Question 12 Marks
Calculate the pH of the resultant mixtures:
10 mL of $0.1 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4+10 \mathrm{~mL}$ of 0.1 M KOH
Answer$\text{Moles of H}_3\text{O}^+=\frac{2\times10\times0.1}{1000}=.002\text{mol}$
$\text{Moles of OH}^-=\frac{10\times.01}{1000}=0.001\text{mol}$
Excess of $\text{H}_3\text{O}^+=.001\text{mol}$
Thus, $[\text{H}_3\text{O}^+]=\frac{.001}{20\times10^{-3}}=\frac{10^{-3}}{20\times10^{-3}}=.05$
$\therefore\ \text{pH}=-\log(0.05)$
$=1.30$
View full question & answer→Question 22 Marks
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
$\text{PCl}_5\text{ (g)}\rightleftharpoons\text{PCl}_3\text{ (g) + }\text{Cl}_2\text{ (g)}$
AnswerThe number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
View full question & answer→Question 32 Marks
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
AnswerFor the concentration of pure solid or pure liquid,
$\text{Molar conc.}=\frac{\text{Molar of the substance}}{\text{Volume of the substance}}$
Since density of pure solid or liquid is constant at constant temperature and molar mass is also constant therefore, their molar concentrations are constant and are included in the equilibrium constant.
View full question & answer→Question 42 Marks
Assuming complete dissociation, calculate the pH of the following solutions:
0.005 M NaOH
Answer0.005M NaOH
$\text{NaOH}_\text{(aq)}\leftrightarrow\text{Na}_\text{(aq)}++\text{HO}^-_\text{(aq)}$
$[\text{H}\text{O}^-]=[\text{NaOH]}$
$\Rightarrow[\text{H}\text{O}^-]=.005$
$\text{pOH}=-\log[\text{H}\text{O}^-]=-\log(.005)$
$\text{pOH}=2.30$
$\therefore\text{pH}=14-2.30$
$=11.70$
Hence, the pH of the solution is 11.70.
View full question & answer→Question 52 Marks
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
What happens when equilibrium is restored finally and what will be the final vapour pressure?
AnswerFinally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.
View full question & answer→Question 62 Marks
Calculate the pH of the resultant mixtures:
10 mL of $0.01 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4+10 \mathrm{~mL}$ of $0.01 \mathrm{M} \mathrm{~Ca}(\mathrm{OH})_2$
Answer$\text{Moles of H}_3\text{O}^+=\frac{2\times10\times0.1}{1000}=.0002\text{mol}$
$\text{Moles of OH}^-=\frac{2\times10\times.01}{1000}=.0002\text{mol}$
Since there is neither an excess of $\text{H}_3\text{O}^+$ or $\text{OH}^-,$the solution is neutral. Hence, pH = 7.
View full question & answer→Question 72 Marks
Assuming complete dissociation, calculate the pH of the following solutions:0.002 M HBr
Answer0.002 M HBr
$\text{HBr}+\text{H}_2\text{O}\leftrightarrow\text{H}_3\text{O}^++\text{Br}^-$
$[\text{H}_3\text{O}^+]=[\text{HBr]}$
$\Rightarrow[\text{H}_3\text{O}^+]=.002$
$\therefore\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log(0.002)$
$=2.69$
Hence, the pH of the solution is 2.69.
View full question & answer→Question 82 Marks
Assuming complete dissociation, calculate the pH of the following solutions:
0.003M HCl
Answer0.003M HCl:
$\text{H}_2\text{O}+\text{HCl}\leftrightarrow\text{H}_3\text{O}^++\text{Cl}^-$
Since HCl is completely ionized,
$[\text{H}_3\text{O}^+]=[\text{HCl].}$
$\Rightarrow[\text{H}_3\text{O}^+]=0.003$
Now,
$\text{pH}=-\log[\text{H}_3\text{O}^+]=-\log(.003)$
$=2.52$
Hence, the pH of the solution is 2.52.
