Ionic Equilibrium JEE Main previous year questions with solutions

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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 (03 Apr, Shift 2)PH of solutions
   10 mL of 2 M NaOH solution is added to 20 mL of 1 M HCl solution kept in a beaker. Now, 10 mL of this mixture is poured into a volumetric flask of 100 mL containing 2 moles of HCl and made the volume upto the mark with distilled water. The solution in this flask is :
   1. A.0.2 M NaCl solution
   2. B.20 M HCl solution
   3. C.10 M HCl solution
   4. D.Neutral solution
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   Answer: (B)

   When $10 \mathrm{ml}, 2 \mathrm{M} \mathrm{NaOH}$ solution is added to 20 ml of 1 M HCl solution : $\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}$ Initial : MV $=2 \times 0.1 \quad \mathrm{MV}=1 \times 0.2$ $=0.2 \text { mole } \quad=0.2 \mathrm{~mole}$ Final $0 \quad 0$ $\therefore$ Resulting solution becomes neutral. Now when 10 mol of above solution is poured into a flask containing 2 mole HCl and made solution 100 ml will distilled water. Molarity of $\mathrm{HCl}=\frac{2}{100} \times 1000=20$
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 (31 Jan, Shift 2)Ionic Product of Water
   Incorrect statement for the use of indicator in acid-base titration is:
   1. A.Methyl orange may be used for a weak acid vs weak bas titration.
   2. B.Phenolphthalein may be used for a strong acid vs strong base titration.
   3. C.Methyl orange is a suitable indicator for a strong acid vs weak base titration.
   4. D.Phenolphthalein is a suitable indicator for a weak acid vs strong base titration.
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   Answer: (A)

   Methyl orange and phenolphthalein indicators work with the change in pH and are termed as acid–base indicators. Methy lorange is sutable indicator for strong acid vs strong base and strong acid vs weak base titration. Phenolphthalein is sutable indicator for strong acid vs strong base and weak acid vs strong base titration. There is no suitable indicator that can be used in the titration of weak acid and weak base. So, the correct option is A.
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)

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