Match the column — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

(P) $[\text{OH}^-] = \sqrt{K_{sp}/[\text{Mg}^{2+}]} = \sqrt{2 \times 10^{-6}/(2 \times 10^{-3})} = \sqrt{10^{-3}} \approx 3.16 \times 10^{-2}\text{ M}$, yielding $\text{pOH} = 1.5$ and $\text{pH} = 12.5$. Matches Left 3. (Q) $[\text{OH}^-] = (K_{sp}/[\text{Al}^{3+}])^{1/3} = (10^{-28}/0.1)^{1/3} = 10^{-9}\text{ M}$, yielding $\text{pOH} = 9$ and $\text{pH} = 5.0$. Matches Left 1. (R) Degree of dissociation $\alpha = \frac{1000}{11}\% = \frac{10}{11}$. $[\text{H}^+] = K_a \frac{1-\alpha}{\alpha} = 10^{-5} \times \frac{1/11}{10/11} = 10^{-6}\text{ M}$, yielding $\text{pH} = 6.0$. Matches Left 4. (S) Since $pK_a = pK_b$, $pK_a = 7$. $[\text{H}^+] = \sqrt{K_a C} = \sqrt{10^{-7} \times 10^{-3}} = 10^{-5}\text{ M}$, yielding $\text{pH} = 5.0$. Matches Left 1. (T) $[\text{A}^-] = 6 \times 10^{-5}\text{ M}$. $K_h = 10^{-14}/(5 \times 10^{-9}) = 2 \times 10^{-6}$. Using quadratic equation for $[\text{OH}^-]^2 + K_h[\text{OH}^-] - K_h C = 0$, we find $[\text{OH}^-] = 10^{-5}\text{ M}$, yielding $\text{pOH} = 5.0$ and $\text{pH} = 9.0$. Matches Left 2. Therefore, correct answer is 1-B,D; 2-E; 3-A; 4-C.