View full question & answer→Question 92 Marks
Assuming complete dissociation, calculate the pH of the following solutions:0.002 M KOH
Answer0.002 M KOH:
$\text{KOH}_\text{(aq)}\leftrightarrow\text{K}_\text{(aq)}+\text{OH}^-_\text{(aq)}$
$[\text{O}\text{H}^-]=[\text{KOH]}$
$\Rightarrow[\text{O}\text{H}^-]=.002$
Now, $\text{pOH}=-\log[\text{OH}^-]$
$=2.69$
$\therefore\text{pH}=14-2.69$
$=11.31$
Hence, the pH of the solution is 11.31.
View full question & answer→Question 102 Marks
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
How do rates of evaporation and condensation change initially?
AnswerOn increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
View full question & answer→Question 112 Marks
The species: $\text{H}_2\text{O}$, $\text{HCO}_3^-,$ $\text{HSO}_4^-$ and $\text{NH}_3$ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.
AnswerThe table below lists the conjugate acids and conjugate bases for the given species.
| $\text{Species Conjugate acid Conjugate base}$ |
| $\text{H}_2\text{O}$ |
$\text{H}_3\text{O}^+$ |
$\text{OH}-$ |
| $\text{HCO}_3^-$ |
$\text{H}_2\text{CO}_3$ |
$\text{CO}_3^{2-}$ |
| $\text{HSO}^-_4$ |
$\text{H}_2\text{SO}_4$ |
$\text{SO}_4^{2-}$ |
| $\text{NH}_3$ |
$\text{NH}_4^+$ |
$\text{NH}_2^-$ |
View full question & answer→Question 122 Marks
Define Lewis acids and bases with example.
AnswerLewis acids are those which can accept a pair of electrons or negatively charged ions, e.g. $\mathrm{BCl}_3$. Lewis bases can donate a pair of electrons or negatively charged ions, e.g. $\mathrm{NH}_3$.
View full question & answer→Question 132 Marks
Write conjugate acid and conjugate base of $\mathrm{H}_2 \mathrm{O}$.
AnswerConjugate acid is $\mathrm{H}_3 \mathrm{O}^{+}$and conjugate base is $\mathrm{OH}^{-}$. Add $\mathrm{H}^{+}$to get conjugate acid and remove $\mathrm{H}^{+}$to get conjugate base.
View full question & answer→Question 142 Marks
Why do we sweat more on humid day?
AnswerIt is because on humid day Water vapours are more in air, therefore sweat does not get evaporated easily and therefore we sweat more.
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At 0°C, ice and water are present in equilibrium.
What will happen on increasing the pressure?
AnswerOn increasing the pressure, ice melts to form water (because water has lesser volume than ice).
View full question & answer→Question 162 Marks
Write the expression for equilibrium constant $K_p$ for the reaction:
$3\text{Fe(s)}+\text{4H}_2\text{O(g)}\rightleftharpoons\text{Fe}_3\text{O}_4(\text{s})+4\text{H}_2(\text{g})$
Answer$\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{pH}_2\right)^4}{\left(\mathrm{pH}_2 \mathrm{O}\right)}\left[\mathrm{Fe}\right.$ and $\mathrm{Fe}_3 \mathrm{O}_4$ are pure solids]
$[\mathrm{Fe}(\mathrm{s})]=\left[\mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})\right]=1$.
View full question & answer→Question 172 Marks
If the value of an equilibrium constant for a particular reaction is $1.6 \times 10^{12}$, then at equilibrium, what will be present in the system?
AnswerMostly products will be formed because value of K is high. It means conc. of products is very high as compared to reactants.
View full question & answer→Question 182 Marks
$\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_4 \mathrm{OH}$ is $1.8 \times 10^{-5}$ and for $\mathrm{CH}_3 \mathrm{NH}_2$ is $4.4 \times 10^{-4}$ Which of them is strongest base and why?
Answer$\mathrm{CH}_3 \mathrm{NH}_2$ is strongest base because it has high value of base dissociation constant ( $\mathrm{K}_{\mathrm{b}}$ ). Higher the value of $K_b$ stronger will be the base.
