Equal volumes of three acid solutions of pH 3,4 and 5 are mixed in a vessel. What will be the H^+ ion… - NEET UG 2008 Chemistry PYQ

NEET UG 2008 — Chemistry Organic Chemistry

2008mcqhard

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?

Official previous-year question

Held on 30 Apr 2008 · Verified 9 Jul 2026.

Options

A
$1.11 \times 10^{-4} \mathrm{M}$
B
$3.7 \times 10^{-4} \mathrm{M}$
C
$3.7 \times 10^{-3} \mathrm{M}$
D
$1.11 \times 10^{-3} \mathrm{M}$

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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 $\begin{aligned} & (M)=\frac{M_1 V_1+M_2 V_2+M_3 V_3}{V_1+V_2+V_3} \\ & =\frac{1 \times 10^3 \times V+1 \times 10^4 \times V+1 \times 10^5 \times V}{V+V+V} \\ & =\frac{1 \times 10^3 \times V(1+0.1+0.01)}{3 V} \\ & =\frac{1.11 \times 10^3}{3}=3.7 \times 10^{-4} M \end{aligned}$

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