Question

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

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

Answer 1

The correct option is D $3.7\times {10}^{-4}\text{M}$

The given $\text{pH}\text{are}3,4,5$ respectively. So the respective concentration of $\left[H^{+}\right]$ ion would be ${10}^{-3}\text{M},{10}^{-4}\text{M},{10}^{-5}\text{M}$ for the given acids.
Let the volume of each solution be $1\text{L}$
So,
$M_{\text{mix}}{V}_{\text{mix}}=M_1{V}_1+M_2{V}_2+M_3{V}_3$
$M_{\text{mix}}\times 3={10}^{-3}\times 1+{10}^{-4}\times 1+{10}^{-5}\times 1$
$M_{\text{mix}}=\frac{{10}^{-5}[100+10+1]}{3}=\frac{111\times {10}^{-5}}{3}=37\times {10}^{-5}\text{M}$
$=3.7\times {10}^{-4}\text{M}$