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+]=10-6molL-1
pH of solution B=4. Hence, [H+]=10-4molL-1
On mixing 1 L of each solution, molar concentration of total H+ is halved.
Total, [H+]=10-6+10-42molL-1
[H+]=1.01×10-42=5.05×10-5molL-1
[H+]=5.0×10-5molL-1
pH=-log[H+]pH=-log(5.0×10-5)
pH=-[log5+(-5log10)]pH=-log5+5
pH=5-log5=5-0.6990pH=4.30104.3
Thus, the pH of resulting solution is 4.3.