Ionic Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

Q 1 :

Given below are two statements:

Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities.

Statement (II) : Blood is naturally occurring buffer solution whose pH is maintained by $H_{2} {CO}_{3} / {HCO}_{3}^{-}$ concentrations.

In the light of the above statements, choose the correct answer from the options given below. [2024]

Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement lI are true

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(3)

Acidic buffer: solution of weak acid and its salt with strong base.

This buffer is effective in the range $0.1 < \frac{[acid]}{[salt]} < 10$

Basic buffer: solution of weak base and its salt with strong acid.

This buffer is effective in the range $0.1 < \frac{[base]}{[salt]} < 10$

Q 2 :

The pH of an aqueous solution containing 1M benzoic acid $({pK}_{a} = 4.20)$ and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ________ mL. (given: log 2 = 0.3) [2024]

(100)

As total volume of mixture is 300mL, let V mL of 1M benzoic acid is mixed with (300-V) mL of 1M sodium benzoate solution.

Moles of benzoic acid $(n_{a c i d})$ = Molarity $\times$ volume

= 1M $\times$ Vml = Vmmol

Moles of sodium benzoate $(n_{s a l t})$ = Molarity $\times$ volume = 1M $\times$ (300 - V)mL = (300 - V) mmol

A mixture of benzoic acid (weak acid) and sodium benzoate is an acidic buffer. For acidic buffer:

$pH = {pK}_{a} + \log \frac{[salt]}{[acid]} = {pK}_{a} + \log \frac{n_{salt}}{n_{acid}}$

$4.5 = 4.20 + \log \frac{300 - V}{V}$

$\log \frac{300 - V}{V} = 0.3$

$\frac{300 - V}{V} = 2$

$V = 100 mL$

Q 3 :

Only litre buffer solution was prepared by adding 0.10 mol each of ${NH}_{3}$ and ${NH}_{4} Cl$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ____ $\times 10^{- 2}$ , (Nearest integer)

(Given : $p K_{b}$ of ${NH}_{3}$ = 4.745 and $\log_{10}$ 3 = 0.477) [2025]

(48)

Before adding HCl:

$p O H = p K_{b} + \log \frac{[N H_{4}^{+}]}{[N H_{3}]}$

$p O H = 4.745 + \log \frac{0.1}{0.1} = 4.745$

After adding 0.05 mol HCl:

$\begin{matrix} N H_{3} & + & H C l & \to & N H_{4} C l \\ t=0 (mol) & 0.1 & 0.05 & 0.1 \\ t =∞ (mol) & 0.1 - 0.05 & 0 & 0.1 + 0.05 \\ = 0.05 & = 0.15 \end{matrix}$

$p O H_{new} = p K_{b} + \log \frac{{[N H_{4}^{+}]}_{new}}{{[N H_{3}]}_{_{new}}}$

$p O H_{new} = p K_{b} + \log \frac{0.15}{0.05}$

$p O H_{new} = 4.745 + \log 3 = 4.745 + 0.477$

Change in pH = Change in pOH

$= p O H_{new} - p O H$

$= (4.745 + 0.477) - 4.745 = 0.477 = 47.7 \times 10^{- 2}$

(Nearest integer = 48)

Q 4 :

A litre of buffer solution contains 0.1 mole of each of ${NH}_{3}$ and ${NH}_{4} Cl$ . On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be ________ $\times 10^{- 3}$ (Nearest integer)

[Given: ${pK}_{b} ({NH}_{3})$ = 4.745

log 2 = 0.301

log 3 = 0.477

T = 298 K] [2023]

(9079)

$\begin{matrix} {NH}_{3} & + & {NH}_{4} Cl \\ 0.1 mole & 0.1 mole \\ {NH}_{3} & + & HCl & \to & {NH}_{4} Cl \\ 0.1 mole & 0.02 mole & 0.1 mole \\ 0.08 mole & __ & 0.12 mole \end{matrix}$

