Question

Calculate the pH of a solution formed by mixing equal volume of two solutions A and B of a strong acid having $pH=6$ and $pH=4$ respectively ?
(Given: log 5.05 = 0.703)

• A. 4.29
• B. 5
• C. 5.29
• D. None of the above

Answer 1

The correct option is A 4.29

pH of solution A = 6
we know that $pH=-log\left[H^{+}\right]$
$6=-log\left[H^{+}\right]$
$\left[H^{+}\right]=antilog\left(6\right)={10}^{-6}mol{L}^{-1}$

pH of solution B = 4
we know that $pH=-log\left[H^{+}\right]$
$4=-log\left[H^{+}\right]$
$\left[H^{+}\right]=antilog\left(4\right)={10}^{-4}mol{L}^{-1}$

Let's take case of $1L$ solution of each.
$\therefore$ total volume $=1+1=2L$
Amount of $H^{+}$ ion in $1L$ of solution A = concentration$\times$ volume $={10}^{-6}\times 1={10}^{-6}mole$
Similarly, Amount of $H^{+}$ ion in 1 L of solution B $={10}^{-4}\times 1={10}^{-4}mole$

Total amount of $H^{+}$ in the solution formed by mixing solution A and B
$=({10}^{-6}+{10}^{-4})=1.01\times {10}^{-4}mole$
Total concentration of $H^{+}=\frac{totalamountof{H}^{+}}{totalvolume}=\frac{1.01\times {10}^{-4}}{2}=5.05\times {10}^{-5}$
$pH=-log\left[H^{+}\right]=-log(5.05\times {10}^{-5})$
$pH=-log\left(5.05\right)+5$
$pH=-0.703+5$
$pH=4.297$