Question

Calculate the pH of $1.0\times {10}^{-3}$ M sodium phenolate, $NaO{C}_6{H}_5\cdot K_a$ for $HO{C}_6{H}_5$ is $0.6\times {10}^{-10}$.

Answer 1

$K_b\times K_a=K_w$ $\therefore$ $K_b=\frac{K_w}{K_a}=\frac{{10}^{-14}}{0.6\times {10}^{-10}}=1.67\times {10}^{-4}$

$K_b=\frac{\left[C_6{H}_{5}OH\right]\left[{OH}^{-}\right]}{\left[C_6{H}_5{O}^{(-)}\right]}=1.67\times {10}^{-4}$

$\Rightarrow \frac{x^2}{0.1-x}=1.67\times {10}^{-4}$

$\Rightarrow x^2-0.167\times {10}^{-4}+1.67\times {10}^{-4}x=0$

$\therefore$ $x=\frac{-1.67\times {10}^{-4}+\sqrt{{(1.67\times {10}^{-4})}^2+\mathrm{4.1.0.167}\times {10}^{-4}}}{2}$

$=4\times {10}^{-3}$

$\therefore$ $\left[C_6{H}_{5}OH\right]=\left[{OH}^{(-)}\right]=4\times {10}^{-3}M$