Answer:
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