View full question & answer→Question 192 Marks
Which will have $\text{CO}_2$ to more extent, hot cold drink bottle or chilled cold drink bottle, why?
AnswerChilled cold drink will dissolve more $\text{CO}_2$ because solubility of gases in liquid increases with decrease in temperature because force of attraction between gas and liquid increases.
View full question & answer→Question 202 Marks
$\text{A}+\text{3B}\rightleftharpoons\text{2X,}\text{K}=\text{x}$
What will be the equilibrium constant for the decomposition of 1mol of X?
Answer$\text{K}=\frac{[\text{X}^2]}{[\text{A}][\text{B}]^3}=\text{x}$
$\text{X}\rightleftharpoons\frac{1}{2}\text{A}+\frac{3}{2}\text{B}$
$\text{K}'=\frac{[\text{A}^{\frac{1}{2}}][\text{B}]^{\frac{3}{2}}}{[\text{X}]}$
$=\frac{1}{\sqrt{\text{K}}}=\frac{1}{\sqrt{\text{x}}}$
View full question & answer→Question 212 Marks
The ionisation constant of formic acid is $1.8 \times 10^{-4}$. Calculate the ratio of sodium formate and formic acid in a buffer of pH 4.25 .
Answer$\text{pK}_{\text{a}}=-\log(1.8\times10^{-4})=3.74$
$\log\frac{\text{[Salt]}}{[\text{Acid}]}=\text{pH}-\text{pK}_{\text{a}}$
$=4.15-3.74=0.51$
$\text{or }\frac{\text{[Salt]}}{[\text{Acid}]}=\text{Antilog }0.51=3.24$
View full question & answer→Question 222 Marks
- Write the conjugate acid for $\text{NH}_2^-\text{ and }\text{NH}_3.$
- What is the relationship between $p K_a$ and $p K_b$ values?
Answer
- The conjugate acid of $\text{NH}_2^-\text{ is }\text{NH}_3.$ whereas conjugate acid of $\text{NH}_3\text{ and }\text{NH}_4^+.$
[Add $\mathrm{H}^{-}$to get conjugate acid]
- $\text{pK}_{\text{a}}+\text{pK}_{\text{b}}=14,\text{pK}_{\text{a}}=14-\text{pK}_{\text{b}}$
View full question & answer→Question 232 Marks
All Bronsted acids are not Lewis acids. Explain.
AnswerBronsted acids can donate $\mathrm{H}^{+}$easily but they may not be able to donate electrons e.g. $\mathrm{HCl}_{,} \mathrm{H}_2 \mathrm{SO}_4$ $\mathrm{HNO}_3$. Therefore, all Bronsted acids are not Lewis acids. They can not accept electrons.
View full question & answer→Question 242 Marks
The ionisation of hydrochloric in water is given below:$\text{HCl}\text{ (aq)}+\text{H}_2\text{O}\text{ (l)}\rightleftharpoons\text{H}_3\text{O}^+\text{ (aq)}+\text{Cl}^-{\text{ (aq)}}$
Label two conjugate acid-base pairs in this ionisation.
Answer
|
$\mathrm{HCl}$
|
(Acid)
|
| $\mathrm{Cl}^{-}$ |
(Conjugate base)
|
| $\mathrm{H}_2 \mathrm{O}$ |
(Base)
|
| $\mathrm{H}_3 \mathrm{O}^{+}$ |
(Conjugate acid)
|
View full question & answer→Question 252 Marks
The concentration of hydrogen ion in a sample of soft drink is: $3.8\times10^{-3}\text{M}.$What is its $\text{pH}(\log 3.8 = 0.58).$
Answer$\text{pH}=-\log[\text{H}^+]$
$=-\log3.8\times10^{-3}$
$\text{pH}=-\log3.8-\log10^{-3}$
$=-0.579+3.0000=2.4202$
View full question & answer→Question 262 Marks
What type of chemical reaction attain a state of equilibrium when carried out in closed vessel?
