Q52 / 180 · KCET 2012 · Chemistry
The $\mathrm{pH}$ of the solution obtained by mixing $100 \mathrm{ml}$ of a solution of $\mathrm{pH}=3$ with $400 \mathrm{~mL}$ of a solution of $\mathrm{pH}=4$ is
- A$7-\log 2.8$
- Correct answerB$4-\log 2.8$
- C$5-\log 2.8$
- D$3-\log 2.8$
Solution
For solution $\mathrm{I}, \mathrm{pH}=3$
$$
\Rightarrow \quad\left[\mathrm{H}^{+}\right]=10^{-3}
$$
$\Rightarrow$ Concentration of solution $\mathrm{M}_{1}=10^{-3} \mathrm{M}$
$$
\mathrm{V}_{1}=100 \mathrm{~mL}
$$
For solution II, pH = 4
$\Rightarrow \quad\left[\mathrm{H}^{+}\right]=10^{-4}$ $\Rightarrow$ Concentration of solution, $\mathrm{M}_{2}=10^{-4} \mathrm{M}$ $\mathrm{V}_{2}=400 \mathrm{~mL}$
Concentration of resulting solution
$$
\begin{aligned}
\mathrm{M} &=\frac{\mathrm{M}_{1} \mathrm{~V}_{1}+\mathrm{M}_{2} \mathrm{~V}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}} \\
&=\frac{10^{-3} \times 100+10^{-4} \times 400}{100+400} \\
&=\frac{0.14}{500} \\
\mathrm{M} &=0.00028=2.8 \times 10^{-4} \\
{\left[\quad\left[\mathrm{H}^{+}\right]\right.} &=2.8 \times 10^{-4} \\
\therefore \mathrm{pH} \text { of resulting solution } \\
&=-\log \left[\mathrm{H}^{+}\right] \\
&=-\log \left(2.8 \times 10^{-4}\right) \\
&=-\log 2.8-\log 10^{-4} \\
&=4-\log 2.8
\end{aligned}
$$
Hence, the correct answer is option: 'B'.