Question

Equal volumes of three acid solutions of $pH3,4,and5$ are mixed in a vessel. What will be the $H^{+}$ ion concentration of the mixture ?
$antilo{g}_b(-x)=b^{-x}$

• A. $3.7\times {10}^{-4}mol{L}^{-1}$
• B. $7.3\times {10}^{-4}mol{L}^{-1}$
• C. $3.7\times {10}^{-3}mol{L}^{-1}$
• D. $7.3\times {10}^{-5}mol{L}^{-1}$

Answer 1

The correct option is A $3.7\times {10}^{-4}mol{L}^{-1}$

Three solutions of equal volume are mixed.
Let's consider, initial volume of each solution= $1L$
we know, $pH=-log\left[H^{+}\right]$
For Ist solution $pH=3\Rightarrow 3=-log\left[H^{+}\right]\Rightarrow \left[H^{+}\right]=antilog(-3)={10}^{-3}mol{L}^{-1}$

For II nd solution $pH=-log\left[H^{+}\right]$
For Ist solution $pH=4\Rightarrow 4=-log\left[H^{+}\right]\Rightarrow \left[H^{+}\right]=antilog(-4)={10}^{-4}mol{L}^{-1}$

For III rd solution $pH=-log\left[H^{+}\right]$
For Ist solution $pH=5\Rightarrow 5=-log\left[H^{+}\right]\Rightarrow \left[H^{+}\right]=antilog(-5)={10}^{-5}mol{L}^{-1}$

But, as the volume of each solution is 1 litre so,
using the formula,
$Moles=Molarity\times Volume$
For Ist solution Moles of $H^{+}={10}^{-3}mol{L}^{-1}\times 1L={10}^{-3}mol$
For II nd solution Moles of $H^{+}={10}^{-4}mol{L}^{-1}\times 1L={10}^{-4}mol$
For III rd solution Moles of $H^{+}={10}^{-5}mol{L}^{-1}\times 1L={10}^{-5}mol$

After the mixing is done,
Total volume = $3L$
Total moles $(={10}^{-3}+{10}^{-4}+{10}^{-5})$
$=111\times {10}^{-5}mol$

So, for: For final solution $\left[H^{+}\right]=\frac{111\times {10}^{-5}}{3}$
$=37\times {10}^{-5}$
$=3.7\times {10}^{-4}mol{L}^{-1}$
hence, the $H^{+}$ ion concentration of the mixture is $3.7\times {10}^{-4}mol{L}^{-1}$