Question

The ionization constant of phenol is $1.0\times {10}^{–10}$. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Answer 1

Ionization of phenol:
$\begin{array}{cccccc}& C_6{H}_{5}OH+H_{2}O& \leftrightarrow & C_6{H}_5{O}^{-}& +& H_3{O}^{+}\\ Initialconc.& 0.05& & 0& & 0\\ Atequilibrium& 0.05-x& & x& & x\end{array}$
$k_a=\frac{\left[C_6{H}_5{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_6{H}_{5}OH\right]}{K}_a=\frac{x\times x}{0.05-x}$
As the value of he ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.
$\therefore x=\sqrt{1\times {10}^{-10}\times 0.05}=\sqrt{5\times {10}^{-12}}=2.2\times {10}^{-6}M=\left[H_3{O}^{+}\right]$
Since $\left[H_3{O}^{+}\right]=\left[C_6{H}_5{O}^{-}\right]$
$\left[C_6{H}_5{O}^{-}\right]=2.2\times {10}^{-6}M$
Now, let $\propto$ be the degree of ionization of phenol in the presence of $0.01M{C}_6{H}_{5}ONa.$
$\begin{array}{cc}{C}_6{H}_{5}ONa\to C_6{H}_5{O}^{-}& +N{a}^{+}\\ Conc.& 0.01\end{array}$
Also,
$\begin{array}{cccccccc}& C_6{H}_{5}OH& +& H_{2}O& \leftrightarrow & C_6{H}_5{O}^{-}& +& H_3{O}^{+}\\ Conc.& 0.05-0.05\alpha & & & & 0.05\alpha & & 0.05\alpha \end{array}$
$\left[C_6{H}_{5}OH\right]=0.05-0.05\alpha ;0.05M\left[C_6{H}_5{O}^{-}\right]=0.01+0.05\alpha ;0.01M\left[H_3{O}^{+}\right]=0.05\alpha K_a=\frac{\left[C_6{H}_5{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_6{H}_{5}OH\right]}{K}_a=\frac{\left(0.01\right)\left(0.05\alpha \right)}{0.05}1.0\times {10}^{-10}=.01\alpha \alpha =1\times {10}^{-8}$