Question

The pH of $HCl$ is $5$. If $10mL$ of this solution is diluted to $100mL$, 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 $10mL$ 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 $\left[H^{+}\right]=10^{-8}+10^{-7}$
$([H{]}^{+}$from water cannot be neglected.)
$=10^{-8}(1+10)=11\times 10^{-8}M$
$pH=-\log \left(11\times 10^{-8}\right)=-(1\cdot 0413-8)=6\cdot 95$