Question

The $pH$ of a solution obtained by mixing of $100\, mL$ of a $HCl$ solution of $pH = 2$ with $400\, mL$ of another $HCl$ solution of $pH = 3$ will be

• A. 2
• B. 3
• C. 2.5
• D. 2.8

Answer 1

Answer: (C)

$100\, mL$ of a $HCl$ solution of $p H=2$
$\Rightarrow \left[H^{+}\right]=10^{-2} 400\, mL$ of a $HCl$
solution of $pH=3$ $\Rightarrow \left[H^{+}\right]=10^{-3}$
Resulting $\left[H^{+}\right]=\frac{100 \times 10^{-2}+400 \times 10^{-3}}{500}$ $=\frac{1+0.4}{500}$
$=\frac{1.4}{500}=0.0028$
$\because p H=-\log \left[H^{+}\right] $
$\therefore p H=-\log (0.0028)$
$=2.5$