What is the pH of aqueous solution of if its first dissociation is 100%, second dissociation is 50% — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

First dissociation (100%): provides $[\text{OH}^-] = 6.67 \times 10^{-3} \text{ M}$. Second dissociation (50%): provides an additional $[\text{OH}^-] = 0.5 \times 6.67 \times 10^{-3} \text{ M} = 3.335 \times 10^{-3} \text{ M}$. Total $[\text{OH}^-] = (6.67 + 3.335) \times 10^{-3} \text{ M} = 10.005 \times 10^{-3} \text{ M} \approx 10^{-2} \text{ M}$. $\text{pOH} = -\log(10^{-2}) = 2$. The $\text{pH} = 14 - \text{pOH} = 14 - 2 = 12$. Therefore, correct answer is B.