Calculate the pH of a 0.021 M NaCN solution. [K_a(HCN) = 4.9 * 10^-10] A) 1.68 B) 3.18 C) 5.49 D) 7.00 E) 10.82

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The dissociation constant of HCN, $K_a$, is $4.9 \times 10^{-10}$. NaCN dissociates completely in water to give Na+ and CN-. The CN- ion then reacts with water to produce HCN and OH-. The concentration of CN- is initially 0.021 M. The $K_b$ for CN- can be calculated using the relation $K_w = K_a \times K_b$, where $K_w = 1.0 \times 10^{-14}$ at 25°C. Thus, $K_b = \frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}} = 2.04 \times 10^{-5}$. Using the formula for $K_b$, $[OH^-] = \sqrt{K_b \times [CN^-]}$, we find $[OH^-]$ and then calculate $pOH$. Finally, $pH = 14 - pOH$, which gives a pH close to 10.82.

Calculate the PH of a 0