Question

The value ${{K}_{a}}$ of ${{C}_{6}}{{H}_{5}}OH$ is $1.0\times {{10}^{-10}}$ what is the $pH$ of a $0.1M\text{ }{{\text{C}}_{6}}{{H}_{5}}{{O}^{-}}$ solution?
A. $10.51$
B. $11.04$
C. $11.50$
D. $12$

Answer 1

Hint: Since we need to find the $pH$ of the anion, first find the $pOH$ of the ion by using the relation $pOH=\sqrt{{{k}_{b}}C}$ and then find the value by the relation $pH=14-pOH$. Using the above relations or the formulas, find the required $pH$ of the anion.

Complete step by step solution:
 Given the value of ${{K}_{a}}$ is equal to $1.0\times {{10}^{-10}}$
We know that ${{k}_{b}}=\dfrac{{{k}_{w}}}{{{k}_{a}}}$ and the value of ${{k}_{w}}$ is equal to $1.0\times {{10}^{-14}}$
By putting the above two values in the above relation, we obtain
${{k}_{b}}=\dfrac{{{k}_{w}}}{{{k}_{a}}}=\dfrac{1.0\times {{10}^{-14}}}{1.0\times {{10}^{-10}}}=1.0\times {{10}^{-4}}$
Given that the concentration of the acid ${{C}_{6}}{{H}_{5}}OH$ is $0.1M\text{ }$
We know that for a base phenoxide ion, the expression for the $pOH$ of the solution is given by the expression
$pOH=-\log \sqrt{{{k}_{b}}C}$
Put the above values in this formula and find out the value of the $pOH$
$pOH=-\log \sqrt{{{k}_{b}}C}=-\log \sqrt{\left( 1.0\times {{10}^{-4}}\times 0.1 \right)}=2.5$
Hence we find the value of the $pOH$ of the phenoxide ion is 2.5
We know the relation between the $pH$ & $pOH$ as
$pH=14-pOH$
Put the above $pOH$ value in the formula and find the required $pH$ of the anion
After substituting the values, we obtain
$pH=14-pOH=14-2.5=11.5$
Therefore the required value of the$pH$ of the phenoxide ion is obtained as
$pH=11.5$

Hence option (C) is the correct answer.

Note: For a cation we can directly find the value of the $pH$ by using a certain relation we are having where as for an anion we cannot find the $pH$ value directly. We are having the formula as $pH=-\log \left[ {{H}^{+}} \right]$, with the cation but not with the anion. But we can find the value of $pOH$ for an anion or a base and using this $pOH$ we need to find the value of $pH$ using the relation $pH=14-pOH$. This is due to the fact that the acid or a cation is having the less $pH$ value whereas for the anion it is having large $pH$ value.