Hydrocyanic acid, HCN, has Ka = 4.9 × 10⁻¹⁰…

For a weak acid, $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} = \dfrac{C\alpha^2}{1 - \alpha}$. When $\alpha$ is small, $\alpha \approx \sqrt{K_a/C}$ and $[\text{H}_3\text{O}^+] \approx \sqrt{K_a C}$. Otherwise solve $x^2 + K_a x - K_a C = 0$. For a weak base, $[\text{OH}^-] \approx \sqrt{K_b C}$. For a conjugate pair, $K_a K_b = K_w$ and $\text{p}K_a + \text{p}K_b = 14$.