MCQ
What will be the $pH$ of a ${10^{ - 8}}\,M\,HCl$ solution
- A$8$
- B$7$
- ✓$6.958$
- D$14.04$
✓
Answer
Correct option: C.
$6.958$
(c) ${H_2}O⇌[{H^ + }]\,\,[O{H^ - }]$
$HCl\, ⇌ \,[{H^ + }]\,\,[C{l^ - }]$
Total $[{H^ + }] = {[{H^ + }]_{{H_2}O}} + {[{H^ + }]_{HCl}}$ $ = {10^{ - 7}} + {10^{ - 8}}$
$ = {10^{ - 7}}[1 + {10^{ - 1}}]$
$[{H^ + }] = {10^{ - 7}} \times \frac{{11}}{{10}}$
$pH=-\log \,[{{H}^{+}}]=-\log \left( {{10}^{-7}}+\frac{11}{10} \right)$ $pH=6.958$
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