What is the pH of solution assuming the first dissociation to be 100% and second dissociation to be — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The first dissociation yields a hydroxide concentration of $4 \times 10^{-3}\text{ M}$. The second dissociation is 50%, producing an additional $0.5 \times 4 \times 10^{-3} = 2 \times 10^{-3}\text{ M OH}^-$. The total $[\text{OH}^-] = (4 + 2) \times 10^{-3} = 6 \times 10^{-3}\text{ M}$. $\text{pOH} = -\log(6 \times 10^{-3}) = 3 - \log 6 = 3 - (0.3 + 0.48) = 3 - 0.78 = 2.22$. The $\text{pH} = 14 - 2.22 = 11.78$.