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Question

how to Calculate the molar concentration of sodium ions when 100 mL of 0.2 mol/L sodium sulphate, Na2SO4, solution is mixed with 400 mL of 0.50 mol/L of sodium nitrate, NaNO3.

Answer 1

Na⁺ from Na₂SO₄:  0.1 L x 0.2 mol/L = 0.02 moles Na₂SO4 x 2 moles Na⁺/mole Na²SO₄ = 0.04 moles Na⁺

Na⁺ from NaNO₃:  0.4 L x 0.50 mol/L = 0.2 moles NaNO₃ x 1 mole Na⁺/mole NaNO₃ = 0.2 moles Na⁺

Total moles Na⁺ = 0.04 + 0.2 = 0.24 moles Na⁺

Total volume = 100 ml + 400 ml = 500 ml = 0.5 L

Final concentration of of sodium ions = 0.24 moles Na⁺/0.5 L = 0.48 M