AnswerReversible reactions attain a state of equilibrium when carried out in a closed vessel because in open container products will escape.
View full question & answer→Question 272 Marks
Classify the following as homogeneous or heterogeneous equilibria:
- $2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$
- $2\text{Mg(s)}+\text{O}_2(\text{g})\rightleftharpoons2\text{MgO(s)}$
Answer
- Homogeneous equilibria because all reactant and products are gases.
- Heterogeneous equilibria because Mg and MgC are solid whereas $\text{O}_2$ is gas.
View full question & answer→Question 282 Marks
$\mathrm{pK}_{\mathrm{a}}$ values of acids $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are 1.5, 3.5, 2.0 and 5.0 .
Which of them is strongest acid? Give reason.
AnswerAcid 'A' with $p K_a=1.5$ is strongest acid, lower the value of $p K_a$ stronger will be the acid. Higher the value of $K_a$, lower will be value of $\mathrm{pK}_{\mathrm{a}}$.
View full question & answer→Question 292 Marks
Which of the following is Lewis acid but not Bronsted acid? $\text{HBrO}_3, \text{SbCl}_3, \text{HSO}_4^-, \text{AlF}_3$
Answer$\mathrm{SbCl}_3$ and $\mathrm{AlF}_3$ are Lewis acids but not Bronsted acids because they connot donate $\mathrm{H}^{+}$(protons).
View full question & answer→Question 302 Marks
A sparingly soluble salt having general formula $\text{A}^{\text{p+}}_{\text{x}}\text{B}^{\text{q-}}_{\text{y}}$ and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.
AnswerA sparingly soluble salt having general formula $\text{A}^{\text{p+}}_{\text{x}}\text{B}^{\text{q-}}_{\text{y}}.$ Its molar solubility is S mol $\text{L}^{-1}$.
View full question & answer→Question 312 Marks
Arrange the following in increasing order of pH. $\text{KNO}_3(\text{aq}),\ \text{CH}_3\text{COONa(aq)},\ \text{NH}_4\text{Cl(aq)},\ \text{C}_6\text{H}_5\text{COONH}_4\text{(aq)}$
AnswerSalts of strong acid and strong base do not hydrolyse and form neutral solution thus, pH will be nearly 7 of $\mathrm{KNO}_3$. In sodium acetate, acetic acid remains unionised this results in increase in $\mathrm{OH}^{\prime}$ concentration and pH will be more than $7 . \mathrm{NH}_4 \mathrm{Cl}$ formed from weak base, NH OH and strong acid, HCl , in water dissociates completely, aq. ammonium ions undergo hydrolysis with water to form $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{H}^{+}$ions resulting in less pH value.
View full question & answer→Question 322 Marks
$\text{H}_2(\text{g})+\text{I}_2(\text{g})\rightleftharpoons2\text{HI}(\text{g}),\text{K}=49$
What is the value of K, for the reaction:
$\text{H}_2(\text{g})\rightleftharpoons\frac{1}{2}\text{H}_2(\text{g})+\frac{1}{2}\text{I}_2(\text{g})?$
Answer$\text{K}=\frac{[\text{HI}^2]}{[\text{H}_2][\text{I}_2]}$
$\text{K}=49$
$\text{K}'=\frac{[\text{H}_2]^{\frac{1}{2}}[\text{I}_2]^{\frac{1}{2}}}{[\text{HI}]}=\frac{1}{\sqrt{\text{K}}}$
$\Rightarrow\text{K}'=\frac{1}{\sqrt{49}}=\frac{1}{7}$
View full question & answer→Question 332 Marks
Glycine is an a-amino acid. It exists in the form of Z witter ion as ${ }^{+} \mathrm{NH}_3 \mathrm{CH}_2 \mathrm{COO}^{-}$
Write the formula of its
- Conjugate acid.
- Conjugate base.
Answer
- Conjugate acid ${ }^{+} \mathrm{NH}_3 \mathrm{CH}_2 \mathrm{COOH}$.
- Conjugate base $\mathrm{NH}_2 \mathrm{CH}_2 \mathrm{COO}^{-}$.
