Ionic Equilibrium JEE Main previous year questions with solutions

5 solved JEE Main questions on Ionic Equilibrium, free to read — no sign-in needed. The full chapter has 155 questions; sign in to attempt the remaining 150 in the exam simulator.

1. Q1JEE Main 2026 (04 Apr, Shift 1)Buffer solutions
   The pH of a solution obtained by mixing $5$ mL of $0.1$ M $NH_4OH$ solution with $250$ mL of $0.1$ M $NH_4Cl$ solution is _____ $\times 10^{-2}$. (Nearest integer) Given: $pK_b(NH_4OH) = 4.74$ $\log 2 = 0.30$ $\log 3 = 0.48$ $\log 5 = 0.70$
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   Answer: 756

   The given mixture forms a basic buffer solution. Millimoles of base $NH_4OH = 5 \times 0.1 = 0.5$ mmol Millimoles of salt $NH_4Cl = 250 \times 0.1 = 25$ mmol Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log \left( \dfrac{[\text{Salt}]}{[\text{Base}]} \right)$ $pOH = 4.74 + \log \left( \dfrac{25}{0.5} \right)$ $pOH = 4.74 + \log (50)$ We know that $\log (50) = \log (5 \times 10) = \log 5 + \log 10 = 0.70 + 1 = 1.70$ $pOH = 4.74 + 1.70 = 6.44$ The pH of the solution is given by: $pH = 14 - pOH$ $pH = 14 - 6.44 = 7.56$ $pH = 756 \times 10^{-2}$ Answer: $756$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Ionic Product of Water
   The equilibrium constant for decomposition of $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ $\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\left(\Delta \mathrm{G}^{\circ}=92.34 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ is $8.0 \times 10^{-3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation $(\alpha)$ of water is ________ $\times 10^{-2}$ (nearest integer value). [Assume $\alpha$ is negligible with respect to 1 ]
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   Answer: 5

   $\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \\ & \mathrm{t}=0 \quad 1 \mathrm{~mole} \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} \quad 1-\alpha \quad \alpha \quad \frac{\alpha}{2} \\ & \mathrm{n}_{\mathrm{T}}=1+\frac{\alpha}{2} \simeq 1(\alpha \ll 1) \\ & \mathrm{k}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{H}_2} \cdot \mathrm{P}_{\mathrm{O}_2}^{1 / 2}}{\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}}=\frac{(\alpha \cdot \mathrm{P})\left(\frac{\alpha}{2} \mathrm{P}\right)^{\frac{1}{2}}}{(1-\alpha) \mathrm{P}} \\ & 8 \times 10^{-3}=\frac{\alpha^{3 / 2}}{\sqrt{2}} \\ & \alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ & \alpha^3=128 \times 10^{-6} \\ & \alpha=\sqrt[3]{128} \times 10^{-2} \\ & =5.03 \times 10^{-2}\end{aligned}$
3. Q3JEE Main 2024 (09 Apr, Shift 2)Solubility
   For a sparingly soluble salt $\mathrm{AB}_2$, the equilibrium concentrations of $\mathrm{A}^{2+}$ ions and $B^{-}$ions are $1.2 \times 10^{-4} \mathrm{M}$ and $0.24 \times 10^{-3} \mathrm{M}$, respectively. The solubility product of $\mathrm{AB}_2$ is :
   1. A.$6.91 \times 10^{-12}$
   2. B.$0.276 \times 10^{-12}$
   3. C.$27.65 \times 10^{-12}$
   4. D.$0.069 \times 10^{-12}$
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   Answer: (A)

   $\begin{aligned} & \mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{+2}+2 \mathrm{~B}_{(\mathrm{aq})}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+2}\right]\left[\mathrm{B}^{-}\right]^2 \\ &=1.2 \times 10^{-4} \times\left(2.4 \times 10^{-4}\right)^2 \\ &=6.91 \times 10^{-12} \mathrm{M}^3\end{aligned}$
4. Q4JEE Main 2023 (12 Apr, Shift 1)PH of solutions
   An analyst wants to convert $1LHCl$ of $pH=1$ to a solution of $HCl$ of $pH=2$. The volume of water needed to do this dilution is _____ $mL$. (Nearest integer)
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   Answer: 9000

   The concentration of ${H}^{+}$ or say $HCl$ in both the given solution is ${10}^{–1}$ and ${10}^{–2}$ mole ${L}^{–1}$ respectively. For the given amount of $HCl$ the concentration depends over dilution hence ${M}_{1}{V}_{1}={M}_{2}{V}_{2} \dfrac{{M}_{1}}{{M}_{2}}=\dfrac{{V}_{2}}{{V}_{1}}$ $\dfrac{{10}^{-1}}{{10}^{-2}}=\dfrac{{V}_{2}}{1L}$ ${V}_{2}=10L=10000mL$ $∆V=10000-1000mL=9000mL$
5. Q5JEE Main 2022 (25 Jul, Shift 1)Salt Hydrolysis
   $20mL$ of $0.1M{NH}_{4}OH$ is mixed with $40mL$ of $0.05MHCl$. The $pH$ of the mixture is nearest to: (Given: ${K}_{b}\left({NH}_{4}OH\right)=1\times {10}^{-5},\log 2=0.30$, $\log 3=0.48,\log 5=0.69,\log 7=0.84$, $\log 11=1.04$)
   1. A.$3.2$
   2. B.$4.2$
   3. C.$5.2$
   4. D.$6.2$
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   Answer: (C)

   ${NH}_{4}OH+HCl\rightarrow {NH}_{4}Cl+{H}_{2}O \begin{matrix}mmole & 2 & 2\end{matrix} afterreaction--2milimoles$ $\left[{NH}_{4}^{+}\right]=\dfrac{2mmole}{60ml}=\dfrac{1}{30}M$ $pH=\dfrac{{pK}_{w}-{pK}_{b}-logC}{2}=\dfrac{14-5+1.48}{2}=5.24$