Question

What is the pH of ${10^{ - 8}}$ M HCl?

Answer 1

Hint: To answer this question, we first need to understand what pH is. The pH of aqueous or other liquid solutions is a quantitative measure of their acidity or basicity. The term, which is frequently used in chemistry, biology, and agronomy, converts hydrogen ion concentrations—which typically vary from around $1 to {10^{ - 14}}$ gram-equivalents per liter—into numbers between 0 and 14.

Complete answer:
As we know that HCl is a strong acid, so its pH will be less than 7.
The concentration of HCl = ${10^{ - 8}}$M
Total $[{H^ + }]$$ = [{H^ + }]$ obtained from HCL $ + [{H^ + }]$ obtained from ${H_2}O[{H^ + }]$
HCl being a strong acid, it completely ionizes.
\[[{H^ + }]HCl = 1.0 \times {10^{ - 8}}\].
The concentration of ${H^ + }$ from ionization is equal to that of $O{H^ - }$ from water, so let us consider it to be $X$.
$[{H^ + }]{H_2}O = [O{H^ - }]{H_2}O = X$.
Total $[{H^ + }] = X + 1.0 \times {10^{ - 8}}$------(1)
But we know
\[[{H^ + }][O{H^ - }] = 1.0 \times {10^{ - 14}}\]
\[(1.0 \times {10^{ - 8}}{\text{ X}} + {\text{ X}}){\text{ }}\left( X \right){\text{ }} = {\text{ }}1.0\; \times {10^{ - 14}}\]
\[{X^2} + {10^{ - 8}}X - {10^{ - 14}} = 0\]
Solving $X$, we get
\[{\text{X }} = {\text{ }}9.5 \times {10^{ - 8}}\;\]
Substituting $X$ value in equation (1)
Total \[[{H^ + }] = 9.5 \times {10^{ - 8}} + 1.0 \times {10^{ - 8}}\]
Total \[[{H^ + }] = 1.05 \times {10^{ - 7}}\].
Consider an equation
\[\Rightarrow pH{\text{ }} = {\text{ }}-{\text{ }}log{\text{ }}[{H^ + }]\]
\[\Rightarrow pH{\text{ }} = {\text{ }}-{\text{ }}log{\text{ }}[1.05 \times {10^{ - 7}}]\]
\[pH{\text{ }} ={\text{ }}6.98\]
So, the final pH value is $6.98$.

Note:
Hydrochloric acid, commonly known as muriatic acid, is a hydrogen chloride aqueous solution. It's a colorless liquid with a strong, pungent odor. It's considered a strong acid. In the digestive processes of most animal species, including humans, it is a component of stomach acid.