33. Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

pH of solutiion A=6. Hence, [H+]=10-6 mol L-1
pH of solution B=4. Hence, [H+]=10-4 mol L-1
On mixing 1 L of each solution, molar concentration of total H+ is halved.
Total, [H+]=10-6+10-42 mol L-1
[H+]=1.01×10-42=5.05×10-5 mol L-1[H+]=5.0×10-5 mol L-1pH=-log [H+]pH=-log (5.0×10-5)pH=-[log 5+(-5 log 10)]pH=-log 5+5pH=5-log 5=5-0.6990pH=4.30104.3
Thus, the pH of resulting solution is 4.3.