Ionic Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

Q 1 :

$K_{a}$ for ${CH}_{3} COOH$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for ${NH}_{4} OH$ is $1.8 \times 10^{- 5} .$ The pH of ammonium acetate solution will be _______ . [2024]

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$As K_{a} of {CH}_{3} COOH = K_{b} of {NH}_{4} OH$

$So, {pK}_{a} of {CH}_{3} COOH = {pK}_{b} of {NH}_{4} OH$

$And p K_{a} - p K_{b} = 0$

For salt of weak acid and weak base:

$pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$

$pH = \frac{1}{2} (14 + 0) = 7$

Q 2 :

pH of water is 7 at 25°C. If water is heated to 80°C, its pH will : [2025]

Remains the same
Decrease
Increase
$H^{+}$ concentration increases, ${OH}^{-}$ concentration decreases

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(2)

$H_{2} O ⇌ H^{+} + {OH}^{-}$

With increase in temperature, dissociation of water increases, hence $[H^{+}]$ increases. Increase of $[H^{+}]$ means decrease of pH $(p H = - \log [H^{+}]) .$

Q 3 :

An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would

(Given log 2 = 0.30) [2025]

reduce to 0.5
increase to 1.3
remain same
increase to 2

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(2)

Initial concentration of $H^{+} ({[H^{+}]}_{initial}) = 10^{- {pH}_{initial}} = 10^{- 1} M$

As equal volume is added, concentration of $H^{+}$ becomes half

$({[H^{+}]}_{initial}) = \frac{10^{- 1}}{2} M$

${pH}_{final} = - \log {[H^{+}]}_{final} = - \log (\frac{10^{- 1}}{2})$

$= - \log (10^{- 1}) + \log 2 = 1 + 0.3 = 1.3$

Q 4 :

A weak acid HA has degree of dissociation $x$ . Which option gives the correct expression of $(pH - p K_{a})$ ? [2025]

$\log (1 + 2 x)$
$0$
$\log (\frac{x}{1 - x})$
$\log (\frac{1 - x}{x})$

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(3)

$\begin{matrix} HA & ⇌ & H^{+} & + & A^{-} \\ (t=0) (moles) & a_{0} & 0 & 0 \\ (t=∞) (moles) & a (1 - x) & a_{0} x & a_{0} x \end{matrix}$

$K_{a} = \frac{[H^{+}] [A^{-}]}{[HA]} = \frac{[H^{+}] \times a_{0} x}{a_{0} (1 - x)}$

$K_{a} = \frac{[H^{+}] \times x}{(1 - x)}$

Taking $- \log$ on both sides

$- \log K_{a} = - \log (\frac{[H^{+}] \times x}{(1 - x)})$

$- \log K_{a} = - \log [H^{+}] - \log (\frac{x}{1 - x})$

$p K_{a} = pH - \log (\frac{x}{1 - x})$

$pH - p K_{a} = \log (\frac{x}{1 - x})$

Q 5 :

If 1 mM solution of ethylamine produces pH = 9, then the ionization constant $(K_{b})$ of ethylamine is $10^{- x}$ . The value of $x$ is ______ (nearest integer).

[The degree of ionization of ethylamine can be neglected with respect to unity.] [2025]

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$\begin{matrix} C_{2} H_{5} {NH}_{2} (a q) & + & H_{2} O & ⇌ & C_{2} H_{5} {NH}_{3}^{+} (a q) & + & O H^{-} (a q) \\ t = 0 & 10^{- 3} M & excess & 0 & 0 \\ t = \infty & 10^{- 3} (1 - α) & 10^{- 3} α M & 10^{- 3} α M \\ ≃ 10^{- 3} M \end{matrix}$

$pOH = 14 - pH = 14 - 9 = 5$

Thus ${[O H]}^{-} = 10^{- 5} M$

Also, $[C_{2} H_{5} {NH}_{3}^{+}] = {[O H]}^{-} = 10^{- 5} M$

$K_{b} = \frac{{[O H]}^{-} [C_{2} H_{5} {NH}_{3}^{+}]}{[C_{2} H_{5} {NH}_{2}]} = \frac{10^{- 5} \times 10^{- 5}}{10^{- 3}} = 10^{- 7}$

Q 6 :

The pH of a 0.01 M weak acid HX $(K_{a} = 4 \times 10^{- 10})$ is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as $x \times 10^{- 4}$ M. The value of $x$ is ______ (nearest integer). [2025]

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New concentration of $H^{+} ({[H^{+}]}_{new}) = 10^{-pH(new)} M = 10^{- 6} M$

$\sqrt{K_{a} C_{new}} = {[H^{+}]}_{new}$

$\sqrt{4 \times 10^{- 10} C_{new}} = 10^{- 6}$

$4 \times 10^{- 10} C_{new} = 10^{- 12}$

$C_{new} = 25 \times 10^{- 4} M$

Note: In the question it should be 0.25 M weak acid instead of 0.01 M weak acid. However, this data is not needed to solve the question.