$pOH = p K_{b} + \log \frac{[{NH}_{4} Cl]}{[{NH}_{3}]}$

$= 4.745 + \log (\frac{0.12}{0.08})$

$= 4.745 + \log (\frac{3}{2})$

$= 4.745 + [0.48 - 0.30]$

$= 4.745 + 0.18 = 4.921$

$pH = 14 - 4.921$

$= 9079 \times 10^{- 3}$

Q 5 :

20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ________ $\times 10^{- 2}$ (Nearest integer)

Given: ${pK}_{a} ({CH}_{3} COOH)$ = 4.76

log 2 = 0.30

log 3 = 0.48 [2023]

(458)

$NaOH$ $+$ ${CH}_{3} COOH \to {CH}_{3} COONa + H_{2} O$

$0.1 M, 20 ml 0.1 M, 50 ml$

$Millimole = 0.1 \times 20 = 2, 0.1 \times 50 = 5$

$L.R. = NaOH$

$So$ $0$ $5 - 2 = (3)$ $(2)$

$So resultant solution is an acidic buffer solution$

So $pH = p K_{a} + \log (\frac{salt}{acid})$

$pH = 4.76 + \log \frac{2}{3}$

$pH = 4.76 + (- 0.18) = 4.58 = 458 \times 10^{- 2}$

Q 6 :

Which of the following mixture gives a buffer solution with pH = 9.25 ?
Given : $p K_{b} (N H_{4} O H)$ = 4.75 [2026]

$0.4 M N H_{4} O H (1 L) + 0.1 M H C l (1 L)$
$0.2 M N H_{4} O H (0.5 L) + 0.1 M H C l (0.5 L)$
$0.2 M N H_{4} O H (0.4 L) + 0.1 M H C l (1 L)$
$0.5 M N H_{4} O H (0.2 L) + 0.2 M H C l (0.5 L)$

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(2)

$pOH = p K_{b} + \log (\frac{Salt}{Base})$

$4.75 = 4.75 + \log (\frac{Salt}{Base})$

$Millimoles of [Salt] = millimoles of [Base]$

$Option (D):$

$\begin{matrix} {NH}_{4} OH & + & HCl & \to & {NH}_{4} Cl & + & H_{2} O \\ 0.2 M, 0.5 L & 0.1 M, 0.5 L \\ 100 mmole & 50 mmole \\ 50 mmole & - & 50 mmole \end{matrix}$

$Millimoles of {NH}_{4} OH = millimoles of {NH}_{4} Cl$

Q 7 :

Consider a weak base ‘B’ of $p K_{b} = 5.699$ .. ‘x’ mL of 0.02 M HCl and ‘y’ mL of 0.02 M weak base ‘B’ are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of ‘x’ and ‘y’ respectively are:
$[Given : \log 2 = 0.3010, \log 3 = 0.4771, \log 5 = 0.699]$ [2026]

$x$ $y$

14.3 85.7
$x$ $y$

11.1 88.9
$x$ $y$

85.7 14.3
$x$ $y$

42.7 57.3

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(1)

$\begin{matrix} HCl & + & B & \to & {BH}^{+} & + & {Cl}^{-} \\ 0.02 M & 0.02 M \\ x ml & y ml \\ t_{f} & 0 & 0.02 y - 0.02 x & 0.02 x \end{matrix}$

$pOH = p K_{b} + \log [\frac{Salt}{Base}]$

$5 = 5.699 + \log [\frac{Salt}{Base}]$

$\frac{x}{y - x} = \frac{1}{5}$

$6 x = y$

$7 x = 100$

$x = \frac{100}{7} ml$

$and y = \frac{600}{7} ml$

Q 8 :

Assertion A: An aqueous solution of ammonium acetate can act as a buffer.

Reason R: Acetic acid is a weak acid and ${NH}_{4} OH$ is a weak base.

Both A and R are true and R is the correct explanation of A
Both A and R are true but R is not the correct explanation of A
A is false but R is true
Both A and R are false

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(2)

Both assertion and reason are true but reason is not correct explanation of assertion.

