Question

The $pH$ of a solution prepared by mixing $2.0mL$ of $HCl$ solution of $pH3.0$ and $3.0mL$ of $NaOH$ of $pH10.0$ is

• A. 2.5
• B. 3.5
• C. 5.5
• D. 6.5

Answer 1

Answer: (B)

$\because pH$ of $HCl$ solution $=3.0$
$\therefore \left[H^{+}\right]$in $HCl$ solution $=1\times 10^{-3}$
$\because pH$ of $NaOH$ solution $=10.0$
$\therefore \left[H^{+}\right]iNaOH$ solution $=\frac{1\times 10^{-14}}{1\times 10^{-10}}=10^{-4}$
Milliequivalents of $HCl=N_1{V}_1=2.0\times 1\times 10^{-3}$
$=2.0\times 10^{-3}$
Milliequivalents of $NaOH=3.0\times 1\times 10^{-4}=3.0\times 10^{-4}$
Since, milliequivalents of $HCl$ are in excess, the
milliequivalents of $\left[H^{+}\right]$in mixture
$=\left(2.0\times 10^{-3}\right)-\left(3.0\times 10^{-4}\right)$
$=1.7\times 10^{-3}$
Concentration of $\left[H^{+}\right]$in mixture
$=\frac{1.7\times 10^{-3}}{3+2}=3.4\times 10^{-4}$
$pH$ of mixture $=-\log \left[H^{+}\right]$
$=-\log \left(3.4\times 10^{-4}\right)=3.5$