Q 7 :

$x$ mg of $M g {(O H)}_{2}$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of $x$ is ____ mg. (Nearest integer)

(Given: $M g {(O H)}_{2}$ is assumed to dissociate completely in $H_{2} O$ ) [2025]

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$pOH = 14 - pH = 14 - 10 = 4$

$[O H^{-}] = 10^{- pOH} = 10^{- 4} M$

$Moles of hydroxide (n_{O H^{-}}) = [O H^{-}] \times volume$

$= 10^{- 4} M \times 1 L = 10^{- 4} mol$

$Moles of Mg(OH)_{2} (n_{Mg(OH)_{2}}) = \frac{n_{O H^{-}}}{2} = \frac{10^{- 4}}{2} mol$

$Mass of Mg(OH)_{2} = n_{Mg(OH)_{2}} \times M_{Mg(OH)_{2}}$

$= \frac{10^{- 4}}{2} \times 58 = 29 \times 10^{- 4} g = 2.9 mg ≈3 g$

Q 8 :

When the hydrogen ion concentration [ $H^{+}$ ] changes by a factor of 1000, the value of pH of the solution ______ [2023]

decreases by 3 units
decreases by 2 units
increases by 2 units
increases by 1000 units

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(1)

${(pH)}_{1} = - \log C$

${(pH)}_{2} = - \log 10^{3} C = - [\log 10^{3} + \log C]$

$= - 3 - \log C$

${(pH)}_{2} - {(pH)}_{1} = - 3 \Rightarrow Decreased by 3$

Q 9 :

Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solutions of pH 12 is $x \times 10^{- 1}$ . The value of $x$ is ______ (Nearest integer). Assume complete dissociation. [2023]

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$pH = 12 \Rightarrow pOH = 2 so [{OH}^{-}] = 10^{- 2} M .$

$Millimole of [{OH}^{-}] = 10^{- 2} \times 100 = 1$

$Ca {(OH)}_{2} (s) ⟶ {Ca}^{2 +} (a q .) + 2 {OH}^{-} (a q .)$

$\frac{1}{2} millimole$ $1 millimole$

${millimole of Ca(OH)}_{2} = \frac{1}{2} = 5 \times 10^{- 1} millimole .$

Q 10 :

600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M $H_{2} {SO}_{4}$ . The pH of the mixture is _____ $\times 10^{- 2}$ . (Nearest Integer)

[Given log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04] [2023]

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$\begin{matrix} HCl & ⟶ & H^{+} & + & {Cl}^{-} \\ Milli moles & 0.01 \times 600 & 6 \\ H_{2} {SO}_{4} & ⟶ & 2 H^{+} & + & {SO}_{4}^{- 2} \\ Milli moles & 0.01 \times 400 & 2 \times 4 = 8 \end{matrix}$

$[H^{+}] = \frac{8 + 6}{1000} = 14 \times 10^{- 3}$

$pH = - \log (14 \times 10^{- 3})$

$= 3 - log 14$

$= 3 - log (2×7)$

$= 3-log 2-log 7$

$= 3 - 0.30 - 0.84$

$= 1.86 = 186 \times 10^{- 2}$

Topic Question Set

Q 1 :

$K_{a}$ for ${CH}_{3} COOH$ is $1.8 \times 10^{- 5}$ and $K_{b}$ for ${NH}_{4} OH$ is $1.8 \times 10^{- 5} .$ The pH of ammonium acetate solution will be _______ . [2024]

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Q 2 :

pH of water is 7 at 25°C. If water is heated to 80°C, its pH will : [2025]

Q 3 :

An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would

(Given log 2 = 0.30) [2025]

Q 4 :

A weak acid HA has degree of dissociation $x$ . Which option gives the correct expression of $(pH - p K_{a})$ ? [2025]

Q 5 :

If 1 mM solution of ethylamine produces pH = 9, then the ionization constant $(K_{b})$ of ethylamine is $10^{- x}$ . The value of $x$ is ______ (nearest integer).

[The degree of ionization of ethylamine can be neglected with respect to unity.] [2025]

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Q 7 :

$x$ mg of $M g {(O H)}_{2}$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of $x$ is ____ mg. (Nearest integer)

(Given: $M g {(O H)}_{2}$ is assumed to dissociate completely in $H_{2} O$ ) [2025]

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Q 8 :

When the hydrogen ion concentration [ $H^{+}$ ] changes by a factor of 1000, the value of pH of the solution ______ [2023]

Q 9 :

Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solutions of pH 12 is $x \times 10^{- 1}$ . The value of $x$ is ______ (Nearest integer). Assume complete dissociation. [2023]

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Q 10 :

600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M $H_{2} {SO}_{4}$ . The pH of the mixture is _____ $\times 10^{- 2}$ . (Nearest Integer)

[Given log 2 = 0.30

log 3 = 0.48

log 5 = 0.69

log 7 = 0.84

log 11 = 1.04] [2023]

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Topic Question Set

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Q 2 : pH of water is 7 at 25°C. If water is heated to 80°C, its pH will : [2025] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 2 :

pH of water is 7 at 25°C. If water is heated to 80°C, its pH will : [2025]