Question

Calculate the pH of a $0.10M$solution of $NaCN\left(aq\right)$. $K_a$ for $HCN$ is $4.9\times {10}^{-10}$ at ${25}^{o}C$.

• A. $11.15$
• B. $2.85$
• C. $8.75$
• D. $7$

Answer 1

The correct option is A $11.15$

$C{N}^{-}+H_{2}O=HCN+O{H}^{-}$

Initial 0.1

Change $(-x)$ $(+x)$

Equilibrium $(0.1-x)$ $x$

$Kb=\left[HCN\right]\left[C{N}^{-}\right]/\left[C{N}^{-}\right]$

Note $K_a\times K_b=K_w=1.0\times {10}^{-14}$

${10}^{-14}/K_a=1.0\times {10}^{-14}/4.9\times {10}^{-10}$

$=\frac{\left[HCN\right]}{\left[C{N}^{-}\right]/\left[C{N}^{-}\right]}=\frac{x^2}{(0.1-x)}$

$2.04\times {10}^{-5}=\frac{x^2}{(0.1-x)}$

Since $\left[NaCN\right]/K_b>100$, we can simplify the above equation to

$2.04\times {10}^{-5}=x^2/0.1$

$x=1.43\times {10}^{-3}M=\left[HCN\right]=\left[O{H}^{-}\right]$

Note: $pOH=-log\left[O{H}^{-}\right]$

$poH=-log(1.43\times {10}^{-3})=2.84$

Note: $pH+pOH=14$

$pH=14-pOH$

$pH=14-2.84=11.15$