Q96 / 200 · NEET 2008 (Screening) · Chemistry
Equal volumes of three acid solutions of $\mathrm{pH} 3,4$ and 5 are mixed in a vessel. What will be the $\mathrm{H}^{+}$ ion concentration in the mixture?
- A$1.11 \times 10^{-4} \mathrm{M}$
- Correct answerB$3.7 \times 10^{-4} \mathrm{M}$
- C$3.7 \times 10^{-3} \mathrm{M}$
- D$1.11 \times 10^{-3} \mathrm{M}$
Solution
Let the volume of each acid \(=V\)
\(p H\) of first, second and third acids \(=3,4\) and 5 respectively
\(\left[H^{+}\right]\)of first acid \(\left(M_1\right)=1 \times 10^{-3}\left[\therefore H^{+}=1 \times 10^{-p H}\right]\)
\(\left[H^{+}\right]\)of second acid \(\left(M_2\right)=1 \times 10^{-4}\)
\(\left[H^{+}\right]\)of third acid \(\left(M_3\right)=1 \times 10^{-5}\)
Total \(\left[H^{+}\right]\)concentrated of mixture
\(\begin{aligned}
& (M)=\frac{M_1 V_1+M_2 V_2+M_3 V_3}{V_1+V_2+V_3} \\
& =\frac{1 \times 10^3 \times V+1 \times 10^4 \times V+1 \times 10^5 \times V}{V+V+V} \\
& =\frac{1 \times 10^3 \times V(1+0.1+0.01)}{3 V} \\
& =\frac{1.11 \times 10^3}{3}=3.7 \times 10^{-4} M
\end{aligned}\)
Hence, the correct answer is option: 'B'.