Question

Calculate the pH of $1.0\times {10}^{-3}M$ sodium phenolate $NaO{C}_6{H}_5$. $K_a$ for $C_6{H}_{5}OH$ is $1\times {10}^{-10}$.

Answer 1

$C_6{H}_5{O}^{-}+H_{2}O\rightleftharpoons C_6{H}_{5}OH+{OH}^{-}$

$C$

$C(1-h)$ $Ch$ $Ch$

$K_h=\frac{\left(Ch\right)\left(Ch\right)}{C(1-h)}$

Let '$h<<1$'

So $K_h=C{h}^2$

We know that for a salt of weak acid and strong base $K_h=\frac{K_w}{K_a}$

So $K_h={10}^{-4}$

$C{h}^2={10}^{-4}$

$h=0.316$

$\left[O{H}^{-}\right]=Ch=3.16\times {10}^{-4}M$

We know that $\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

So $\left[H^{+}\right]=3.16\times {10}^{-11}M$

$pH=-\log \left(\right[H^{+}\left]\right)=10.43$