Question

Find the pH of the solution by mixing equal volumes of three acid solutions of $pH4,5,and6$ ?
$log\left(3.7\right)=0.57$

• A. 4.43
• B. 6
• C. 5.43
• D. 5

Answer 1

The correct option is A 4.43

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

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

For III rd solution $pH=-log\left[H^{+}\right]$
For Ist solution $pH=6\Rightarrow 6=-log\left[H^{+}\right]\Rightarrow \left[H^{+}\right]=antilog\left(6\right)={10}^{-6}mol{L}^{-1}$

But, as the volume of each solution is 1 litre so,
using the formula,
$Moles=Molarity\times Volume$
For Ist solution Moles of $H^{+}={10}^{-4}mol{L}^{-1}\times 1L={10}^{-4}mol$
For II nd solution Moles of $H^{+}={10}^{-5}mol{L}^{-1}\times 1L={10}^{-5}mol$
For III rd solution Moles of $H^{+}={10}^{-6}mol{L}^{-1}\times 1L={10}^{-6}mol$

After the mixing is done,
Total volume = $3L$
Total moles $(={10}^{-4}+{10}^{-5}+{10}^{-6})$
$=111\times {10}^{-6}$ mol

So, for :
For final solution $\left[H^{+}\right]=\frac{111\times {10}^{-6}}{3}mol{L}^{-1}$
$=37\times {10}^{-6}mol{L}^{-1}$
$=3.7\times {10}^{-5}mol{L}^{-1}$
hence, the $H^{+}$ ion concentration of the mixture is $3.7\times {10}^{-5}mol{L}^{-1}$
$pH=-log(3.7\times {10}^{-5})$
$pH=5-log\left(3.7\right)$
$pH=(5-0.57)$
$pH=4.43$