
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa} $
- B
$ \mathrm{HCOOH}+\mathrm{HCOOK}$
- C
$ \mathrm{CH}_3 \mathrm{COONH}4 $
- ✓
$ \mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl} $
AnswerCorrect option: D. $ \mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl} $
View full question & answer→MCQ 311 Mark
Production of ammonia according to the reaction,
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})2\text{NH}_3(\text{g}); \Delta\text{H}=-92.38\text{kJ mol}^{-1}$ is an exothermic process. At low temperature, the reaction shifts in :
- ✓
- B
- C
Either forward or backward direction.
- D
View full question & answer→MCQ 321 Mark
The equibrium constant $\ce{K_c}$ for the reaction : $\text{H}_2(\text{g})+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}(\text{g})\text{ at }700\text{K}\text{ is }49$ The $'K\ '$ for reaction $\text{HI(g)}\rightleftharpoons\frac{1}{2}\text{H}_2(\text{g})+\ \frac{1}{2}\text{I}_2(\text{g}).$
- A
$49$
- B
$0.02$
- ✓
$\frac{1}{7}$
- D
$1.43$
AnswerCorrect option: C. $\frac{1}{7}$
$\text{K}=\frac{[\text{KI}^2]}{[\text{H}_2][\text{I}_2]}=49$
$\Rightarrow\text{K}'=\frac{[\text{H}_2]^{\frac{1}{2}}[\text{I}^2]^{\frac{1}{2}}}{[\text{HI}]}$
$=\frac{1}{\sqrt{\text{K}}}=\frac{1}{\sqrt{49}}=\frac{1}{7}$
View full question & answer→MCQ 331 Mark
The dissociation constant of water is represented by $\text{K}=\frac{[\text{H}_3\text{O}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$$\text{or }=[\text{H}^+][\text{OH}^-]\text{K}_{\text{w}}$ is called:
- A
- ✓
- C
Ionisation constant of water.
- D
Ionisation constant of acid and base.
AnswerThe dissociation constant is represented by,
$\text{K}=\frac{[\text{H}_3\text{O}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$
The concentration of water is omitted from the denominator as water is a pure liquid and its concentration remains constant. $\ce{[H_2O)}$ is incorporated within the equilibrium constant to give a new constant, $K_w$, which is called the ionic product of water,
$\text{K}_{\text{w}}=[\text{H}^+][\text{OH}^-]$
View full question & answer→MCQ 341 Mark
The chemical equilibrium of reversible reaction is not influenced by:
- A
- ✓
- C
Concentration of the reactants.
- D
AnswerChemical equilibrium of reversible reaction is not influenced by the catalyst. Catalyst decreases the activation barrier for a reaction so the reaction proceeds fast. In the presence of the catalyst, the equilibrium reaches faster but it doesn't affect the thermodynamic properties.
View full question & answer→MCQ 351 Mark
At $55^{\circ}C$, autoprotolysis constant of water is $4 × 10^{-14}$. If a given sample of water has a $pH$ of $6.9$, then it is:
View full question & answer→MCQ 361 Mark
Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as the:
- ✓
$pH$ scale.
- B
$\ce{pOH}$ scale.
- C
- D
Both $(a)$ and $(b).$
AnswerCorrect option: A. $pH$ scale.
View full question & answer→MCQ 371 Mark
$1\ mL$ of $\frac{\text{N}}{100}\text{HCl} $ is added to $1L$ of buffer having $pH = 5$. The $pH$ of the solution will be:
- A
Become $7$
- B
Become $3$
- C
Become $6$
- ✓
View full question & answer→MCQ 381 Mark
We know that the relationship between $K_c$ and $K_p$ is $\text{K}_\text{p}=\text{K}_\text{c}\text{(RT})^{\Delta\text{n}}$ What would be the value of $\Delta\text{n}$ for the reaction $\text{NH}_4\text{Cl(s)}\rightleftharpoons\text{NH}_3\text{(g)}+\text{HCl(g)}$
AnswerThe relationship between $K_p$ and $K_c$ is
$\text{K}_\text{p}=\text{K}_\text{c}\text{(RT)}^{\Delta\text{n}}$
Where $\Delta\text{n} = ($number of moles of gaseous products$) - ($number of moles of gaseous reactants$)$
For the reaction,
$\text{NH}_4\text{C1(s)}\leftrightharpoons\text{NH}_3\text{(g)}+\text{HCl(g)}$
$\Delta\text{n}=2-0=2$
View full question & answer→MCQ 391 Mark
For the reaction, $\text{H}_2(\text{g})+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}(\text{g}),$ the standard free energy is $\Delta\text{G}^{\ominus}>0$ The equilibrium constant $(K)$ would be:
- A
$K = 0$
- B
$K > 1$
- C
$K = 1$
- ✓
$K < 1$
AnswerCorrect option: D. $K < 1$
View full question & answer→MCQ 401 Mark
Ammonia dissociates to give nitrogen and hydrogen. What happens if the pressure is increased on the system at equilibrium?
- A
- ✓
- C
Backward reaction is exothermic.
- D
Answer$2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g})$
upon increase in pressure the equilibrium will shift to the side of lesser gas molecules i.e. left. and thereby a decrease in number of moles of gas and volume of the system.
View full question & answer→MCQ 411 Mark
$\mathrm{PCl}_5, \mathrm{PCl}_3$ and $\mathrm{Cl}_2$ are at equilibrium at $500\ K$ in a closed container and their concentrations are $0.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}, 1.2 \times 10^{-3}$ $\mathrm{mol} \mathrm{L}^{-1}$ and $1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$ respectively.
The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction $\text{PCl}_5\text{g}\rightleftharpoons\text{PCl}_3\text{(g)}+\text{Cl}_2\text{g}$ will be.
AnswerCorrect option: B. $1.8 \times 10^{-3}$
$\text{PCl}_5\text{g}\rightleftharpoons\text{PCl}_3\text{(g)}+\text{Cl}_2\text{g}$
$\text{K}_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}=\frac{1.2\times10^{-3}\times1.2\times10^{-3}}{0.8\times10^{-3}}$
$[=1.8\times10^{-3}]\text{moL L}^{-1}$
View full question & answer→MCQ 421 Mark
In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?
- A
$\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\rightleftharpoons\text{2HI(g)}$
- B
$\text{PCl}_5\text{(g)}\rightleftharpoons\text{PCl}_3\text{(g)}+\text{Cl}_2\text{(g)}$
- C
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$
- ✓
The equilibrium will remain unaffected in all the three cases.
AnswerCorrect option: D. The equilibrium will remain unaffected in all the three cases.
At constant volume equilibrium remains unaffected with the addition of inert gas.
View full question & answer→MCQ 431 Mark
Buffer Solution is prepared by mixing $.........$
- ✓
Weak acid and its salt of strong base.
- B
Strong acid $+$ its salt of strong base.
- C
Weak acid $+$ its salt of weak base.
- D
Strong base $+$ its salts of strong acid.
AnswerCorrect option: A. Weak acid and its salt of strong base.
A solution that resists change in $pH$ value upon addition of a small amount of strong acid or base $($less than $1 \%)$ or when the solution is diluted is called buffer solution.
An acidic buffer solution consists of a solution of a weak acid and its salt with a strong base. while basic buffer solution consists of a mixture of a weak base and its salt with strong acid.
View full question & answer→MCQ 441 Mark
The solubility product of $\mathrm{CaSO}_4$ is $6.4 \times 10^{-5}$. The solubility of salt in mol $L^{-1}$ is:
- A
$8.10^{-16}$
- ✓
$8.10^{-2}$
- C
$8.10^{-3}$
- D
$1.6^{-3}$
AnswerCorrect option: B. $8.10^{-2}$
$\text{CaSO}_4(\text{s})\rightleftharpoons\text{Ca}^{2+}(\text{aq})+\text{SO}^{2-}_4$
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]$
$\Rightarrow6.4\times10^{-5}\times=\text{s}\times\text{s}$
$\Rightarrow\text{s}^2=\sqrt{64\times10^{-6}}$
$\Rightarrow\text{s}=8\times10^{-3}\text{mol L}^{-1}$
View full question & answer→MCQ 451 Mark
$100\ ml$ of a buffer of $1 \mathrm{MNH}_3(\mathrm{aq})$ and $1 \mathrm{MNH}_4^{+}(\mathrm{aq})$ are placed in two half cells separately. A current of $1.5\ amp$ is passed through both half cells for $20$ min, if electrolysis of only water take place. The $pH$ of :
- ✓
Anode half cell will decrease.
- B
Cathode half cell will decrease.
- C
Both half cell will decrease.
- D
Both half cell remain same.
AnswerCorrect option: A. Anode half cell will decrease.
As only the electrolysis of water take place so the conc. of $H^+$ ion in Anode compartment increases.
View full question & answer→MCQ 461 Mark
- A
$ \mathrm{NH}_4 \mathrm{C} 1+\mathrm{HC} 1 $
- B
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{CO}_3$
- ✓
$40\ ml$ of $\ce{0.1 M NaCN + 20ml}$ of $\ce{0.1 MHCN}$
- D
$\ce{NaC1 + NaOH}$
AnswerCorrect option: C. $40\ ml$ of $\ce{0.1 M NaCN + 20ml}$ of $\ce{0.1 MHCN}$
Buffer solution is a mixture of weak acid and its salt of weak acid.
