100mL of solution contains 0.2mol of H2SO4, what is the pH?

In lower forms, it is always assumed that H₂SO₄ is strong diprotic (dibasic acid), and 1 mol of H₂SO₄ dissociates in water to give 2 mol of hydrogen ions.

Concentration = (No. of moles) / (Volume in L)
[H₂SO₄] = (0.2 mol) / (100/1000 L) = 2 mol/L

1 mol of H₂SO₄ dissociates in water to give 2 mol of hydrogen ions.
[H₂⁺] = (2 mol/L) × 2 = 4 mol/L

pH = -log[H⁺] = -log(4) = -0.6

Note : The range of pH can be -1 to 15. The useful range of pH is 0 to 14.


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In higher forms, the 1st and 2nd dissociations of H₂SO₄ are considered.

Concentration = (No. of moles) / (Volume in L)
[H₂SO₄] = (0.2 mol) / (100/1000 L) = 2 mol/L

For H₂SO₄, K₁   and K₂ = 0.012 mol/L

1st ionization of H₂SO₄ :
H₂SO₄(aq) ⇌ H⁺(aq) + HSO₄⁻(aq)

[H⁺] produced in 1st ionization = 2 mol/L

2nd ionization of H₂SO₄ :
HSO₄⁻(aq) ⇌ H⁺(aq) + SO₄²⁻(aq)
Assume that y mol/L of HSO₄⁻ is dissociated.
[HSO₄⁻] = (2 - y) mol/L, [H⁺] = (2 + y) mol/L (aq), [SO₄²⁻] = y mol/L

K₂ = y(2 + y)/(2 - y) = 0.012
2y + y² = 0.024 - 0.012y
y² + 2.012y - 0.024 = 0
y = 0.01 or y = -2.0 (rejected)

[H⁺] = (2 + 0.01) mol/L = 2.01
pH = -log[H⁺] = -log(2.01) = -0.3

Note:
1. The range of pH can be -1 to 15. The useful range of pH is 0 to 14.
2. K₁ and K₂ are only applied for dilute acid solutions. In concentrated solution, due to the high concentrations of ions, the hydrogen ions and anions would combine to form acid molecules more readily.