ANS: pH of -2

The reaction for this neutralisation is:

H₂SO₄ + Ba(OH)₂ -> BaSO₄ + 2H₂O

n(H₂SO₄) = c x V = 2.000 x 10^-3 x 0.1000 = 2.000 x 10^-4 mol

n(Ba(OH)₂) = c x V = 4.825 x 10^-3 x 0.0400 = 1.93 x 10^-4 mol

n(H₂SO₄) left-over after neutralisation is equal to:

n(H₂SO₄) - n(Ba(OH)₂)

= 2.000 x 10^-4 - 1.93 x 10^-4

= 7.00 x 10^-6 mol

n(H⁺) = 2 x n(H₂SO₄) = 2 x 7.00 x 10^-6 mol = 1.40 x 10^-5 mol

concn(H⁺) = n / V = 1.40 x 10^-5 / (0.1000 + 0.0400) = 10 M

pH = -log[H⁺] = -2