If 200 ml of 0.031 M solution of `H _(2) SO_(4)` is added to 84 ml of a 0.150 M KOH solution. What is the pH of the resulting solution? (log 7 = 0.845)

Answer

10.85

Explanation

M mol of `H^(+)` (initial) `= 200 xx 0.031 xx 2 = 12.4`
M mol of `OH^(-)` (initial `= 8.4 xx 0.15 = 12.6`
M mol of `OH^(-)` left after neutralization = 0.2
`[OH^(-)]_("final") = (0.2)/(284) = 7 xx 10^(-4)M , pOH = 3.15` & pH = 10.85