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Lelechka [254]
3 years ago
15

Calculate the concentrations of all species in a 1.15 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants for sulfur

ous acid are K a 1 = 1.4 × 10 − 2 and K a 2 = 6.3 × 10 − 8 .
Chemistry
1 answer:
sesenic [268]3 years ago
3 0

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

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