$\ce{NaCN} \rightarrow$ salt of strong base and weak acid
$\ce{HCN} \rightarrow$ weak acid
$\therefore$ They can act as buffer.
While no other options follow definition of buffer solution.
View full question & answer→MCQ 471 Mark
The product of molar concentrations of hydrogen ions and hydroxide ions in a $0.01\ M$ aqueous solution of sodium chloride is known as:
- A
Hydrolysis constant of salt.
- B
Dissociation constant of acid.
- C
Dissociation constant of base.
- ✓
AnswerSodium chloride is a salt of strong base $\text{NaOH}$ and strong acid $\text{HCl}$. In its aqueous solution, following equilibrium is observed.
$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$
View full question & answer→MCQ 481 Mark
In the gaseous equilibrium $A + 2B ⇌ C +$ Heat, the forward reaction is favoured:
- A
Low $P$, High $T$
- B
Low $P$, Low $T$
- ✓
High $P$, Low $T$
- D
High $P$, High $T$
AnswerCorrect option: C. High $P$, Low $T$
$A + 2B ⇌ C +$ Heat
The equation shows, that it is exothermic reaction.
Since the heat is released in the reaction, so the reaction is favoured in forward direction at low temperature.
$\triangle n = 1 − (2 + 1) = −2$
View full question & answer→MCQ 491 Mark
An equimolar solution of $\ce{NaNO_2}$ and $\ce{HNO_2}$ can act as a:
Answer$\ce{HNO_2}$ is a weak acid.
$\ce{NaNO_2}$ is a salt of weak acid $\ce{HNO_2}$ with strong base $\ce{NaOH}.$
View full question & answer→MCQ 501 Mark
At $500K$, equilibrium constant, $K_c$ , for the following reaction is $5.\ \frac{1}{2}\text{H}_2\text{(g)}+\frac{1}{2}\text{I}_2\text{(g)}\rightleftharpoons\text{HI(g)}$ What would be the equilibrium constant $K_c$ for the reaction $2\text{HI}\text{(g)}\rightleftharpoons\text{H}_2\text{(g)}+\text{I}_2\text{(g)}$
AnswerCorrect option: A. $0.04$
For the reaction,
$\frac{1}{2}\text{H}_2\text{(g)}+\frac{1}{2}\text{I}_2\text{(g)}\rightleftharpoons\text{HI(g)}\ ;\text{K}_\text{c}=5$
For $\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI}\text{(g)}\ ;\text{K}_\text{c}=(5)^2=25$
For $2\text{HI}\text{(g)}\rightleftharpoons\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\ ;\text{K}_\text{c}=\frac{1}{25}=0.04$
View full question & answer→MCQ 511 Mark
For the reactipon, $\text{SO}_2(\text{g})+\frac{1}{2}\text{O}_2\text{(g)}\rightleftharpoons\text{SO}_3(\text{g})$ if $K_p=K_c(R T)^X$ where, the symbols have usual meaning, then the value of $x$ is $($assuming ideality$)$
- A
$-1$
- ✓
$-\frac{1}{2}$
- C
$\frac{1}{2}$
- D
$1$
AnswerCorrect option: B. $-\frac{1}{2}$
View full question & answer→MCQ 521 Mark
$\ce{ {SO}_2 + {O}_2 \rightleftharpoons 2 {SO}_3} +$ Heat The equilibrium reaction proceeds in forward direction by:
- ✓
Addition of $O_2$
- B
Removal of $O_2$
- C
- D
AnswerCorrect option: A. Addition of $O_2$
According to Le$-$Chatelier's principle, equilibrium shift towards forward direction by addition of reactant.
View full question & answer→MCQ 531 Mark
Equal volume of following $Ca^{2+}$ and $F$ solution are mixed. In which of the Solutions will precipitation occur? $[K_{SP}$ of $\ce{CaF_2} = 1.7 × 10^{-10}$
- A
$10^{-2}\text{M Ca}^{2+}+10^{-5}\text{MF}^-$
- B
$10^{-3}\text{M Ca}^{2+}$ and $10^{-3}\text{MF}^-$
- ✓
$10^{-2}\text{M Ca}^{2+}+10^{-3}\text{MF}^-$
- D
$10^{-3}\text{M Ca}^{2+}$ and $10^{-5}\text{MF}^-$
AnswerCorrect option: C. $10^{-2}\text{M Ca}^{2+}+10^{-3}\text{MF}^-$
$\because\text{I.P}=(\text{Ca}^{2+})(\text{F}^-)^2$
$=10^{-2}\times(10^{-3})^2=10^{-8}>\text{K}_{\text{SP}}$
$\therefore$ precipitation will occur.
View full question & answer→MCQ 541 Mark
What will happen when $\ce{CH_3COONa}$ is added to an aqueous solution of $\ce{CH_3COOH}$?
- A
The $pH$ of the solution decreases.
- ✓
The $pH$ of the solution increases.
- C
The $pH$ of the solution remains unaltered.
- D
An acidic salt is produced.
AnswerCorrect option: B. The $pH$ of the solution increases.
$\text{CH}_3\text{COOH}⇋\text{CH}_3\text{COO}^−+\text{H}^+$
If $\ce{CH_3COONa}$ is added, equilibrium will shift backward due to the common ion effect. $H^+$ will decrease.
View full question & answer→MCQ 551 Mark
In the equilibrium, $\text{AB}\rightleftharpoons\text{A}+\text{B}$ if the equilibrium concentration of $A$ is double, then equilibrium concentration of $B$ will be:
- ✓
- B
- C
$\frac{1}{4}^{th}$
- D
$\frac{1}{8}^{th}$
Answer$\text{AB}\rightleftharpoons\text{A}+\text{B}$
or $\text{K}=\frac{[\text{A}][\text{B}]}{[\text{AB}]}$
If concentration of $A$ is doubled, the equilibrium concentration of $B$ becomes half to maintain $K$ constant.
View full question & answer→MCQ 561 Mark
In the melting of ice, which one of the conditions will be more favourable?
- ✓
High temperature and high pressure.
- B
Low temperature and low pressure.
- C
Low temperature and high pressure.
- D
High temperature and low pressure.
AnswerCorrect option: A. High temperature and high pressure.
Since ice melts with the absorption of heat and decreases in volume, hence both temperature and pressure affect the melting of ice. Since the change of ice into water is an endothermic process hence with rising of temperature ice melts into water. Since the volume of ice is more than that of water so an increase of pressure favour melting.
View full question & answer→MCQ 571 Mark
Which of the following is correct regarding buffer sol?
AnswerCorrect option: C. It shows little change in $pH$ on adding small amount of acid or base.
A buffer solution shows little change in $pH$ on adding small amount of acid or base.
View full question & answer→MCQ 581 Mark
The addition of $\text{NaCl}$ to $\text{AgCl}$ decreases the solubility of $\text{AgCl}$ because:
- A
Solubility product decreases.
- ✓
Solubility product remains constant.
- C
Solution becomes unsaturated.
- D
Solution becomes supersaturated.
AnswerCorrect option: B. Solubility product remains constant.
$[Cl^-]$ increases but $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right][\mathrm{Cl}]$remains constant. So, $[Ag^+]$ decreases.
View full question & answer→MCQ 591 Mark
Amongst the following hydroxides, the one which has the highest value of $\text{Ksp}$ at $25^\circ C$ is:
- A
$\text{KOH}$
- ✓
$\text{CsOH}$
- C
$\text{LiOH}$
- D
$\text{RbOH}$
AnswerCorrect option: B. $\text{CsOH}$
Going down Group$-I$ in the periodic nature, the ionic nature of the hydroxides increases. As ionic nature increases, the solubility also increases.
View full question & answer→MCQ 601 Mark
In the formation of nitric acid $N_2$ and $O_2$ are made to combine. Thus,$N_2 + O_2⇌ 2NO−$Heat which of the following condition will favour the formation of $NO$?
View full question & answer→MCQ 611 Mark
The equilibrium $\ce{2SO_2(g) + O_2(g) ⇌ 2SO_3(g)}$ shifts forward if:
AnswerCorrect option: B. An adsorbent is used to remove $SO_3$ as soon as it is formed.
Removal of any product or adding of reactants favours forward reaction i.e.,$ SO_3$ formation. This is according to Le$-$ Chatelier's principle.
View full question & answer→MCQ 621 Mark
The rate of the reaction $\ce{2NO + Cl_2 \rightarrow 2NOCls}$ given by the rate equation. rate $=\ce{ k[NO]^2[Cl_2]}$ The value of the rate constant can be increased by:
AnswerCorrect option: C. Increasing the temperature.
Rate constant is only affected by the temperature, it does not get affected by the increase in concentration of $NO$ and $\ce{Cl_2}$ .
View full question & answer→MCQ 631 Mark
When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer.
$[\text{Co(H}_2\text{O})_6]^{3+}\text{(aq)}+4\text{Cl}^-\text{(aq)}\rightleftharpoons[\text{CoCl}_4]^{2-}\text{(aq)}+6\text{H}_2\text{O (l)}\\ \ \ \ \ \ \ \ ^\text{(pink)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(blue)}$
- ✓
$\Delta\text{H}>0$ for the reaction.
