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.
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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.
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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.
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"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.
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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.
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