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.

Calculate the pH of a solution formed by mixing equal volumes of two s

1.	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.
Answer» Solution :pH of solution 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^(-4))/2 "mol L"^(-1)` `[H^+]=(0.01+10^(-4))/2 5.05 XX 10^(-5) "mol L"^(-1)` `[H^+] = 5.0xx10^(-5) "mol L"^(-1)` `pH=-log [H^+] rArr pH-log (5.0xx10^(-5))` `pH=-[log 5+ (-5 log 10 )] rArr pH =- log 5+5` `pH=5-log 5 =5-0.6990 rArr pH` 4.3010 = 4.3 Thus, the pH of RESULTING solution is 4.3