- B
$\Delta\text{H}<0$ for the reaction.
- C
$\Delta\text{H}=0$ for the reaction.
- D
The sign of $\Delta\text{H}$ cannot be predicted on the basis of this information.
AnswerCorrect option: A. $\Delta\text{H}>0$ for the reaction.
On cooling the mixture reaction moves towards backwards direction it means it is an endothermic reaction i.e., $\Delta\text{H}>0.$
View full question & answer→MCQ 641 Mark
Which of the following has maximum $pK_a$:
- A
$ \mathrm{CH}_2 \mathrm{FCOOH}$
- B
$\mathrm{CH}_3 \mathrm{ClCOOH} $
- ✓
$\mathrm{CH}_3 \mathrm{COOH} $
- D
$ \mathrm{HCOOH} $
AnswerCorrect option: C. $\mathrm{CH}_3 \mathrm{COOH} $
View full question & answer→MCQ 651 Mark
In the following reaction, $\begin{matrix}\text{H}_2\text{O}(\text{l})&+&\text{H}_2\text{O}(\text{l})&\rightleftharpoons&\text{H}_3\text{O}^+(\text{aq})&+&\text{OH}^-(\text{aq})\\\text{Acid}&&\text{ X}&&\text{Conjugate Acid}&&\text{Y}\end{matrix} X$ and $Y$ respectively are:
View full question & answer→MCQ 661 Mark
$\text{A}+\text{B}\rightleftharpoons\text{C}+\text{D}.$ If the concentration of $A$ and $B$ are equal at equilibrium and concentration of $D$ will be twice that of $A$, then what will be the equilibrium constant of the reaction?
- ✓
$4$
- B
$6$
- C
$\frac{4}{5}$
- D
$\frac{6}{7}$
View full question & answer→MCQ 671 Mark
Which of the following options will be correct for the stage of half completion of the reaction $\text{A}\rightleftharpoons\text{B}.$
- ✓
$\Delta\text{G}^\ominus=0$
- B
$\Delta\text{G}^\ominus>0$
- C
$\Delta\text{G}^\ominus<0$
- D
$\Delta\text{G}^\ominus=-\text{RT}$ ln $2$
AnswerCorrect option: A. $\Delta\text{G}^\ominus=0$
$\text{A}\rightleftharpoons\text{B}$
$\Delta\text{G}^\ominus=-\text{RT}$ ln $K$
At the stage of half completion of reaction $[A] = [B],$
Therefore, $K = 1$. Thus, $\Delta\text{G}^\ominus=0$
View full question & answer→MCQ 681 Mark
Strong acid dissociates completely in water, the resulting base formed would be very weak. The reason is that:
- A
Strong acids have strong conjugate bases.
- B
Strong acids have strong conjugate acids.
- ✓
Strong acids have very weak conjugate bases.
- D
Strong acids have very weak conjugate acids.
AnswerCorrect option: C. Strong acids have very weak conjugate bases.
View full question & answer→MCQ 691 Mark
In the equilibrium reaction, $\mathrm{N}_2+3 \mathrm{H}_2 \Leftrightarrow 2 \mathrm{NH}_3$, the sign of $\triangle H$ accompanying the reaction is:
- A
- ✓
- C
May be positive or negative.
- D
AnswerThe reaction for the formation of ammonia is $\mathrm{N}_2+3 \mathrm{H}_2 \Leftrightarrow 2 \mathrm{NH}_3+92.2 \mathrm{~kJ}$.
View full question & answer→MCQ 701 Mark
Acidity of $BF_3$ can be explained on the basis of which of the following concepts?
- A
- B
- ✓
- D
Bronsted Lowry as well as Lewis concept.
AnswerAccording to Lewis concept, a positively charged or an electron deficient species acts as Lewis acid. $BF_3$ is an electron deficient compound with B having 6 electrons only.
View full question & answer→MCQ 711 Mark
Which one does not give a buffer solution?
- ✓
Ammonia and sodium hydroxide in water.
- B
Sodium acetate and acetic acid in water.
- C
Ammonia and ammonium chloride in water.
- D
Sodium acetate and hydrochloric acid in water.
AnswerCorrect option: A. Ammonia and sodium hydroxide in water.
Buffer solution is a mixture of weak acid and its conjugate base or weak base and its conjugate acid. So ammonia $($strong base$)$ and sodium hydroxide does not give a buffer.
View full question & answer→MCQ 721 Mark
The addition of $\text{NaCl}$ to $\text{AgCl}$ decreases the solubility of $\text{AgCl}$ because $.........$
- A
Solubility product decreases.
- B
Solubility product remains constant.
- C
Solution becomes unsaturated.
- ✓
Solution becomes super saturated.
AnswerCorrect option: D. Solution becomes super saturated.
$\text{NaCl}$ is highly soluble and when it is added to $\text{AgCl}$ it decreases the solubility of $\text{AgCl}$ because of common ion $\ce{Cl^-}$ and solution become super saturated.
View full question & answer→MCQ 731 Mark
Which of the following will produce a buffer solution when mixed in equal volumes?
- A
$0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
- B
$0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
- ✓
$0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
- D
$0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{CH}_4 \mathrm{COONa}$ and $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaOH}$.
AnswerCorrect option: C. $0.1 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NH}_4 \mathrm{OH}$ and $0.05 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{HCl}$.
In $(c)$, all $\text{HCl}$ will be neutralized and $\mathrm{NH}_4 \mathrm{Cl}$ will be formed. Also some $\mathrm{NH}_4 \mathrm{OH}$ will be left unneutralized. Thus, the final solution will contain $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{NH}_4 \mathrm{Cl}$ and hence will form a buffer.
View full question & answer→MCQ 741 Mark
$\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_3 \mathrm{COOH}$ is $1.8 \times 10^{-5}$ and $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ . The $pH$ of ammonium acetate will be
- A
$7.005$
- B
$4.75$
- ✓
$7.0$
- D
Between $6$ and $7$
AnswerAmmonium acetate is a salt of weak acid and weak base.
$\text{pH}=\frac{1}{2}\big[\text{pK}_\text{w}+\text{pK}_\text{a}+\text{pK}_\text{b}\big]$
$=\frac{1}{2}\big[14-\log(1.8\times10^{-5})+\log(1.8\times10^{-5})\big]=7.0$
View full question & answer→MCQ 751 Mark
A solution of an acid has $pH = 4.70.$ Find out the concentration of $OH^-$ ions $(pK_w = 14).$
- ✓
$ 5 \times 10^{-10} $
- B
$ 6 \times 10^{-10} $
- C
$ 2 \times 10^{-5} $
- D
$ 9 \times 10^{-5} $
AnswerCorrect option: A. $ 5 \times 10^{-10} $
View full question & answer→MCQ 761 Mark
Given the chemical equilibrium,$ A ⇌ B + C$, where $\triangle H_{rxn}$ is negative, what effect increasing the temperature $($at constant pressure$)$ would have on the system at equilibrium?
Answer$A \rightarrow B + C\ ; \triangle H < 0$
Here, forward reaction is exothermic. So, if temperature is increased then as per Le Chatelier's principle, the equilibrium shifts to left side $($backward direction$)$.
View full question & answer→MCQ 771 Mark
Which of the following statements is incorrect?
- A
In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time.
- ✓
The intensity of red colour increases when oxalic acid is added to a solution containing iron $(III)$ nitrate and potassium thiocyanate.
- C
On addition of catalyst the equilibrium constant value is not affected.
- D
Equilibrium constant for a reaction with negative $\Delta\text{H}$ value decreases as the temperature increases.
AnswerCorrect option: B. The intensity of red colour increases when oxalic acid is added to a solution containing iron $(III)$ nitrate and potassium thiocyanate.
$\text{Fe}^{3+}+\text{SCN}^-\rightleftharpoons\text{FeSCN}^{2+}\text{(Red)}.$
When oxalic acid is added to a solution containing iron nitrate and potassium thiocyanate, oxalic acid reacts with $\text{Fe}{3+}$ ions to form a stable complex ion $[\text{Fe}\text{(C}_2\text{O}_4)3]^{3-},$ thus, decreasing the concentration of free $\text{Fe}^{3+}$ ions which in mm decreases the intensity of red colour.