Ammonium acetate is a salt of weak acid $({CH}_{3} COOH)$ and weak base $({NH}_{4} OH)$ .

Q 9 :

At $25 ° C$ , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are:

Given: $K_{a} = 5 \times 10^{- 4}$ , $p K_{a} = 3.3$ , $α ≪ 1$ [2026]

$\begin{matrix} (a) \\ 0.7 \end{matrix}$ $\begin{matrix} (b) \\ 2.0 \end{matrix}$
$\begin{matrix} (a) \\ 2.0 \end{matrix}$ $\begin{matrix} (b) \\ 3.3 \end{matrix}$
$\begin{matrix} (a) \\ 1.1 \end{matrix}$ $\begin{matrix} (b) \\ 2.2 \end{matrix}$
$\begin{matrix} (a) \\ 3.0 \end{matrix}$ $\begin{matrix} (b) \\ 2.2 \end{matrix}$

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(2)

(a) When no NaOH is added, pH is just due to weak acid. pH of weak acid (if $α ≪ 1$ )

$= \frac{1}{2} [p K_{a} - \log C] = \frac{1}{2} [- \log K_{a} - \log C]$

$= \frac{- 1}{2} \log C K_{a} = \frac{- 1}{2} \log 0.2 \times 5 \times 10^{- 4} = 2$

(b) When 10 mL of 0.2 M NaOH is added:

$\begin{array}{lccccc} HX & + & NaOH & \to & NaX & + & H_{2} O \\ t = 0 (mmol) & 0.2 \times 20 = 4 & 0.2 \times 10 = 2 & 0 & excess \\ t = \infty (mmol) & 4 - 2 = 2 & 0 & 2 & excess \end{array}$

The resulting solution is an acidic buffer.

$pH of the acidic buffer = p K_{a} + \log \frac{[Salt]}{[Acid]}$

$pH = 3.3 + \log \frac{2}{2} = 3.3$

Q 10 :

The pH of a solution obtained by mixing 5 mL of 0.1 M ${NH}_{4} OH$ solution with 250 mL of 0.1 M ${NH}_{4} Cl$ solution is _____ $\times 10^{- 2}$ . (Nearest integer).

Given $p K_{b} ({NH}_{4} OH) = 4.74$

log 2 = 0.30

log 3 = 0.48

log 5 = 0.70 [2026]

(756)

Millimoles of ${NH}_{4} OH = Molarity \times volume in mL$

$= 0.1 \times 5 mmol = 0.5 mmol$

Millimoles of ${NH}_{4} Cl = 0.1 \times 250 mmol = 25 mmol$ .

Total volume of solution $= (250 + 5) mL = 255 mL$

Concentration of ${NH}_{4} OH$ in final solution $= (\frac{0.5}{255}) M$

Concentration of ${NH}_{4} Cl$ in final solution $= (\frac{25}{255}) M$

The resulting solution is a basic buffer

$pOH = p K_{b} + \log \frac{[{NH}_{4} Cl]}{[{NH}_{4} OH]}$

$pOH = 4.74 + \log \frac{25 / 255}{0.5 / 255}$

$pOH = 4.74 + \log 50$

$pOH = 4.74 + \log 5 + \log 10$

$pOH = 4.74 + 0.7 + 1$

$pOH = 6.44$

$pH = 14 - pOH = 14 - 6.44 = 7.56 = 756 \times 10^{- 2}$

Topic Question Set

Q 2 :

The pH of an aqueous solution containing 1M benzoic acid $({pK}_{a} = 4.20)$ and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ________ mL. (given: log 2 = 0.3) [2024]

Q 5 :

20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ________ $\times 10^{- 2}$ (Nearest integer)

Given: ${pK}_{a} ({CH}_{3} COOH)$ = 4.76

log 2 = 0.30

log 3 = 0.48 [2023]

Q 6 :

Which of the following mixture gives a buffer solution with pH = 9.25 ?
Given : $p K_{b} (N H_{4} O H)$ = 4.75 [2026]

Q 8 :

Assertion A: An aqueous solution of ammonium acetate can act as a buffer.

