Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh

1.60.

Explanation:

The no. of millimoles of HCl = MV = (0.15 M) (20.0 mL) = 3.0 mmol. The no. of millimoles of KOH = MV = (0.10 M) (20.0 mL) = 2.0 mmol.

Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.

∴ M of remaining HCl [H⁺] remaining = (NV) HCl - (NV) KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

∵ pH = - log[H⁺]

∴ pH = - log[H⁺] = - log (0.025) = 1.602 ≅ 1.60.