Calculate the pH of a solution obtained by mixing 2 mL of HCl of pH 2 and 3 mL of solution of KOH of pH = 12
11.30
H+ in HCl solution (pH=2) = 10−2M; [OH−] in KOH solution (pOH = 14 - 12 = 2) = 10−2M
Excess m mol of OH− in 5 mL mixture = (3 × 10−2) −(2 × 10−2)=(1.0 × 10−2)
[OH−] in mixture = (1.0 × 10−2)5=2.0 × 10−3M;
pOH = −log2 ×10−3=3 −log2;
pH=14 − (3−log2)=11.30