Question

Calculate the pH of the solution formed by mixing equal volume of two solutions of $pH=3$ and $pH=4$ repectively ?
( log 5.5 = 0.740 )

• A. 7
• B. 4.26
• C. 5.26
• D. 3.26

Answer 1

The correct option is D 3.26

Let first solution is A
pH of solution A = 3
we know that $pH=-log\left[H^{+}\right]$
$3=-log\left[H^{+}\right]$
$\left[H^{+}\right]=antilog\left(3\right)={10}^{-3}mol{L}^{-1}$

Let first solution is B
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=${10}^{-3}\times 1={10}^{-3}mol$
Similarly, Amount of $H^{+}$ ion in 1 L of solution B $={10}^{-4}\times 1={10}^{-4}mol$

Total amount of $H^{+}$ in the solution formed by mixing solution A and B
$=({10}^{-3}+{10}^{-4})=1.1\times {10}^{-3}mol$
Total concentration of $H^{+}=\frac{totalamountof{H}^{+}}{totalvolume}=\frac{1.1\times {10}^{-3}}{2}=5.5\times {10}^{-4}$
$pH=-log\left[H^{+}\right]=-log(5.5\times {10}^{-4})$
$pH=-log\left(5.5\right)+4$
$pH=-0.740+4$
$pH=3.26$