Question

Determine the pH of a 0.034 solution of H2SO4. The dissociation occurs in two steps. Ka1 is extremely large; Ka2 is 1.2 * 10^-2

Please show work thanks

Answer 1

K_a1 is large so the first proton is fully dissociated

This gives [H⁺] of 3.4 x10^-2 = 0.034

The second proton is weakly dissociated so

K_a2 = [H⁺][SO₄^2-]/[HSO₄^-)] = 1.2x10^-2 = 0.012

We know that the total concentration of HSO₄^- present is the same as the initial concetration of H₂SO₄ because the first dissociation is complete.

Let the amount of dissociation of HSO₄^- be x.

After dissociation [HSO₄^- ]= (0.0032 - x) and [H⁺] = [SO₄^2-] = x

But from K_a1 we already have [H⁺] of 0.0032 so total [H⁺] = 0.0032 + x

From K_a2:

0.012 = (0.0032 + x) x/(0.0032 - x)

0.012 (0.0032 - x) = (0.0032 + x) x

0.0000384 - 0.012x = 0.0032x + x²

x² + 0.0152x - 0.0000384 = 0

Solve this quadratic for x using the quadratic formula.

on calculation we get, x = 0.0022

This is the concentration of H⁺ from the second dissociation

Total [H⁺] = 0.0032 + 0.0022 = 0.0054

pH = - log(.0054) = 2.27