## General Chemistry: Principles and Modern Applications (10th Edition)

$$[CN^-]= 3.6 \times 10^{-3} \space M$$
$Na^+$ does not hydrolyzes. $CN^-$ act as a weak base. $$K_b(CN^-) = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-10}} = 1.6 \times 10^{-5}$$ $$[H_3O^+] = 10^{-pH} = 10^{-10.38} = 4.2 \times 10^{-11} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 4.2 \times 10^{-11} } = 2.4 \times 10^{-4} \space M$$ 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ CN^- ]& [ OH^- ]& [ HCN ]\\ Initial& y & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& y -x& 0 +x& 0 +x\\ \end{vmatrix}$$ y is the initial concentration of the ion. $x = [OH^-] = 2.4 \times 10^{-4}$ $$K_b = \frac{[OH^-][HCN]}{[CN^-]} = \frac{(x)(x)}{[CN^-]}$$ $$1.6 \times 10^{-5} = \frac{(2.4 \times 10^{-4})^2}{[CN^-]}$$ $$[CN^-]= 3.6 \times 10^{-3} \space M$$