View full question & answer→Question 342 Marks
$\mathrm{BF}_3$ does not have proton but still acts as an acid and reacts with $\mathrm{NH}_3$. Why is it so? What type of bond is formed between the two?
AnswerIn $\mathrm{BF}_3$, the octet of boron is incomplete, therefore in order to complete its octet, it accepts a lone pair of electron. Any species which is capable of accepting a lone pair of electron is acidic in nature and known as Lewis acid. Hence $\mathrm{BF}_3$ is a lewis acid.
When $\mathrm{NH}_3$, which is having a lone pair of electron act as lewis base, reacts with $\mathrm{BF}_3$ by donating its lone pair of electron forms an adduct. The bond form between the two species is known as coordinate bond,
View full question & answer→Question 352 Marks
What is the expression for $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
Answer$\text{Ag}_2\text{CrO}_4(\text{s})\rightleftharpoons2\text{Ag}^+(\text{aq})+\text{CrO}_4^{2-}(\text{aq});$
$\text{K}_{\text{sp}}=[\text{Ag}^+]^2[\text{CrO}_4^{2-}]$
$\because[\text{Ag}_2\text{Cr}_2\text{O}_4(\text{s})]=1.$
View full question & answer→Question 362 Marks
$\text{A}+\text{B}\rightleftharpoons\text{AB};\text{K}=1\times10^2$
$\text{E}+\text{F}\rightleftharpoons\text{EF};\text{K}=1\times10^{-3}$
Out of AB and EF, which one is more stable AB or EF?
AnswerAB is more stable. Higher the value of K more will be the stability of product formed. More stable product is formed to more extent.
View full question & answer→Question 372 Marks
Predict the nature of solution when $\mathrm{NH}_4 \mathrm{NO}_3$ undergo hydrolysis.
Answer$\mathrm{NH}_4 \mathrm{NO}_3$ on hydrolysis gives $\mathrm{NH}_4 \mathrm{OH}$ (weak base), $\mathrm{H}^{+}$and $\mathrm{NO}_3^{-}$( $\mathrm{HNO}_3$ is strong acid) since $\mathrm{H}^{+}$ions are more than $\mathrm{OH}^{-}$.
$\therefore$ Solution is acidic.
View full question & answer→Question 382 Marks
The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation and how is it affected by concentration of sodium chloride?
Answer
- Sugar being a non - electrolyte does not ionize in water, whereas NaCl ionizes completely in water and produces $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions which help in the conduction of electricity.
- When concentration of NaCl is increased, more $\mathrm{Na}^{+}$and Cl ions will be produced. Hence, conductance increases.
View full question & answer→Question 392 Marks
'All Lewis bases are also Bronsted bases'. Is it true? If yes, Why?
AnswerYes, it is true. It is because Lewis bases are- vely charged or electron rich. They are Bronsted bases also because they can accept $\mathrm{H}^{+}$easily, e.g., $\mathrm{NH}_3$ can donate electron (Lewis base) and accept $\mathrm{H}^{+}$ (Bronsted base).
View full question & answer→Question 402 Marks
Ionisation constant of a weak base MOH, is given by the expression.$\text{K}_\text{b}=\frac{[\text{M}^+][\text{OH}^-]}{[\text{MOH}]}$
Values of ionisation constant of some weak bases at a particular temperature are given below:
|
Base
|
Dimethylamine
|
Urea
|
Pyridine
|
Ammonia
|
|
$K_b$
|
$5.4 \times 10^{–4}$
|
$1.3 \times 10^{–14}$
|
$1.77 \times 10^{–9}$
|
$1.77 \times 10^{–5}$
|
Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest? AnswerGreater is the ionization constant ($K_b$) of a base, greater is the ionization of the base. Order of extent of ionization at equilibrium is dimethylamine > ammonia > pyridine > urea. Dimethylamine is the strongest base due to maximum value of $K_b$.
View full question & answer→Question 412 Marks
What is the effect of increasing pressure on the equilibrium?
