Question

The $pH$ of a solution prepared by mixing $2.0 \,mL$ of $HCl$ solution of $pH \,3.0$ and $3.0 \,mL$ of $NaOH$ of $pH\, 10.0$ 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 $=10.0$
$\therefore \left[H^{+}\right] i N a O H$ 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 $=-\log \left[H^{+}\right]$
$=-\log \left(3.4 \times 10^{-4}\right)=3.5$