Question

The pH of $HCl$ is $5$. If $10 \,mL$ of this solution is diluted to $100\, mL$, the pH of the resultant solution is

• A. 5.1
• B. 6.9
• C. 11
• D. 12

Answer 1

Answer: (B)

$pH$ of $HCl =5$
So, $\left[ H ^{+}\right]=10^{-5} mol / L$
$H ^{+}$ions in $10\, mL$ of the solution $=\frac{10^{-5} \times 10}{1000}$
$=10^{-7} mol$
Now as solution is diluted 10 times
$No \cdot$ of $H ^{+}$ions $=\frac{10^{-7}}{10}=10^{-8}$
Total $\left[ H ^{+}\right]=10^{-8}+10^{-7}$
$\left([ H ]^{+}\right.$from water cannot be neglected.)
$=10^{-8}(1+10)=11 \times 10^{-8} M $
$ pH =-\log \left(11 \times 10^{-8}\right)=-(1 \cdot 0413-8)=6 \cdot 95$