Calculate the pH of a 0.035 M sulfuric acid solution

Question

0.89

1.15

1.46

1.89

None of the above

Answer 1

Sulfuric acid = H₂SO₄

H₂SO₄ <==> H⁺ + HSO₄^-

HSO₄^- <==> H⁺ + SO₄^2-

Since the problem does not provide the Ka for the 2nd ionization, we might assume that only the first contributes significantly to the [H⁺]. With this assumption, we can find the pH as follows:

0.035 M H₂SO₄ produces 0.035 M H⁺

pH = -log [H⁺] = -log 0.035

pH = 1.46

If we were to include the 2nd ionization, we would proceed as follows:

H₂SO₄ <==> H⁺ + HSO₄^- and [H⁺] = 0.035 M and [HSO4-] = 0.035 M

HSO₄^- <==> H⁺ + SO₄^2-

0.035.........0.035.......0.035......Initial

-x................+x..........+x............Change

0.035-x........0.035+x...0.035+x.....Equilibrium

Ka = [H⁺][SO₄^2-] / [HSO₄^-]

Look up Ka for HSO₄^- and plug in equilibrium values.

Solve for [H+]. Add this to 0.035 M from 1st ionization. Take negative log to obtain pH.

The answer would have to be a pH less than 1.46, but not much less.

Answer 2

It might be the case that additional information is required to adequately answer this question because we know that H₂SO₄ is a strong acid in solution and thus would fully dissociate its first proton and only partially dissociate its second proton meaning a RICE table solving a quadratic formula given a hydrogen ion concentration and Ka value.