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$
AnswerOn increasing pressure on the equilibrium:
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$
The reaction will shift in the forward direction, i.e. towards lesser number of moles because number of moles are decreasing from reactants to products.
View full question & answer→Question 422 Marks
What is the relation between $\mathrm{K}_p$ and $\mathrm{K}_{\mathrm{c}}$ ?
Answer$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\left(\mathrm{RT}^{\Delta \mathrm{n}}\right)$
where $\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\Delta \mathbf{n}=$ number of moles of gaseous products - number of moles of gaseous reactants.
View full question & answer→Question 432 Marks
What will be the pH of 1M $\mathrm{Na}_2 \mathrm{SO}_4$ solution?
Answer$\mathrm{Na}_2 \mathrm{SO}_4$ is salt of strong acid and strong base, thus its aqueous solution will be neutral. Therefore, its pH will be 7.
View full question & answer→Question 442 Marks
What could be temperature $15^{\circ} \mathrm{C}$ or $100^{\circ} \mathrm{C}$ for $\mathrm{K}_{\mathrm{w}}=7.5 \times 10^{-14}$. What happens to ionic product if some acid is added to it?
AnswerTemperature will be $100^{\circ} \mathrm{C} . \mathrm{K}_{\mathrm{w}}$ increases with increase in temperature $\mathrm{K}_{\mathrm{w}}$ (ionic product) does not change if same acid is added to it.
View full question & answer→Question 452 Marks
$\text{Hb(s)}+\text{O}_2\text{(g)}\rightleftharpoons\text{HbO}_2\text{(s)}$
Predict the direction in which equilibrium gets shifted it partial pressure of $\text{O}_2$ is lowered.
AnswerThe reaction will shift to backward direction, e.g., in tissues partial pressure of $\text{O}_2$ is less, oxy-haemoglobin releases oxygen.
View full question & answer→Question 462 Marks
$\mathrm{pK}_{\mathrm{a}}$ value of acids A, B, C, D are 1.5, 3.5, 2.0 and 5.0. Which of them is strongest acid?
AnswerAcid A with $\mathrm{pK}_{\mathrm{a}}=1.5$ is strongest acid, lower the value of $\mathrm{pK}_{\mathrm{a}}$ stronger will be the acid.
View full question & answer→Question 472 Marks
Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases?
$\text{OH}^–,\text{ RO}^–,\text{ CH}_3\text{COO}^–,\text{ Cl}^–$
AnswerConjugate acids of given bases are:
$\text{H}_2\text{O},\text{ ROH},\text{ CH3COOH},\text{ HCl}.$
Their acidic strength is in the order.
$\text{HCl}>\text{CH}_3\text{COOH}>\text{H}_2\text{O}>\text{ROH}$ Hence, basic strength is in the order $\text{RO}^->\text{OH}^->\text{CH}_3\text{COO} \xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{Cl}^- > $
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$\text{C(diamond)}\rightleftharpoons\text{C(graphite)}\\\text{d}=3.5\text{g cm}^{-3}\ \ \ \ \text{d}=2.3\text{g cm}^{-3}$
What will be effect of increasing pressure in this equilibrium?
AnswerIt will shift to backward direction because high pressure will load to formation more denser diamond.
View full question & answer→Question 492 Marks
Why is ammonia termed as a base though it does not contain $\text{OH}^-$ ions?
AnswerThe basic nature of ammonia is due to its tendency to donate electron pair. Therefore it is a Lewis base.
View full question & answer→Question 502 Marks
What will be the pH of 0.1 M ammonium acetate solution? $\mathrm{pK}_{\mathrm{a}}=\mathrm{pK}_{\mathrm{b}}=4.74$.
AnswerFor a salt of ammonium acetate,
$\text{pH}=\frac{1}{2}\text{pK}_{\text{w}}+\frac{1}{2}\text{pK}_{\text{a}}-\frac{1}{2}\text{pK}_{\text{b}}$
$=\frac{1}{2}(14)+\frac{1}{2}(4.47)-\frac{1}{2}(4.74)=7$
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