Ionic Equilibrium JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Ionic Equilibrium, free to read — no sign-in needed. The full chapter has 20 questions; sign in to attempt the remaining 15 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   At $25^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}^{+}$ions in $1.00 \times 10^{-3} \mathrm{M}$ aqueous solution of a weak monobasic acid having acid dissociation constant $\left(K_a\right)=4.00 \times 10^{-11}$ is $\boldsymbol{X} \times 10^{-7} \mathrm{M}$. The value of $\boldsymbol{X}$ is _____ . Use: Ionic product of water $\left(K_w\right)=1.00 \times 10^{-14}$ at $25^{\circ} \mathrm{C}$
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   Answer: 2.23

   Because concentration of $\mathrm{H}+$ from weak acid is less we need to consider self ionization of $\mathrm{H}_2 \mathrm{O}$ also. $\begin{array}{rl}\mathrm{HX}(\mathrm{aq}) \rightleftharpoons & \mathrm{H}^{+}+\mathrm{X}^{-}(\mathrm{aq}) \\ 10^{-3}-\mathrm{x} \quad & \mathrm{x}+\mathrm{y} \quad \mathrm{x} \\ \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons & \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ & \mathrm{x}+\mathrm{y} \quad \quad \mathrm{y}\end{array}$ Approximation : $\left(10^{-3}-\mathrm{x}\right) \simeq 10^{-3}$ $\begin{aligned} & \Rightarrow \quad \frac{x(x+y)}{10^{-3}}=K_a=4 \times 10^{-11}....(1) \\ & \Rightarrow \quad y(x+y)=K w=10^{-14}.....(2) \end{aligned}$ $\begin{array}{ll} \text { Add }(1)+(2) \\ \end{array}$ $\begin{array}{ll}\Rightarrow & (\mathrm{x}+\mathrm{y})^2=5 \times 10^{-14} \\ \Rightarrow & \mathrm{x}+\mathrm{y}=\left[\mathrm{H}^{+}\right]=\sqrt{5} \times 10^{-7} \\ \Rightarrow & \mathrm{x}=\sqrt{5}=2.236\end{array}$ Answer 2.23 or 2.24.
2. Q2JEE Advanced Adv 2023 (Paper 1)
   On decreasing the $pH$ from $7$ to $2$, the solubility of a sparingly soluble salt $(MX)$ of a weak acid $(HX)$ increased from ${10}^{-4}mol{L}^{-1}$ to ${10}^{-3}mol{L}^{-1}$. The ${pK}_{a}$ of $HX$ is
   1. A.$3$
   2. B.$4$
   3. C.$5$
   4. D.$2$
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   Answer: (B)

   At $pH=7$, i.e., the solubility in water is $S$, The solubility product, ${K}_{sp}={S}^{2}-----\gt 1$ Assume the solubility of sparingly soluble salt is $atpH=2$ $MX\rightleftharpoons {M}^{⊕}+{X}^{⊖} {S}_{1}{S}_{1}-x$ The weak dissociation reaction is as follows, ${X}^{⊖}+{H}^{⊕}\rightleftharpoons HX {S}_{1}-x{10}^{-2}x$ Assume the acid dissociation constant is ${K}_{a}$. Now, $\dfrac{1}{{K}_{a}}=\dfrac{\left[HX\right]}{\left[{H}^{+}\right]\left[{X}^{-}\right]}$ ${K}_{sp}={S}_{1}\left({S}_{1}-x\right)$ $\dfrac{1}{{K}_{a}}=\dfrac{{S}_{1}}{\left[{H}^{+}\right]{S}_{1}-x}$ assume ${S}_{1}≃x$ ${S}_{1}-x=\dfrac{{K}_{sp}}{{S}_{1}}$ $\dfrac{1}{{K}_{a}}=\dfrac{{{S}_{1}}^{2}}{\left[{H}^{+}\right]{K}_{sp}} \Rightarrow {{S}_{1}}^{2}=\dfrac{{10}^{-2}{K}_{sp}}{{K}_{a}}-------\gt 2$ Now from 1 and 2, $\dfrac{{{S}_{1}}^{2}}{{S}^{2}}=\dfrac{{10}^{-2}}{{K}_{a}} \Rightarrow {K}_{a}=\dfrac{{10}^{-2}\times {10}^{-8}}{{10}^{-6}} \Rightarrow {K}_{a}={10}^{-4} \Rightarrow {pK}_{a}=4$
3. Q3JEE Advanced Adv 2022 (Paper 2)
   Concentration of ${H}_{2}{SO}_{4}$ and ${Na}_{2}{SO}_{4}$ in a solution is $1M$ and $1.8\times {10}^{-2}M$, respectively. Molar solubility of ${PbSO}_{4}$ in the same solution is $X\times {10}^{-Y}M$ (expressed in scientific notation). The value of $Y$ is [Given: Solubility product of ${PbSO}_{4}\left({K}_{sp}\right)=1.6\times {10}^{-8}$. For ${H}_{2}{SO}_{4},{K}_{a1}$ is very large and ${K}_{a2}=1.2\times {10}^{-2}$] If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
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   Answer: 6