Reason R: Acetic acid is a weak acid and ${NH}_{4} OH$ is a weak base.

Q 10 :

The pH of a solution obtained by mixing 5 mL of 0.1 M ${NH}_{4} OH$ solution with 250 mL of 0.1 M ${NH}_{4} Cl$ solution is _____ $\times 10^{- 2}$ . (Nearest integer).

Given $p K_{b} ({NH}_{4} OH) = 4.74$

log 2 = 0.30

log 3 = 0.48

log 5 = 0.70 [2026]

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$x$	$y$
14.3	85.7

$x$	$y$
11.1	88.9

$x$	$y$
85.7	14.3

$x$	$y$
42.7	57.3

Topic Question Set

Q 2 : The pH of an aqueous solution containing 1M benzoic acid (pKa=4.20) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ________ mL. (given: log 2 = 0.3) [2024]

Q 3 : Only litre buffer solution was prepared by adding 0.10 mol each of NH3 and NH4Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ____ ×10-2, (Nearest integer) (Given : pKb of NH3 = 4.745 and log10 3 = 0.477) [2025]

Q 4 : A litre of buffer solution contains 0.1 mole of each of NH3 and NH4Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be ________ ×10-3 (Nearest integer) [Given: pKb(NH3) = 4.745 log 2 = 0.301 log 3 = 0.477 T = 298 K] [2023]

Q 5 : 20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ________ ×10-2 (Nearest integer) Given: pKa(CH3COOH) = 4.76 log 2 = 0.30 log 3 = 0.48 [2023]

Q 6 : Which of the following mixture gives a buffer solution with pH = 9.25 ? Given : pKb(NH4OH) = 4.75 [2026]

Q 7 : Consider a weak base ‘B’ of pKb=5.699.. ‘x’ mL of 0.02 M HCl and ‘y’ mL of 0.02 M weak base ‘B’ are mixed to make 100 mL of a buffer of pH 9 at 25°C. The values of ‘x’ and ‘y’ respectively are: [Given: log2=0.3010, log3=0.4771, log5=0.699] [2026]

Q 8 : Assertion A: An aqueous solution of ammonium acetate can act as a buffer. Reason R: Acetic acid is a weak acid and NH4OH is a weak base.

Q 9 : At 25°C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are: Given: Ka=5×10-4, pKa=3.3, α≪1 [2026]

Q 10 : The pH of a solution obtained by mixing 5 mL of 0.1 M NH4OH solution with 250 mL of 0.1 M NH4Cl solution is _____ ×10-2. (Nearest integer). Given pKb(NH4OH)=4.74 log 2 = 0.30 log 3 = 0.48 log 5 = 0.70 [2026]

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Q 2 :

The pH of an aqueous solution containing 1M benzoic acid $({pK}_{a} = 4.20)$ and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ________ mL. (given: log 2 = 0.3) [2024]

Q 5 :

20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ________ $\times 10^{- 2}$ (Nearest integer)

Given: ${pK}_{a} ({CH}_{3} COOH)$ = 4.76

log 2 = 0.30

log 3 = 0.48 [2023]

Q 6 :

Which of the following mixture gives a buffer solution with pH = 9.25 ?
Given : $p K_{b} (N H_{4} O H)$ = 4.75 [2026]

Q 8 :

Assertion A: An aqueous solution of ammonium acetate can act as a buffer.

Reason R: Acetic acid is a weak acid and ${NH}_{4} OH$ is a weak base.

Q 10 :

The pH of a solution obtained by mixing 5 mL of 0.1 M ${NH}_{4} OH$ solution with 250 mL of 0.1 M ${NH}_{4} Cl$ solution is _____ $\times 10^{- 2}$ . (Nearest integer).

Given $p K_{b} ({NH}_{4} OH) = 4.74$

log 2 = 0.30

log 3 = 0.48

log 5 = 0.70 [2026]