Question

Calculate the pH of the solution formed by mixing equal volume of two solutions A and B having $pH=10$ and $pH=12$ repectively ?
(Given log (5.05) = 0.713)

• A. 11.7
• B. 10.29
• C. 12
• D. None of the above

Answer 1

The correct option is A 11.7

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

pH of solution B = 12
we know that $pH=-log\left[H^{+}\right]$
$12=-log\left[H^{+}\right]$
$\left[H^{+}\right]=antilog\left(12\right)={10}^{-12}mol{L}^{-1}$

Assuming volume of $1L$ solution of each.
$\therefore$ total volume $=1+1=2L$
Amount of $H^{+}$ ion in $1L$ of solution A = concentration$\times$ volume $={10}^{-10}\times 1={10}^{-10}mol$
Similarly, Amount of $H^{+}$ ion in 1 L of solution B $={10}^{-12}\times 1={10}^{-12}mol$

Total amount of $H^{+}$ in the solution formed by mixing solution A and B
$=({10}^{-10}+{10}^{-12})=1.01\times {10}^{-10}mol$
Total concentration of $H^{+}=\frac{totalamountof{H}^{+}}{totalvolume}=\frac{1.01\times {10}^{-10}}{2}=5.05\times {10}^{-11}mol{L}^{-1}$
$pH=-log\left[H^{+}\right]=-log(5.05\times {10}^{-11})$
$pH=-log\left(5.05\right)+11$
$pH=-0.703+11$
$pH=10.297$