Calculate the concentrations of all species in a 1.15 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants for sulfurous acid are K a 1 = 1.4 * 10 - 2 and K a 2 = 6.3 * 10 - 8.

Na₂SO₄ = 1.15 M

Na⁺ = 2.3 M

SO₃²⁻ = 1.15

Explanation:

First, the concentration of ions in solution is dependent on the mole ratio of the substance and the dissolved ions in solution. Let's take sodium sulfite in the question:

Na₂SO₃ (s) → 2Na²⁺ (aq) + SO₃²⁻ (aq)

From the stoichiometrical ratios:

1 mole of Na₂SO₃ gives:

2 moles of Na²

1 mole of SO₃²⁻

The concentration of Na²⁺ will be 2 * 1.15 = 2.3 M

The concentration of SO₃²⁻ will be = 1.15 M

The concentration of Na₂SO₃ will be = 1.15 M