   $\begin{matrix}{H}_{2}{SO}_{4} & \rightarrow _{} & {H}^{+} & + & {HSO}_{4}^{2-} \\ & & 1M & & 1M\end{matrix}$ (${K}_{{a}_{1}}$ is very large) ${HSO}_{4}^{-}\rightleftharpoons ^{}{H}^{+}+{SO}_{4}^{2-}{K}_{{a}_{2}}=1.2\times {10}^{-2}$ $\left[{SO}_{4}^{2-}\right]$ coming from ${Na}_{2}{SO}_{4}=1.8\times {10}^{-2}$ $\dfrac{\left[{SO}_{4}^{2-}\right]\left[{H}^{+}\right]}{\left[{HSO}_{4}^{-}\right]}=\dfrac{1.8\times {10}^{-2}\times 1}{1}\gt {K}_{{a}_{2}}$ $∴$ Rather than dissociation of ${HSO}_{4}^{-}$ into ${H}^{+}$and ${SO}_{4}^{2-}$ ions, association between already present ${H}^{+}$ and ${SO}_{4}^{2-}$ will take place. Assuming '$x$' $mol/L$ of ${SO}_{4}^{2-}$ and ${H}^{+}$combines to form ${HSO}_{4}^{-}$ $∴\left[{SO}_{4}^{2-}\right]=1.8\times {10}^{-2}-x$ $\left.\begin{matrix}\left[{H}^{+}\right]=1-x\approx 1 \\ \left[{HSO}_{4}^{-}\right]=1+x\approx 1\end{matrix}\right\}$ (assuming $x\lt \lt 1$) $\dfrac{\left(1.8\times {10}^{-2}-x\right)1}{1}=1.2\times {10}^{-2}$ $\Rightarrow x=0.6\times {10}^{-2}$ $\left[{SO}_{4}^{2-}\right]=1.2\times {10}^{-2}M$ ${PbSO}_{4}\left(s\right)\rightleftharpoons {Pb}^{2+}\left(aq\right)+{SO}_{4}^{2-}\left(aq\right)$ If solubility of ${PbSO}_{4}=sM$ $∴\left[{Pb}^{2+}\right]=s$ $\left[{SO}_{4}^{2-}\right]=s+1.2\times {10}^{-2}\approx 1.2\times {10}^{-2}$ (assuming $s≪1.2\times {10}^{-2}$) $∴s\times 1.2\times {10}^{-2}=1.6\times {10}^{-8}$ $s=\dfrac{1.6}{1.2}\times {10}^{-6}=1.33\times {10}^{-6}$ On comparing with $X\times {10}^{-Y}$ $Y=6$
4. Q4JEE Advanced Adv 2022 (Paper 1)
   A solution is prepared by mixing $0.01mol$ each of ${H}_{2}{CO}_{3},{NaHCO}_{3},{Na}_{2}{CO}_{3}$, and $NaOH$ in $100 \mathrm{~mL}$ of water. $pH$ of the resulting solution is [Given: ${pK}_{{a}_{1}}$ and ${pK}_{{a}_{2}}$ of ${H}_{2}{CO}_{3}$ are $6.37$ and $10.32$, respectively. $\log 2=0.30$]
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   Answer: 10.02

   First acid base reaction between ${H}_{2}{CO}_{3}$ and $NaOH$ takes place. ${H}_{2}{CO}_{3}{}_{0.01mole}+NaOH_{0.01mole}⟶{NaHCO}_{3}{}_{-}{}_{0.01mole}+{H}_{2}O$ After the acid base reaction, we have $0.01mole{Na}_{2}{CO}_{3}$ and $0.02$ moles of ${NaHCO}_{3}$. Here, This will form an acidic buffer of ${NaHCO}_{3}$ and ${Na}_{2}{CO}_{3}$. $∴pH={pK}_{{a}_{2}}+\log \dfrac{\left[Salt\right]}{\left[Acid\right]}$ $=10.32+\log \dfrac{\left(\dfrac{0.01}{0.1}\right)}{\left(\dfrac{0.02}{0.1}\right)}$ $=10.32+\log \dfrac{1}{2}$ $=10.32-\log 2$ $=10.32-0.3$ $=10.02$ $∴pH=10.02$
5. Q5JEE Advanced Adv 2020 (Paper 2)
   An acidified solution of $0.05M{Zn}^{2+}$ is saturated with $0.1M{H}_{2}S$. What is the minimum molar concentration $(M)$ of ${H}^{+}$ required to prevent the precipitation of $ZnS$? Use ${K}_{sp}(ZnS)=1.25\times {10}^{-22}$ and overall dissociation constant of ${H}_{2}S,{K}_{NET}={K}_{1}{K}_{2}=1\times {10}^{-21}$
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   Answer: 0.2

   For $ZnS$ ${K}_{sp}(ZnS)=\left[{Zn}^{2+}\right]\times \left[{S}^{2-}\right]=1.25\times {10}^{-22}$ $=0.05\times \left[{S}^{2-}\right]=1.25\times {10}^{-22}$ $\left[{S}^{2-}\right]=25\times {10}^{-22}$ For ${H}_{2}S$ ${\text{H}}_{2}\text{S}_{0.1 M}\rightleftharpoons 2{\text{H}}^{+}+{\text{S}}^{2-}_{25\times {10}^{-22}}$ ${K}_{a}=\dfrac{\left[{H}^{'}\right]\left[{S}^{2-}\right]}{\left[{H}_{2}S\right]}=1\times {10}^{-21}$ $=\dfrac{{\left[{H}^{+}\right]}^{2}\times 25\times {10}^{-22}}{0.1}=1\times {10}^{-21}$ ${\left[{H}^{+}\right]}^{2}=\dfrac{1}{25}$ $\left[{H}^{+}\right]=\dfrac{1}{5}$ $\left[{H}^{+}\right]=0.2M$