$\text{Fe}^{3+}+\text{SCN}^-\rightleftharpoons\text{[Fe(SCN)}]^{2+}\text{(Red)}.$
View full question & answer→MCQ 781 Mark
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g}),$ if the degree of dissociation is a at equilibrium pressure$ 'p',$ then the equilibrium constant for the reaction is:
- A
$\text{K}_{\text{p}}=\frac{\alpha^2}{1+\alpha^2\text{P}}$
- B
$\text{K}_{\text{p}}=\frac{\alpha^2\text{P}^2}{1-\alpha^2}$
- C
$\text{K}_{\text{p}}=\frac{\alpha\text{P}^2}{1-\alpha^2}$
- ✓
$\text{K}_{\text{p}}=\frac{\alpha^2\text{P}}{1-\alpha^2}$
AnswerCorrect option: D. $\text{K}_{\text{p}}=\frac{\alpha^2\text{P}}{1-\alpha^2}$
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2(\text{g})$
$\begin{matrix}\text{Initial}&1&0&0\\\text{At equilibrium}&1-\alpha&\alpha&\alpha\end{matrix}$
Total no. of moles $=1-\alpha+\alpha+\alpha=1+\alpha$
$\text{pPCl}_5=\frac{1-\alpha}{1+\alpha}$
$\text{pCl}_2=\text{pCl}_3=\frac{\alpha}{1+\alpha}$
$\text{K}_{\text{p}}=\frac{(\text{PCl}_3)(\text{Cl}_2)}{(\text{PCl}^-_3)}$
$=\frac{\frac{\alpha}{1+\alpha}\times\frac{\alpha}{1+\alpha}\times\text{p}^2}{\frac{1-\alpha}{1+\alpha}\times\text{p}}$
$=\frac{\alpha^2}{1-\alpha^2}\times\text{p}$
View full question & answer→MCQ 791 Mark
In a reversible reaction $\mathrm{H}_2+\mathrm{I}_2 \leftrightharpoons 2 \mathrm{HI}$, if the concentration of $H_2$ and $I_2$ are increased, the value of $Kc:$
AnswerThe magnitude of the equilibrium constant is not affected by the changes in concentrations of reactants and products, pressure and volume. Thus, when the concentrations of hydrogen and iodine are increased, the value of the equilibrium constant remains the same.
View full question & answer→MCQ 801 Mark
In the presence of a common ion $($incapable of forming complex ion$)$, the solubility of salt $.........$ in solution.
Answer$ \mathrm{AB} \rightarrow \mathrm{A}^{+}+\mathrm{B}^{-} $
$ \mathrm{BC} \rightarrow \mathrm{B}^{+}+\mathrm{C}^{-} $
Since $B^+$ is incapable of forming a complex salt it tends to decrease the solubility by $Le-$Chatelier's principle.
View full question & answer→MCQ 811 Mark
Degree of ionisation does not depend on:
- A
- B
Ature of the electrolyte.
- C
- ✓
Molecular mass of the electrolyte.
AnswerCorrect option: D. Molecular mass of the electrolyte.
Degree of ionization$(\alpha)$ depends on$-$
$(1)$ Concentration of solute.
$(2)$ Temperature.
$(3)$ Nature of electrolysis.
$(4)$ Nature of solvent.
$(5)$ Dilution.
$"α"$ does not depend on molecular mass of electrolyte.
View full question & answer→MCQ 821 Mark
Strong electrolyte of the following is?
- A
$\ce{01M HAc}$
- ✓
$\ce{0.1M HCl}$
- C
$\ce{0.1M KCl}$
- D
$\ce{0.1M NaCl}$
AnswerCorrect option: B. $\ce{0.1M HCl}$
View full question & answer→MCQ 831 Mark
Calculate the molar solubility $(S)$ of a salt like zirconium phosphate of molecular formula $(\text{Zr}^{4+})_3(\text{PO}^{3-}_4)_4.$
- A
$\Big(\frac{\text{K}_{\text{sp}}}{9612}\Big)^{\frac{1}{8}}$
- ✓
$\Big(\frac{\text{K}_{\text{sp}}}{6912}\Big)^{\frac{1}{7}}$
- C
$\Big(\frac{\text{K}_{\text{sp}}}{5348}\Big)^{\frac{1}{6}}$
- D
$\Big(\frac{\text{K}_{\text{sp}}}{8435}\Big)^{\frac{1}{7}}$
AnswerCorrect option: B. $\Big(\frac{\text{K}_{\text{sp}}}{6912}\Big)^{\frac{1}{7}}$
$[\text{Zr}^{4+}]=3\text{S}$ and $\text{PO}^{3-}_4]=4\text{S}$
and $\text{K}_{\text{sp}}=(3\text{S})^3(4\text{S})^4=6912(\text{S})^7$
or $\text{S}=\Big\{\ \frac{\text{K}_{\text{sp}}}{(3^3\times4^4)}\Big\}^{\frac{1}{7}}$
$=\Big(\frac{\text{K}_{\text{sp}}}{6912}\Big)^{\frac{1}{7}}$
View full question & answer→MCQ 841 Mark
In the reaction,$\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})-180.7 \mathrm{~kJ}$, on increasing the temperature, the production of $NO$ :
AnswerGiven $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})-180.7 \mathrm{~kJ}$
For this reaction, $\ce{\triangle H = +180.7 kJ}$
Positive value of $\triangle H$ shows that the reaction is endothermic.
According to Le$-$Chatelier's principle, in case of endothermic reaction, increase in temperature, shifts the reaction towards product side.
View full question & answer→MCQ 851 Mark
Predict which of the following reaction will have appreciable concentration of reactants and products?
- A
$\text{Cl}_2(\text{g})\rightleftharpoons2\text{Cl(g)};\text{K}_{\text{c}}=5\times10^{-39}$
- B
$\text{Cl}_2(\text{g})+\text{2NO}(\text{g})\rightleftharpoons2\text{NOCl}(\text{g});\text{K}_{\text{c}}=37\times10^8$
- ✓
$\text{Cl}_2\text{(g)}+2\text{NO}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\text{K}_{\text{c}}=1.8$
- D
All have appreciable concentration of reactants and products.
AnswerCorrect option: C. $\text{Cl}_2\text{(g)}+2\text{NO}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\text{K}_{\text{c}}=1.8$
View full question & answer→MCQ 861 Mark
The following reaction goes to completion in lime kiln because : $\mathrm{CaCO}_3 \rightleftharpoons \mathrm{CaO}+\mathrm{CO}_2(\mathrm{~g})$
- A
- B
$\ce{CaO}$ is more stable than $\ce{CaCO_3}$.
- C
$\ce{CaO}$ is not dissociated.
- ✓
$\ce{CO_2}$ escapes continuously.
AnswerCorrect option: D. $\ce{CO_2}$ escapes continuously.
In lime kilns, $\ce{CO_2}$ formed continues to escape into the atmosphere and equilibrium is never established.
View full question & answer→MCQ 871 Mark
Dissociation events in $\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}$ is termed as ionization because:
- ✓
The electron is initially shared between both atoms, thus the dissociation event into ions involves the transfer of an electron from one atom to the other.
- B
It does not involve any electron transfer.
- C
- D
No charge species formed.
AnswerCorrect option: A. The electron is initially shared between both atoms, thus the dissociation event into ions involves the transfer of an electron from one atom to the other.
$\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-}$
This dissociation is termed as ionization because initially there is a polar covalent compound $\mathrm{H}_2 \mathrm{O}$.
This dissociation leads to the formation of ions $H^+$ and $OH^-$
This dissociation involves a transfer of an electron from one atom to another leading to the formation of charged species.
View full question & answer→MCQ 881 Mark
A buffer solution is a solution whose $\ce{pH}$ value on keeping in the air :
- A
- B
- C
May increase or decrease.
- ✓
View full question & answer→MCQ 891 Mark
The ionic product of water $ .......... $ if a few drops of acid or base are added to it.
AnswerThe ionic product of water at a particular temperature is constant and has no effect of acid or base addition.
View full question & answer→MCQ 901 Mark
Which of the following species is amphoteric in nature.
- A
$\text{H}_3\text{O}^+$
- ✓
$\text{Cl}^-$
- C
$\text{HSO}^-_4$
- D
$\text{CO}^{2-}_3$
AnswerCorrect option: B. $\text{Cl}^-$
$\text{HSO}^-_3$ because if can gain $\ce{H}^+$ as well as lose $\ce{H}^+$.
View full question & answer→MCQ 911 Mark
The solubility product $K_{sp}$ of the sparingly soluble salt $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $4 \times 10^{-12}$. The molar solubility of the salt is :
- ✓
$1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} $
- B
$ 2 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} $
- C
$ 1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} $
- D
$ 2 \times 10^{-12} \mathrm{~mol} \mathrm{~L}^{-1} $
AnswerCorrect option: A. $1.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} $
View full question & answer→MCQ 921 Mark
Which of the following factors will favour the reverse reaction in a chemical equilibrium?
- A
Increase in concentration of one of the reactants.
- ✓
Increase in concentration of one of the products.
- C
Removal of one of the products regularly.
- D
AnswerCorrect option: B. Increase in concentration of one of the products.
$\ce{A + B \rightleftharpoons C + D}$
According to Le ChateIier's principle, as the concentration of the products are increased, the reaction proceeds in the backward direction.
View full question & answer→MCQ 931 Mark
$\mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_5(\mathrm{~g})+$ energy For the reaction above, what is the effect of increasing the pressure?
- ✓
Increased production of products.
- B
Wild fluctuations in the amounts of reactants and products.
- C
Increased production of reactants.
- D
No impact on the equilibrium.
AnswerCorrect option: A. Increased production of products.
$\mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_5(\mathrm{~g})+$ energy
In this reaction, the forward reaction is accompanied by a decrease of moles of gaseous species. So, if pressure on the system is increased then, as per Le Chatelier's principle, the equation shifts in direction in which a decrease in total no. of moles takes place i.e. in formation of products in this case.
View full question & answer→MCQ 941 Mark
Addition of $\text{HCl}$ will not suppress the ionization of :
AnswerAny acid weaker than $\text{HCl}$ will be suppressed by $\text{HCl}$. Hence, among the given options, only sulphuric acid is an acid with comparable acidity strength to $\text{HCl}$. The same can also be verified using $\text{Ka}$ values from the data.
