Question 9

For a very weak acid where the $[H^+]$ from acid is comparable to that from water ($10^{-7}\ M$), we must consider the auto-ionization of water. $[H^+]_{total} = [A^-] + [OH^-]$ From $K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{[H^+][A^-]}{C}$, we have $[A^-] = \frac{K_a C}{[H^+]}$. From $K_w = [H^+][OH^-]$, we have $[OH^-] = \frac{K_w}{[H^+]}$. Substituting: $[H^+] = \frac{K_a C}{[H^+]} + \frac{K_w}{[H^+]} \Rightarrow [H^+]^2 = K_a C + K_w$ $[H^+]^2 = (4.00 \times 10^{-11} \times 1.00 \times 10^{-3}) + 1.00 \times 10^{-14} = 4 \times 10^{-14} + 1 \times 10^{-14} = 5 \times 10^{-14}$ $[H^+] = \sqrt{5 \times 10^{-14}} = \sqrt{5} \times 10^{-7} \approx 2.236 \times 10^{-7}\ M$. Thus, $X \approx 2.24$.