Question

Calculate the pH of the solution formed by mixing equal volume of two solutions A and B at $25{}^{o}C$ having $pH=10$ and $pH=12$ repectively ?
(Given log (5.05) = 0.703)

• 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
then pOH = 14 - 10 =4
we know that $pOH=-log\left[O{H}^{-}\right]$
$4=-log\left[O{H}^{-}\right]$
$\left[O{H}^{-}\right]=antilog\left(10\right)={10}^{-4}mol{L}^{-1}$

pH of solution B = 12
pOH of the solution = 14 -12 = 2
we know that $pOH=-log\left[O{H}^{-}\right]$
$2=-log\left[O{H}^{-}\right]$
$\left[O{H}^{-}\right]=antilog\left(2\right)={10}^{-2}mol{L}^{-1}$

Assuming volume of $1L$ solution of each.
$\therefore$ total volume $=1+1=2L$
Amount of $O{H}^{-}$ ion in $1L$ of solution A = concentration$\times$ volume $={10}^{-4}\times 1={10}^{-4}mol$
Similarly, Amount of $O{H}^{-}$ ion in 1 L of solution B $={10}^{-2}\times 1={10}^{-2}mol$

Total amount of $O{H}^{-}$ in the solution formed by mixing solution A and B
$=({10}^{-4}+{10}^{-2})=1.01\times {10}^{-2}mol$
Total concentration of $O{H}^{-}=\frac{totalamountofO{H}^{-}}{totalvolume}=\frac{1.01\times {10}^{-2}}{2}=5.05\times {10}^{-3}mol{L}^{-1}$
$pOH=-log\left[O{H}^{-}\right]=-log(5.05\times {10}^{-3})$
$pOH=-log\left(5.05\right)+3$
$pOH=-0.703+3$
$pOH=2.3$
so $pH=14-2.3=11.7$