Question

50.0 mL of 0.050 M hydrochloric acid is titrated with 0.025 M aqueous sodium hydroxide. Calculate the pH of the solution after 51.3 mL of sodium hydroxide solution has been added.

Answer 1

n(HCl) = C_m * V
n(HCl) = 0.05*0.05 = 0.0025mol
n(NaOH) = 0.0513 * 0.0025 = 0.00012825 mol
HCl + NaOH = NaCl + H2O
n(HCl)_residue = 0.0025 - 0.00012875 = 0.00237175mol
V_solution = 50 + 51.3 = 101.3 ml = 0.1013 L
C_m(HCl) = 0.00237175 / 0.1013 = 0.002375 M
pH = -lg C(HCl)
pH = 2.62