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?

Question

Aicharya · Accepted Answer

Let the volume of each acid be $V$.
$pH$ of first, second, and third acids are $3$, $4$, and $5$ respectively.
Concentration of $H^{+}$ for the first acid $(M_1)$ is $1 \times 10^{-3}$ (since $H^{+}=1 \times 10^{-pH}$).
Concentration of $H^{+}$ for the second acid $(M_2)$ is $1 \times 10^{-4}$.
Concentration of $H^{+}$ for the third acid $(M_3)$ is $1 \times 10^{-5}$.
Total $H^{+}$ concentration of the mixture is given by $\begin{aligned} (M) &= \frac{M_{1} V_{1}+M_{2} V_{2}+M_{3} V_{3}}{V_{1}+V_{2}+V_{3}} \\ &= \frac{1 \times 10^{-3} \times V+1 \times 10^{-4} \times V+1 \times 10^{-5} \times V}{V+V+V} \\ &= \frac{1 \times 10^{-3} \times V(1+0.1+0.01)}{3 V} \\ &= \frac{1.11 \times 10^{-3}}{3}=3.7 \times 10^{-4} M \end{aligned}$

Equal volumes of three acid solutions of $\mathrm{pH} 3,4$ and 5 are mixed in a vessel. What will be the…

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