MCQ
What will be the $pH$ of a ${10^{ - 8}}\,M\,HCl$ solution
- A$8$
- B$7$
- ✓$6.958$
- D$14.04$
$HCl\, ⇌ \,[{H^ + }]\,\,[C{l^ - }]$
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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Assertion $(A) :$ Treatment of bromine water with propene yields $1-$bromopropan$-2-$ol.
Reason $(R):$ Attack of water on bromonium ion follows Markovnikov rule and results in $1-$bromopropan$-2-$ol.
In the light of the above statements, choose the most appropriate answer from the options given below:
$C{{H}_{3}}CN+2H\underset{\text{Ether}}{\mathop{\xrightarrow{HCl}}}\,X\underset{{{H}_{2}}O}{\mathop{\xrightarrow{\text{Boiling}}}}\,Y;$ the term $Y$ is: