Question

Calculate the pH of 0.1 mol/L H2SO4 ; you are given the second dissociation constant of the acid to be 0.01.

Answer 1

H₂SO₄ (aq) + H₂O(l) ---> H₃O⁺(aq) + HSO₄^-(aq)

  HSO₄^-(aq) + H₂O(l) ---> H₃O⁺(aq) + SO₄^-2(aq)

Ka2 = 0.01 = [SO₄^-2] [H₃O⁺]/[HSO₄^-]

As H₂SO₄ is a strong acid then the molar concentration of H₃O⁺ and HSO₄^- are equivalent to 0.1 M

HSO₄^-(aq) + H₂O(l) ---> H₃O⁺(aq) + SO₄^-2(aq)

Initial    0.1 0.1 0

change -x    +x    +x

   Equilibrium 0.1-x    0.1+x x

Ka2 = 0.01 = x (0.1+x)/(0.1-x)

0.01 (0.1-x) = 0.1x+x²

-0.001+0.01x+0.1x+x² = 0

x² + 0.11x-0.001 = 0

roots of quadratic equation are calculated as follows :

x1 = - b+b²-4ac/2a and x2 = - b-b²-4ac/2a

b²-4ac = 0.11²-4x1x(-0.001)= 0.0121+0.004 = 0.0161

0.0161 = 0.1269

x1 = -0.11+0.1269/2 = 0.00845 M

x2 = -0.11-0.1269/2 = -0.11845 M

[H₃O⁺] = 0.1+x = 0.1+0.00845 = 0.10845 M

pH = - log [H₃O⁺] = - log (0.10845) = 0.965