Question

During a titration experiment, a student titrated 25.00 mL of a 0.100 M sodium hydroxide solution with 5.10 mL of a 0.250 M sulphuric acid solution. What is the pH of the resulting solution?

Answer 1

25mL (=0.025L) of 0.1 M NaOH = 0.025 L x 0.1 M = 0.0025 mole NaOH = 0.0025mole OH^-

5.10mL (=0.0051 L) of 0.250M H₂SO₄ = 0.0051 L x 0.250 M =0.001275 mole = 0.0013 mole H₂SO₄

H₂SO₄ dissociates to give 2mole of H⁺ ion and one mole of SO₄^2- ion

Thus, moles of H⁺ = 2 x 0.0013 = 0.0026 mole

Hence it is apparant that, 0.0025 mole of acid gets neutralized by 0.0025 mole base. what remains un reacted is = 0.0026 -0.0025 = 0.0001 mole H⁺ ion

Thus,

pH of the resulting solution = - log[H⁺ ] = - log(0.0001) = 4