View full question & answer→MCQ 951 Mark
The $\ce{pH}$ value of blood does not appreciably change by a small addition of an acid or a base, because the blood :
- A
- B
Can be easily coagulated.
- C
Contains iron as a part of the molecule.
- ✓
Contains serum protein which acts as buffer.
AnswerCorrect option: D. Contains serum protein which acts as buffer.
The buffer system present in serum is $\mathrm{H}_2 \mathrm{CO}_3+\mathrm{NaHCO}_3$ and as we know that a buffer solution resist the change in $\ce{pH}$ therefore $\ce{pH}$ value of blood does not change by a small addition of an acid or a base.
View full question & answer→MCQ 961 Mark
A solution which maintains constant $\ce{pH}$ when small amounts of acid or alkali are added is known as $ .........$
AnswerA buffer solution is one which resists changes in $\ce{pH}$ when small quantities of an acid or an alkali are added to it.
View full question & answer→MCQ 971 Mark
If the temperature of the system of equilibrium is increased, the equilibrium will shift in the direction which $ ......... $ heat.
AnswerIf a chemical system at equilibrium experiences a change in concentration, temperature, volume or pressure then, the equilibrium shifts to counteract the imposed change and a new equilibrium is established.
Therefore, If the temperature of the system of equilibrium is increased, the equilibrium will shift in the direction which absorbs heat. It is a direct implication of the Le Chatelier's principle $($effect of pressure and temperature$).$
View full question & answer→MCQ 981 Mark
For the reaction $\text{H}_2\text{(g)}+\text{I}_2\text{(g)}\rightleftharpoons2\text{HI(g)},$ the standard free energy is $\Delta\text{G}^\ominus > 0.$ The equilibrium constant $(K)$ would be $ .........$
- A
$K = 0.$
- B
$K > 1.$
- C
$K = 1.$
- ✓
$K < 1.$
AnswerCorrect option: D. $K < 1.$
$\Delta\text{G}^\circ=-\text{R}\text{T }1\text{nK}$
If $\Delta\text{G}^\ominus > 0,$ then $-\Delta\text{G}^\ominus/\text{RT}$ is negative, and $\text{e}^{\Delta\text{G}^\ominus/\text{RT}} < 1.$ That is $K < 1,$ which implies a non $-$ spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.
View full question & answer→MCQ 991 Mark
$2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2$ The equilibrium constant of the above reaction is $6.4$ at $300K$. If $0.25$ mole each of $\ce{H_2}$ and $\ce{I_2}$ are added to the system, the equilibrium constant will be :
View full question & answer→MCQ 1001 Mark
If $\text{HCl}$ is added to pure water at $25^\circ C$ the ionic product of water will be :
- A
$ > 10^{-14} $
- B
$ < 10^{-14} $
- ✓
$ 10^{-14} $
- D
$ > 10^{-10} $
AnswerCorrect option: C. $ 10^{-14} $
For ionic product of pure water :
$ {\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{M}} $
$ {\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}} $
$\text { Product, } \mathrm{K}_{\mathrm{W}}=10^{-14}$
lonic Product $\text{Kw}$ of water is dependent only on temperature so it will remain same,
$ \mathrm{K}_{\mathrm{W}}=10^{-14}$.
View full question & answer→MCQ 1011 Mark
Which of the following solutions are acidic?
AnswerCorrect option: C. $\mathrm{AlCl}_3$
$(a)$ and $(c)$ are acidic.
Because they are salts of strong acid $\mathrm{H}_2 \mathrm{SO}_4$ and $\text{HCl}$ and weak bases $\mathrm{Cu}(\mathrm{OH})_2$ and $\mathrm{Al}(\mathrm{OH})_2$ respectively.
View full question & answer→MCQ 1021 Mark
The state of equilibrium refers to :
Answerhe state of equilibrium refers to dynamic state. Both forward and backward reactions proceed in the opposite directions with equal rates. Neither the forward reaction nor the backward reaction has ceased.
View full question & answer→MCQ 1031 Mark
The ionisation constant of an acid, $\mathrm{K}_{\mathrm{a}}$, is the measure of strength of an acid. The $\mathrm{K}_{\mathrm{a}}$ values of acetic acid, hypochlorous acid and formic acid are $1.74 \times 10^{-5}, 3.0 \times 10^{-8}$ and $1.8 \times 10^{-4}$ respectively. Which of the following orders of $\ce{pH}$ of $0.1 \mathrm{~mol} \mathrm{~dm}^{-3}$ solutions of these acids is correct?
- A
Acetic acid $ > $ hypochlorous acid $ > $ formic acid.
- B
Hypochlorous acid $ > $ acetic acid $ > $ formic acid.
- C
Formic acid $ > $ hypochlorous acid $ > $ acetic acid.
- ✓
Formic acid $ > $ acetic acid $ > $ hypochlorous acid.
AnswerCorrect option: D. Formic acid $ > $ acetic acid $ > $ hypochlorous acid.
$[\text{H}_3\text{O}^+]=\sqrt{\text{K}_\text{a}\cdot\text{C}}$ for the same concentration, $[\text{H}_3\text{O}^+]\propto\sqrt{\text{K}_\text{a}\cdot}$
But $\text{pH}=-\log[\text{H}_3\text{O}^+]$
Larger the value of $\ce{Ka},$ larger will be $[\text{H}_3\text{O}^+]$ and lower will be $\ce{pH}.$
View full question & answer→MCQ 1041 Mark
Cottrell precipitator acts on which of the following principle?
- A
Hardy $-$ Schulze rule.
- B
- C
Le Chatelier's principle.
- ✓
Neutralization of charge on the colloidal particles.
AnswerCorrect option: D. Neutralization of charge on the colloidal particles.
It is related with neutralisation of charge on colloidal particlr.
View full question & answer→MCQ 1051 Mark
Dissolution of sodium sulphate is an exothermic process. If a saturated solution of sodium sulphate containing extra undissolved sodium sulphate is heated, then :
- A
More of sodium sulphate will dissolve.
- ✓
Some sodium sulphate will be precipitated out.
- C
Concentration of the solution will not change.
- D
The solution will become supersaturated.
AnswerCorrect option: B. Some sodium sulphate will be precipitated out.
View full question & answer→MCQ 1061 Mark
$0.1 \mathrm{M} \mathrm{~CH} \mathrm{COOH}_3$ and $1.01 \mathrm{CH}_3 \mathrm{COONa}$ are mixed togethere, what will be $\ce{pH}$ of buffer solution if $\text{pK}_{\text{a}} = 4.75 [\log 10-1 = -1]$
- ✓
$3.75$
- B
$4.75$
- C
$5.75$
- D
$6.75$
AnswerCorrect option: A. $3.75$
$\text{pH}=\text{pK}_{\text{a}}+\log\frac{(\text{Salt})}{\text{(Acid)}}$
$=4.75+\log\frac{10^{-2}}{10^{-1}}$
$=4.75-1=3.75$
View full question & answer→MCQ 1071 Mark
Concentration of $\mathrm{Ag}^{+}$ ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$ Solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}$ is :
- A
$ 2.66 \times 10^{-12} $
- B
$ 4.5 \times 10^{-11} $
- ✓
$ 5.3 \times 10^{-12} $
- D
$ 2.42 \times 10^{-8} $
AnswerCorrect option: C. $ 5.3 \times 10^{-12} $
$\text{Ag}_2\text{C}_2\text{O}_4(\text{aq})\rightleftharpoons2\text{Ag}^{+}+ \text{C}_2\text{O}^{2-}_4\ 2.2\times10^{-4}\ 1.1\times10^{-4}$
$\text{K}_{\text{sp}}=(\text{Ag}^+)^2(\text{C}_2\text{O}^{2-}_4)$
$=(2.2\times10^{-4})^2=(1.1\times10^{-4})$
$5.3\times10^{-12}$
View full question & answer→MCQ 1081 Mark
When no more solute can be dissolved in solution at a given temperature the solution is known as :
- A
- ✓
- C
- D
Both $(a)$ and $(b).$
View full question & answer→MCQ 1091 Mark
$\ce{pH}$ of water is $7$ at $25^\circ C$. If water is heated at $80^\circ C,$ its $\ce{pH}$ will :
AnswerHand $\mathrm{OH}^{-}$both will increase, therefore $\ce{pH}$ will decrease due to increase in $\mathrm{H}^{+}$ as $\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$ will increase with increase in temperature.
View full question & answer→MCQ 1101 Mark
When ammonium chloride is added to ammonia solution, the $\ce{pH}$ of the resulting solution will be :
AnswerCommon ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte which provides an ion common to that provided by a weak electrolyte.
Ammonium hydroxide is a weak electrolyte and ammonium chloride is a strong electrolyte.
Ammonium chloride provides ammonium ion which is common to that provided by ammonium hydroxide.
Thus, the pair $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$ shows common ion effect. Ammonium chloride suppresses the ionization of ammonium hydroxide.
View full question & answer→MCQ 1111 Mark
The strength of acid is highest in :
- A
$ \mathrm{pK}_{\mathrm{a}}=6$
- B
$\mathrm{pK}_{\mathrm{a}}=5$
- C
$\mathrm{pK}_{\mathrm{a}}=10$
- ✓
$ \mathrm{pK}_{\mathrm{a}}=1 $
AnswerCorrect option: D. $ \mathrm{pK}_{\mathrm{a}}=1 $
$\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}$
Higher the $\ce{Ka},$ higher is the strength of the acid.
