Sulfuric acid dissolves in an aqueous solution producing two protons. If the initial concentration of sulfuric acid is 0.82 M, what would be the concentration of the protons?



Answer :

Sulfuric acid (H₂SO₄) is a strong diprotic acid, meaning it can donate two protons (H⁺) per molecule. When it dissolves in water, it dissociates completely in two steps:

1. \( \text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \)

2. \( \text{HSO}_4^- \rightarrow \text{H}^+ + \text{SO}_4^{2-} \)

Initially, every mole of H₂SO₄ produces one mole of H⁺ and one mole of HSO₄⁻.

The second dissociation of HSO₄⁻ is also complete for the purpose of initial concentration calculations, so it produces an additional mole of H⁺ for every mole of H₃SO₄⁻ ionized.

Thus, for an initial concentration of 0.82 M H₂SO₄, each molecule produces a total of 2 moles of H⁺:

\[

[\text{H}^+] = 0.82 \, \text{M} \times 2 = 1.64 \, \text{M}

\]

So, the concentration of protons (H⁺) in the solution would be 1.64 M.