Calculate pH of solution. for . () — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Because $K_a$ is very small, we must consider the autoionization of water. Total $[\text{H}^+] = \sqrt{K_a C + K_w} = \sqrt{(2 \times 10^{-12})(0.02) + 10^{-14}} = \sqrt{4 \times 10^{-14} + 1 \times 10^{-14}} = \sqrt{5 \times 10^{-14}} = 2.236 \times 10^{-7}\text{ M}$. $\text{pH} = -\log(2.236 \times 10^{-7}) = 7 - \log(\sqrt{5}) = 7 - \frac{1}{2}\log(10/2) = 7 - \frac{1}{2}(1 - 0.3) = 7 - 0.35 = 6.65$.