For higher $\text{Ka, pKa}$ value is smaller.
View full question & answer→MCQ 1121 Mark
In which condition, the reaction proceeds in the forward direction?
- A
$\text{Q}_{\text{C}}=\text{K}_{\text{C}}$
- B
$\text{Q}_{\text{C}} > \text{K}_{\text{C}}$
- ✓
$\text{Q}_{\text{C}} < \text{K}_{\text{C}}$
- D
$\text{Q}_{\text{C}}\neq\text{K}_{\text{C}}$
AnswerCorrect option: C. $\text{Q}_{\text{C}} < \text{K}_{\text{C}}$
When $\text{Q}_{\text{C}} < \text{K}_{\text{C}}$ then the reaction proceeds in the forward direction.
View full question & answer→MCQ 1131 Mark
Which among the following factors changes the value of ionic product of water?
- ✓
- B
- C
- D
Addition of either acid and base.
AnswerIonic product depends only on temperature.
View full question & answer→MCQ 1141 Mark
For a system in equilibrium, $\triangle\text{G}=0$ under conditions of constant :
- ✓
Temperature and pressure.
- B
- C
- D
AnswerCorrect option: A. Temperature and pressure.
When a system is at equilibrium under constant temperature and pressure, its free energy change is zero $(\triangle\text{G}=0)$
View full question & answer→MCQ 1151 Mark
If little heat is added to ice $\rightleftharpoons$ liquid, equilibrium in a sealed container, then :
- A
- B
- C
- ✓
No change in pressure and temperature.
AnswerCorrect option: D. No change in pressure and temperature.
View full question & answer→MCQ 1161 Mark
- A
Rapidly changes $\text{pH}$ due to addition of an acid.
- ✓
Resists changes in $\text{pH}.$
- C
Does not change $\text{pH}$ at all.
- D
Changes $\text{pH}$ only with the addition of a strong base.
AnswerCorrect option: B. Resists changes in $\text{pH}.$
The solution of buffer resists changes in $\text{pH}.$
A buffer solution is defined as a solution which resists drastic changes in $\text{pH}$ upon the addition of a small amount of either an acid or a base.
View full question & answer→MCQ 1171 Mark
The mass of acetic acid present in $500\ ml$ of solution in which it is $1\%$ ionised $(\ce{Ka}$ of $\mathrm{CH}_3 \mathrm{COOH}=1.8 \times 10^{-5})$
- ✓
$5.4g$
- B
$12.6g$
- C
$6.4g$
- D
$10.8g$
AnswerCorrect option: A. $5.4g$
View full question & answer→MCQ 1181 Mark
What do you mean by buffer solution?
- A
Buffer solution have no $\ce{pH}.$
- ✓
Its $\ce{pH}$ changes very little when a small amount of acid or base is added to the it.
- C
Its $\ce{pH}$ changes very largely when a small amount of acid or base is added to the it.
- D
All solutions are buffer.
AnswerCorrect option: B. Its $\ce{pH}$ changes very little when a small amount of acid or base is added to the it.
It's $\ce{pH}$ changes very little when a small amount of strong acid or base is added to it.
Buffer solutions are used as a means of keeping $\ce{pH}$ at a nearly constant value in a wide variety of chemical applications. In nature, there are many systems that use buffering for $\ce{pH}$ regulation. For example, the bicarbonate buffering system is used to regulate the $\ce{pH}$ of blood.
View full question & answer→MCQ 1191 Mark
What will be the value of $\ce{pH}$ of $0.01 \text{ mol}\ \mathrm{ dm}^{-3} \mathrm{CH}_3 \mathrm{COOH}\left(\mathrm{Ka}=1.74 \times 10^{-5}\right) \ ?$
Answer$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{COOH}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{CH}_3\text{COOH}^-$
$^\text{Initial conc.} \ \ \ \ \ \ \ \ \ \ \ \ 0.01 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$
$^\text{At equilibrium} \ \ \ \ \ \ \ \ \ 0.01-\text{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x}$
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH]}}=\frac{\text{x}^2}{0.01-\text{x}}$
Since $\text{x} << 0.01,$
Therefore, $0.01-\text{x}\approx0.01$
$\frac{\text{x}^2}{0.01}=1.74\times10^{-5}$
$\text{x}^2=1.74\times10^{-7}\text{ or x}=4.2\times10^{-4}$
$\text{pH}=-\log(4.2\times10^{-4})=3.4$
View full question & answer→MCQ 1201 Mark
The $\ce{pH}$ of a dilute solution of acetic acid was found to be $4.3$ The addition of a small crystal of sodium acetate will cause $\ce{pH}$ to:
- A
Become less than $4.3$
- ✓
Become more than $4.3$
- C
Remain equal to $4.3$
- D
AnswerCorrect option: B. Become more than $4.3$
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion.The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.Due to this common ion effect, when we add sodium acetate dissociation of acetic acid decreases and solution will have less number of hydrogen ion and so, $\ce{pH}$ increases. $($as $\ce{pH} = −\log[H^+])$
View full question & answer→MCQ 1211 Mark
For the equilibrium reaction, $\mathrm{H}_2 \mathrm{O}(\mathrm{I}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g})$, what happens, if pressure is applied :
- A
- ✓
The boiling point of water is increased.
- C
No effect on boiling point.
- D
AnswerCorrect option: B. The boiling point of water is increased.
The equilibrium reaction is $\mathrm{H}_2 \mathrm{O}(\mathrm{I}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g},)$ when pressure is applied, the equilibrium will shift to left as the value of $\triangle n$ is positive. Hence, for boiling to occur more temperature is required. So, it will increase the boiling point.
View full question & answer→MCQ 1221 Mark
The solutions which tend to keep the concentration of hydrogen ions constant, even when small amounts of strong acid or strong base are added to them, are known as :
AnswerBuffer solutions have the capacity to react with small amounts of added acid or base without affecting the hydrogen ion concentration of the solution.
View full question & answer→MCQ 1231 Mark
The equilibrium constant of a reaction at $298K$ and $1000K$ is $5 \times 10^{-3}$ and $2 \times 10^{-3}$ respectively. The $\triangle\text{H}$ for the reaction is:
- A
- ✓
- C
Either positive or negative.
- D
AnswerWith the increase in temperature from $298K$ to $1000K,$ the value of the equilibrium constant decreases from $5 \times 10^{-3}$ to $2 \times 10^{-3}$.
Thus, as the temperature increases, the equilibrium shifts to the left direction. This is possible for an exothermic reaction.
View full question & answer→MCQ 1241 Mark
Which of the following will supress the ionisation of acetic acid in aqueous solution?
- A
$\ce{NaCl}$
- ✓
$\ce{HCI}$
- C
$\ce{KCI}$
- D
AnswerCorrect option: B. $\ce{HCI}$
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
View full question & answer→MCQ 1251 Mark
The $\ce{pH}$ of boiling water is $6.4$. This implies that boiling water is :
AnswerThe $\ce{pH}$ of boiling water is $6.4$. This implies that boiling water is neutral.
When water is boiled, both hydrogen ion and hydroxide ion concentration increases to same extent.
Hence it is neutral. With increase in the hydrogen ion concentration, $\ce{pH}$ decreases from $7$ to $6.4.$
Also, the value of $\ce{Kw}$ also increases as the degree of dissociation of water increases with increase in temperature.
View full question & answer→MCQ 1261 Mark
The ionisation of weak base $\text{MOH}$ can be represented by equation, $\text{MOH}(\text{aq})\rightleftharpoons\text{M}^+(\text{aq})+\text{OH}^-(\text{aq});$
$\text{K}_{\text{b}}=\frac{[\text{M}^+][\text{OH}^-]}{[\text{MOH}]}$ where, $K_b$ is represented as:
- A
Acidic ionisation constant.
- ✓
Basic ionisation constant.
- C
- D
Both $(a)$ and $(b).$
AnswerCorrect option: B. Basic ionisation constant.
The equilibrium constant for basic ionization called basic ionization constant and is represented by $\ce{K_b}$.
$\text{K}_{\text{b}}=\frac{[\text{M}^+][\text{OH}^-]}{[\text{MOH}]}$
View full question & answer→MCQ 1271 Mark
The equilibrium constants for the reaction, $\text{Zn(s)}+\text{Cu}^{2+}(\text{aq})\rightleftharpoons\text{Zn}^{2+}(\text{aq})+\text{Cu}(\text{s})$ and
$\text{Cu(s)}+2\text{Ag}^+(\text{aq})\rightleftharpoons\text{Cu}^{2+}(\text{aq})+2\text{Ag(s)}$ are $\text{K}_1$ and $\text{K}_2$ respectively. The equilibrium constant for the combined reaction is,
AnswerCorrect option: A. $\text{K}_1\times\text{K}_2$
As we are combining the two equations.
$\therefore$ Equilibrium constant for combined reaction, i.e. $\text{K = K}_1 \times \text{K}_2$
View full question & answer→MCQ 1281 Mark
Which of the following is false of a binary solution showing false deviation from Roult's law?
