Question

Equal volumes of three acid solutions of pH $3,4$ and $5$ are mixed in a vessel. What will be the $H^{+}$ ion concentration in the mixture?

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

Answer 1

Answer: (B)

Let the volume of each acid $=V$
$pH$ of first, second and third acids $=3,4$ and $5$ respectively
$\left[H^{+}\right]$of first acid $\left(M_1\right)=1\times 10^{-3}\left[\therefore H^{+}=1\times 10^{-pH}\right]$
$\left[H^{+}\right]$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
$(M)=\frac{M_1{V}_1+M_2{V}_2+M_3{V}_3}{V_1+V_2+V_3}$
$=\frac{1\times 10^{-3}\times V+1\times 10^{-4}\times V+1\times 10^{-5}\times V}{V+V+V}$
$=\frac{1\times 10^{-3}\times V(1+0.1+0.01)}{3V}$
$=\frac{1.11\times 10^{-3}}{3}=3.7\times 10^{-4}M$