Question

The pH of \[{10^{ - 8}}\] M HCl solution is:
A. 8
B. 1
C. Close to 7
D. 0

Answer 1

Hint: The pH of the solution is defined as the negative logarithm of hydrogen ion concentration. When the pH is below 7, then the solution is considered as acid. When the pH is above 7 then the solution is considered as base. When the pH is 7, then the solution is neutral. When dilute acidic solution is used, then hydrogen ion concentration of hydrochloric acid and hydrogen ion concentration of water are comparable. The concentration of the hydrogen ion of water is not neglected.

Complete step by step answer:
Concentration of hydrochloric acid solution is \[{10^{ - 8}}\]M.
As it is a dilute solution, hydrochloric acid is mixed in water.
The total hydrogen ion concentration is,
${[{H^ + }]_{Total}} = {[{H^ + }]_{acid}} + {[{H^ + }]_{water}}$………(i)
Hydrochloric acid is a strong acid, on dissolving in water it completely ionizes into constituent ions.
$ \Rightarrow {[{H^ + }]_{HCl}} = 1 \times {10^{ - 8}}$
The concentration of ${H^ + }$ is equal to the concentration of $O{H^ - }$ from the water.
${[{H^ + }]_{{H_2}O}} = {[O{H^ - }]_{{H_2}O}}$
Let $O{H^ - }$ be x.
Substitute in the value in equation (i).
$ \Rightarrow {[{H^ + }]_{Total}} = [1 \times {10^{ - 8}}] + x$
But we know,
$[{H^ + }][O{H^ - }] = 1.0 \times {10^{ - 14}}$
Substitute the value of total hydrogen ion concentration in the above equation.
$ \Rightarrow [1 \times {10^{ - 8}} + x][x] = 1.0 \times {10^{ - 14}}$
$ \Rightarrow ({x^2} + {10^{ - 8}})(x - {10^{ - 14}}) = 0$
$ \Rightarrow x = 9.5 \times {10^{ - 8}}$
Substitute the value of x in the equation.
$ \Rightarrow {[{H^ + }]_{Total}} = [1 \times {10^{ - 8}}] + [9.5 \times {10^{ - 8}}]$
$ \Rightarrow {[{H^ + }]_{Total}} = [10.5 \times {10^{ - 8}}]$
$ \Rightarrow {[{H^ + }]_{Total}} = [1.05 \times {10^{ - 7}}]$
The pH of the solution is defined as the negative logarithm of hydrogen ion concentration.
$pH = - \log [{H^ + }]$
Substitute the value of hydrogen ion in the above equation.
$ \Rightarrow pH = - \log [1.05 \times {10^{ - 7}}]$
$ \Rightarrow pH = 6.98$
Thus, the pH of \[{10^{ - 8}}\] M HCl solution is 6.98.
Therefore, the correct option is C.

Note:
When we calculate the pH using the relation $pH = - \log [{H_3}{O^ + }]$, the pH obtained is 8 but the value of acid solution cannot be more than 7.