- A
$\text{P}_\text{A} < \text{X}_\text{A}\text{P}\mathring{\text{A}}$
- B
$\triangle \text{H}_\text{mix}<0$
- C
$\triangle \text{V}_\text{mix}<0$
- ✓
View full question & answer→MCQ 1291 Mark
In the reaction, $\text{C(s)}+\text{CO}_2\text{(g)}\rightleftharpoons2\text{CO(g)},$ when pressure is increased, the reaction goes in the :
View full question & answer→MCQ 1301 Mark
The equilibrium constant $K$ for the reaction : $\ce{2HI(g) \rightleftharpoons H_2(g) + I_1(g)}$ at room temp is $2.85$ and that at $698 K$ is $1.4 \times 10 − 2$. This implies that the forward reaction is:
AnswerWith the increase of temperature, $k$ value decreases, so that forward reaction decreases with increase of temperature. This implies that reaction will proceed in forward direction with decrease of temperature, i.e., heat is liberated and hence forward reaction is exothermic.
View full question & answer→MCQ 1311 Mark
If $\ce{PCl_5}$ is heated in two seperate vessels of volume $5$ lit and $10$ lit respectively at $27^\circ C$ The extent of dissociation of $\ce{PCl_5}$ will be :
- A
More in $5$ lit vessel.
- ✓
More in $10$ lit vessel.
- C
- D
AnswerCorrect option: B. More in $10$ lit vessel.
For the reaction $\mathrm{PCl}_5 \Leftrightarrow \mathrm{PCl}_3+\mathrm{Cl}_2$ , the forward reaction occurs with increase in the number of moles from $1$ to $2$. Also the reverse reaction occurs with decrease in the number of moles from $2$ to $1$. When the pressure of the system is increased, the backward reaction will be favoured as the reverse reaction occurs with decrease in the number of moles. Thus the dissociation of $\ce{PCl_5}$ is suppressed. Hence, the degree of dissociation decreases. Pressure will be more in $5 L$ vessel than in $10 L$ vessel. Thus the extent of dissociation of $\ce{PCl_5}$ will be more in $10L$ vessel.
View full question & answer→MCQ 1321 Mark
The solubility of $\ce{AgCl}$ in $\ce{NaCl}$ solution is less than that in pure water, because of the $ .........$
AnswerSince, $\ce{NaCl}$ is soluble to a very significant extent, when $\ce{AgCl}$ is added to $\ce{NaCl}$ solution, the common ion $\left[\mathrm{Cl}^{-}\right]$ increases in the solution. To have the solubility product or $\mathrm{K}_{\mathrm{sp}}$ of $\ce{AgCl}$ constant, $\left[\mathrm{Ag}^{+}\right]$will decrease or $\ce{AgCl}$ will percipitate out from the solution. This is common ion effect.
View full question & answer→MCQ 1331 Mark
What will be the molar solubility $S$ of a solid salt with general formula $\text{M}^{\text{p+}}_{\text{x}}\text{X}^{\text{q-}}_{\text{y}}\ ?$
- A
$\Big(\frac{\text{K}_{\text{sp}}}{\text{x}^{\text{y}}.\text{y}^{\text{x}}}\Big)^{\frac{1}{\text{x}+\text{y}}}$
- B
$\Big(\frac{\text{K}_{\text{sp}}}{\text{x}^{\text{x}}.\text{y}^{\text{y}}}\Big)^{\text{x}+\text{y}}$
- ✓
$\Big(\frac{\text{K}_{\text{sp}}}{\text{x}^{\text{x}}.\text{y}^{\text{y}}}\Big)^{\frac{1}{\text{x}+\text{y}}}$
- D
$\Big(\frac{\text{K}_{\text{sp}}}{\text{x}^{\text{y}}.\text{y}^{\text{x}}}\Big)^{\text{x}+\text{y}}$
AnswerCorrect option: C. $\Big(\frac{\text{K}_{\text{sp}}}{\text{x}^{\text{x}}.\text{y}^{\text{y}}}\Big)^{\frac{1}{\text{x}+\text{y}}}$
View full question & answer→MCQ 1341 Mark
Point out the correct relation between $\mathrm{K}_{\mathrm{a}}, \mathrm{K}_{\mathrm{b}}$ and $\mathrm{K}_{\mathrm{w}},$
- A
$\text{K}_{\text{a}}+\text{K}_{\text{b}}=\text{K}_{\text{w}}$
- B
$\text{K}_{\text{a}}-\text{K}_{\text{b}}=\text{K}_{\text{w}}$
- ✓
$\text{K}_{\text{a}}\times\text{K}_{\text{b}}=\text{K}_{\text{w}}$
- D
$\frac{\text{K}_{\text{a}}}{\text{K}_{\text{b}}}=\text{K}_{\text{w}}$
AnswerCorrect option: C. $\text{K}_{\text{a}}\times\text{K}_{\text{b}}=\text{K}_{\text{w}}$
In case of a conjugate acid $-$ base pair,
$\text{K}_{\text{a}}\times\text{K}_{\text{b}}=\text{K}_{\text{w}}.$
View full question & answer→MCQ 1351 Mark
What will be the conjugate bases for the following Bronsted acids ? $\text{HF}, \text{H}_2\text{SO}_4 $ and $\text{HCO}^-_3$
- A
$\text{F}^-,\text{SO}_4^{2-}$ and $\text{CO}_3^{2-}$
- B
$\text{F}^-,\text{SO}^{2-}_4$ and $\text{H}_2\text{CO}_3$
- C
$\text{F}^-,\text{HSO}^-_4$ and $\text{H}_2\text{CO}_3$
- ✓
$\text{F}^-,\text{HSO}^-_4$ and $\text{CO}^{2-}_3$
AnswerCorrect option: D. $\text{F}^-,\text{HSO}^-_4$ and $\text{CO}^{2-}_3$
The conjugate bases should have one proton less in each case and therefore, the corresponding conjugate bases are $\text{F}^-,\text{HSO}^-_4$ and $\text{CO}^{2-}_3$ respectively.
View full question & answer→MCQ 1361 Mark
Which one of the following mixture does not act as a buffer solution?
- A
- B
Sodium phosphate & disodium hydrogen phosphate.
- C
Sodium propionate and propionic acid.
- ✓
Sodium acetate and sodium propionate.
AnswerCorrect option: D. Sodium acetate and sodium propionate.
An acidic buffer contains equimolar quantities of weak acid and its salt with strong base. A basic buffer contains equinolar quantities of weak base and its salt with strong acid. Sodium acetate is a salt with strong base but sodium propionate is not weak acid, it is also a salt.
View full question & answer→MCQ 1371 Mark
Which of the following statements is true about buffer solution?
- ✓
It keeps the $\ce{pH}$ value constant in a chemical reaction.
- B
It decreases the $\ce{pH}$ value in a chemical reaction.
- C
It increases the $\ce{pH}$ value in a chemical reaction.
- D
It first increases and then decreases the $\ce{pH}$ value in a chemical reaction.
AnswerCorrect option: A. It keeps the $\ce{pH}$ value constant in a chemical reaction.
Buffer solutions have the capacity to react with small amounts of added acid or base without affecting the hydrogen ion concentration of the solution. Thus, the buffer solutions help to keep the $\ce{pH}$ value constant in a chemical reaction.
View full question & answer→MCQ 1381 Mark
When $\mathrm{NH}_4 \mathrm{Cl}$ is added to $\mathrm{NH}_4 \mathrm{OH}$ solution, the dissociation of ammonium hydroxide is reduced. It is due to :
AnswerWhen $\mathrm{NH}_4 \mathrm{Cl}$ is added to $\mathrm{NH}_4 \mathrm{OH}$ solution, concentration of $\mathrm{NH}_4^{+}$ ions increases so the equilibrium shift towards left.
So the dissociation of ammonium hydroxide is reduced.
View full question & answer→MCQ 1391 Mark
$\text{K}_{\text{a}_1},\text{K}_{\text{a}_2}$ and $\text{K}_{\text{a}_3}$ are the respective ionisation constants for the following reactions.
$\text{H}_2\text{S}\rightleftharpoons\text{H}^++\text{HS}^-$
$\text{H}\text{S}^-\rightleftharpoons\text{H}^++\text{S}^{2-}$
$\text{H}_2\text{S}\rightleftharpoons\text{2H}^++\text{S}^{2-}$
The correct relationship between $\text{K}_{\text{a}_1},\text{K}_{\text{a}_2}$ and $\text{K}_{\text{a}_3}$ is
- ✓
$\text{K}_{\text{a}_3}=\text{K}_{\text{a}_1}\times\text{K}_{\text{a}_2}$
- B
$\text{K}_{\text{a}_3}=\text{K}_{\text{a}_1}+\text{K}_{\text{a}_2}$
- C
$\text{K}_{\text{a}_3}=\text{K}_{\text{a}_1}-\text{K}_{\text{a}_2}$
- D
$\text{K}_{\text{a}_3}=\text{K}_{\text{a}_1}-\text{ K}_{\text{a}_2}$
AnswerCorrect option: A. $\text{K}_{\text{a}_3}=\text{K}_{\text{a}_1}\times\text{K}_{\text{a}_2}$
For the reaction, $\text{H}_2\text{S}\rightleftharpoons\text{H}^++\text{HS}^-$
$\text{K}_{\text{a}_1}=\frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]}$
For the reaction, $\text{H}\text{S}^-\rightleftharpoons\text{H}^++\text{S}^{2-}$
$\text{K}_{\text{a}_2}=\frac{[\text{H}^+][\text{S}^{2-}]}{[\text{H}\text{S}^-]}$
When the above two reaction are added, their equilibrium constants are multiplied. Thus
$\text{K}_{\text{a}_3}=\frac{[\text{H}^+]^2[\text{S}^{2-}]}{[\text{H}_2\text{S}]}=\text{K}_{\text{a}_1}\times\text{K}_{\text{a}_2}$
Hence, $\text{K}_{\text{a}_3}=\text{K}_{\text{a}_1}\times\text{K}_{\text{a}_2}$
View full question & answer→MCQ 1401 Mark
In which of the following solvents is silver chloride most soluble?
