# How do you calculate the pH of a 0.050 M sodium cyanide solution?

May 20, 2016

$p O H = p {K}_{w} - p H$ $=$ $14 - 3$.

#### Explanation:

""^(-)C-=N(aq) + H_2O rarr H-C-=N + HO^-

K_b=([HC-=N][HO^-])/([""^(-)C-=N]) $=$ $2.1 \times {10}^{-} 5$

Let the concentration of cyanide that associates $=$ $x$, and given this stoichiometry:

${x}^{2} / \left(0.05 - x\right)$ $=$ $2.1 \times {10}^{-} 5$

This is a quadratic in $x$, which is solvable, but we can make the reasonable assumption that $\left(0.05 - x\right) \cong 0.05$

$x$ $\cong$ $\sqrt{2.1 \times {10}^{-} 5 \times 0.05}$ $=$ $0.001$

Thus $\left[H {O}^{-}\right]$ $=$ $0.001 \cdot m o l \cdot {L}^{-} 1$, $p O H$ $=$ $3$, and $p H$ $=$ $11$

Successive approximations would not markedly change ths value. Our assumption that $\left(0.05 - x\right) \cong 0.05$ was entirely justified.