Question

If $p{K}_b$ for $C{N}^{-}$ at ${25}^{o}C$ is 4.7. The $pH$ of 0.5 M aqueous $NaCN$ solution is:

• A. 12
• B. 10
• C. 11.5
• D. 11

Answer 1

The correct option is C 11.5

$NaCN⟶N{a}^{+}+C{N}^{-}$

Given, $p{K}_b=4.7$

$-\log K_b=4.7$

$\Rightarrow K_b={10}^{-4.7}=1.99\times {10}^{-5}$

Now, $K_b$=$\frac{\left[N{a}^{+}\right]\left[C{N}^{-}\right]}{\left[NaCN\right]}$

as, $\left[NaCN\right]=0.5M$

$\therefore \left[Na\right]\left[CN\right]=K_b\times \left[NaCN\right]$

=$1.99\times {10}^{-5}\times 0.5$

=$0.995\times {10}^{-5}$

as, $\left[CN\right]=\left[Na\right]$

$\therefore [CN{]}^2=9.95\times {10}^{-6}$

Hence, the concentration of $\left[C{N}^{-}\right]=\sqrt{9.95\times {10}^{-6}}$

=$3.15\times {10}^{-3}$

$pOH=-\log [3.15\times {10}^{-3}]$

$pOH=2.5$

and, $pH=14-2.5=11.5$