JEE Main Chemistry: Ionic Equilibrum High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)

Correct Answer: (a)

Explanation: For an acidic buffer containing weak acid $\text{CH}_3\text{COOH}$ and its salt $\text{CH}_3\text{COONa}$, we use Henderson equation: $ \text{pH} = \text{p}K_a + \log \frac{[\text{salt}]}{[\text{acid}]} $ Here, $ [\text{CH}_3\text{COOH}] = 0.1\,M $ $ [\text{CH}_3\text{COONa}] = 0.1\,M $ So, $ \frac{[\text{salt}]}{[\text{acid}]} = \frac{0.1}{0.1} = 1 $ Since $\log 1 = 0$, $ \text{pH} = \text{p}K_a $ Now, $ \text{p}K_a = -\log K_a $ $ \text{p}K_a = -\log(1.8 \times 10^{-5}) $ $ \text{p}K_a = 5 - \log 1.8 $ Using $\log 1.8 \approx 0.255$, $ \text{p}K_a = 5 - 0.255 $ $ \text{pH} = 4.745 $ So, the correct option is: $ \boxed{\text{Option A } \; 4.745} $

Correct Answer: (d)

Explanation: NH$_4$NO$_3$ is a salt of strong acid, HNO$_3$ and weak base, NH$_4$OH.

Correct Answer: (d)

Explanation: $\begin{aligned} & \begin{aligned} & \mathrm{S}=7.2 \times 10^{-7} \mathrm{~mol} \mathrm{~dm}^{-3}, \mathrm{~K}_{\mathrm{sp}}=? \\ & \text { For } \mathrm{AgCl}, \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\ &=\left(7.2 \times 10^{-7}\right)^2=5.18 \times 10^{-13}\end{aligned}\end{aligned}$

Correct Answer: (a)

Explanation: For acidic medium, we know that $\mathrm{pH}$ values range is between $0-6,9$, for neutral it is 7 and for bases it is $7.1-14$. So, $\mathrm{HCl}$ is acidic and $\mathrm{NaOH}$ is basic. Thus, we can straight forward infer that $\mathrm{NaOH}$ has a higher $\mathrm{pH}$ value. Now, 1 mole of sodium hydroxide produces 1 mole of $\mathrm{Na}^{+}$ and 1 mole of $\mathrm{OH}^{-}$ ions. This means that the concentration of the hydroxide ions will be equal to $\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=1 \mathrm{M}$ For case A, $\begin{aligned} \mathrm{pOH} & =-\log \left[\mathrm{OH}^{-}\right]=-\log [1]=0 \\ \therefore \quad \mathrm{pH} & =14-\mathrm{pOH}=14-0=14 \end{aligned}$ For case B, $\begin{gathered} {\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=0.1 \mathrm{~M}} \\ \mathrm{pOH}=-\log [0.1]=1 \\ \therefore \quad \mathrm{pH}=14-1=13 \end{gathered}$ So, $\mathrm{pH}$ of $1 \mathrm{~M} \mathrm{~NaOH}$ is highest among the given solution.

Correct Answer: (d)

Explanation: $\begin{aligned} &\mathrm{K}_{\mathrm{a}}=3.2 \times 10^{-4} ; \mathrm{c}=0.04 \mathrm{M}\\ &\text { For a monobasic acid, }\\ &\alpha=\sqrt{\frac{K_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{3.2 \times 10^{-4}}{0.04}}=\sqrt{8 \times 10^{-3}}=0.089 \end{aligned}$