The pH of solution A is 3. It is mixed with an equal volume of another solution B having pH 2. What is the resultant pH of the solution?

Options

A. 3.2
B. 2.26
C. 2.5
D. 3.5

Answer

B

Explanation

For solution A, `pH = 3 rArr [H^(+)] = 10^(-3) M` and for solution B, `pH = 2 rArr [H^(+)] = 10^(-2)M`.
For the resultant solution, `[H^(+)] = (10^(-3)M + 10^(-2)M)/(2) = 5.5 xx 10^(-3)`
`pH = -log[H^(+)] = -log[5.5 xx 10^(-3)] = 2.26`