- A
$0.1 \mathrm{~mol}\ \mathrm{dm}^{-3} \ \mathrm{AgNO}_3$ solution.
- B
$0.1 \mathrm{~mol}\ \mathrm{dm}^{-3} \ \mathrm{HCl}$ solution.
- C
$\mathrm{H}_2 \mathrm{O} \text {. }$
- ✓
AnswerAqueous ammonia will absorb chloride ions and thus the equilibria will shift in forward direction and solubility of silver chloride will increase.
View full question & answer→MCQ 1411 Mark
The equilibrium $\mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ is attained at $25^{\circ} \mathrm{C}$ in a closed container and an inert gas, helium, is introduced. Which of the following statements is/ are correct?
- A
The concentrations of $\mathrm{SO}_2, \mathrm{Cl}_2$, and $\mathrm{SO}_2 \mathrm{Cl}_2$ change.
- B
- C
The concentration of $\ce{SO_2}$ is reduced.
- ✓
AnswerAt constant volume, there is no effect of addition of inert gas to a reaction in equilibrium.
View full question & answer→MCQ 1421 Mark
Milk of magnesia used as a medicine for treating indigestion is a substance that $-$
- A
Helps in disintegration of food products leading to their facile metabolism.
- B
Combines with gastric hydrochloric acid thereby enhancing the latter's efficiency.
- C
Improves the enzymatic activities inside the stomach.
- ✓
Neutralises excess acidity, providing a buffered medium inside the stomach.
AnswerCorrect option: D. Neutralises excess acidity, providing a buffered medium inside the stomach.
Milk of magnesia used as a medicine for treating indigestion is a substance that neutralises excess acidity, providing a buffered medium inside the stomach.
View full question & answer→MCQ 1431 Mark
Of the following, which change will shift the reaction towards the product at equilibrium? $\text{I}_2(\text{g})\rightleftharpoons2\text{I}(\text{g});\Delta\text{H}^\circ(298\text{K})=+150\text{KJ}$
AnswerThe given reaction is endothermic, so on increasing the temperature, it will shift in forward direction.
View full question & answer→MCQ 1441 Mark
One litre of a buffer solution containing $0.01M\ \ce{NH_4Cl}$ and $0.1M\ \ce{ NH_4OH }$ having $\ce{pK_b}$ of $5$ has $\ce{pH}$ of :
Answer$\text{pOH}=\text{pK}_{\text{b}}+\log\frac{[\text{salt}]}{[\text{base}]}$
$=5+\log\frac{0.01}{0.1}5+\log10^{-1}$
$=5-1=4$
$\therefore\text{pH}=14-4=10$
View full question & answer→MCQ 1451 Mark
Which of the following is a biodegradable polymer?
- ✓
- B
$\text{PVC}$
- C
Nylon $-6$
- D
AnswerThe biodegradable polymer is cellulose because many types of microorganisms are known to biodegrade cellulose. Cellulose is the main constituent of plant cell walls which make the leaves, branches strong and hard. Bacteria and fungi are of particular interest because they are the most widely available degrading microorganisms.
View full question & answer→MCQ 1461 Mark
Which of the following reactions is correct regarding homogeneous equilibria?
- A
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$
- B
$\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}(\text{l})$
$\rightleftharpoons\text{CH}_3\text{COOH}(\text{aq})+\text{C}_2\text{H}_5\text{OH(aq)}$
- C
$\text{Fe}^{3+}(\text{aq})+\text{SCN}^-(\text{aq})\rightleftharpoons\text{[Fe(SCN)}]^{2+}(\text{aq})$
- ✓
AnswerIn the gaseous reaction,
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}),$
reactants and products are in the homogeneous phase.
Similarly, for the reactants,
$\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}(\text{l})$
$\rightleftharpoons\text{CH}_3\text{COOH}(\text{aq})+\text{C}_2\text{H}_5\text{OH(aq)}$
and $\text{Fe}^{2+}(\text{aq})+\text{SCN}^-(\text{aq})\rightleftharpoons\text{[Fe(SCN)}]^{2+}(\text{aq})$
all the reactants and products are in the homogeneous solution phase.
View full question & answer→MCQ 1471 Mark
Buffer solution is prepared by mixing :
- A
Strong acid $+$ its salt of strong base.
- ✓
Weak acid $+$ its salt of strong base.
- C
Strong acid $+$ its salt of weak base.
- D
AnswerCorrect option: B. Weak acid $+$ its salt of strong base.
Acidic buffer solution is prepared by mixing a weak acid and its salt with strong base.
Fro example, a mixture of acetic acid $($a weak acid$)$ and sodium acetate $($salt with strong base sodium hydroxide$)$ acts as buffer.
View full question & answer→MCQ 1481 Mark
"An acid is a substance that is capable of donating a hydrogen ion $H^+$ and bases are substances capable of accepting a hydrogen ion, $H^+$". The above statement is justified by
AnswerCorrect option: B. Bronsted $-$ Lowry theory.
View full question & answer→MCQ 1491 Mark
Which of the following salts will give highest $\ce{pH}$ in water?
AnswerCorrect option: C. $ \mathrm{Na}_2 \mathrm{CO}_3 $
It will give highest $\ce{pH}$ in water because its is basic in nature, $ \mathrm{CuSO}_4$ is acidic, $\text{KCl}$ and $\text{NaCl}$ are neutral.
View full question & answer→MCQ 1501 Mark
Pure ammonia is placed in a vessel at a temperature where its dissociation constant $(\alpha )$ is appreciable. At equilibrium :
- ✓
$\ce{K_p}$ does not change significantly with pressure.
- B
$\alpha$ does not change with pressure.
- C
The concentration of $\ce{NH_3}$ does not change with pressure.
- D
The concentration of hydrogen is less than that of nitrogen.
AnswerCorrect option: A. $\ce{K_p}$ does not change significantly with pressure.
The equilibrium constant does not change at all with changes in concentrations, volume, pressure, presence of catalyst, etc. It changes only with changes in temperature of the system. For endothermic reaction, the value of $K$ increases with increase in temperature and vice versa. For exothermic reaction, the value of $K$ decreases with increase in temperature and vice versa.
View full question & answer→MCQ 1511 Mark
Fear or excitement generally causes one to breathe rapidly and it results in the decrease of concentration of $\ce{CO_2}$ in blood. In what way, it will change the $\ce{pH}$ of blood?
- A
$\ce{pH}$ will increase
- B
$\ce{pH}$ will decrease
- ✓
- D
$\ce{pH}$ will be $7$
AnswerAs we know, $\ce{CO_2}$ is acidic in nature. So, a decrease in its concentration should increase the $\ce{pH}.$ But, blood contains serum which has proteins in it. This acts as a buffer solution. We already know that buffer solutions resist a change in the $\ce{pH}$ even on addition of a small amount of acid or small amount of base or dilution, etc.
View full question & answer→MCQ 1521 Mark
$0.10M \ \ce{CH_3COOH}$ is $1.34\%$ ionised, calculate its $\ce{K_a}$ :
- ✓
$ 1.8 \times 10^{-5} $
- B
$ 1.8 \times 10^{-4} $
- C
$ 5 \times 10^{-4} $
- D
$ 4 \times 10^{-5} $
AnswerCorrect option: A. $ 1.8 \times 10^{-5} $
$\text{K}_{\text{a}}=\text{C}\alpha^2=0.1\times\frac{1.34}{100}\times\frac{1.34}{100}=1.795\times10^{-5}$
View full question & answer→MCQ 1531 Mark
Which of the following is not a buffer solution?
- A
$ \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CH}_3 \mathrm{COONa} $
- B
$ \mathrm{H}_3 \mathrm{BO}_3+\mathrm{Na}_3 \mathrm{BO}_3 $
- ✓
$\mathrm{HClO}_4+\mathrm{NaClO}_4 $
- D
$ \mathrm{NH}_4 \mathrm{OH}+\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 $
AnswerCorrect option: C. $\mathrm{HClO}_4+\mathrm{NaClO}_4 $
A buffer solution is a mixture of a weak acid or a weak base and its salt.
In the given question, $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{H}_3 \mathrm{BO}_3$ are weak acids whereas $\mathrm{NH}_4 \mathrm{OH}$ is a weak base. $\mathrm{HClO}_4$ is a strong acid, its mixture with its salt cannot be a buffer solution according to the definition of buffer solution.
View full question & answer→