concentration of the final solution? Chemistry question?

2016-11-02 5:07 am

If 100.0mL of 0.100M NazSO4 (aq) is combined with 200.0mL of 0.150M NaCl, what is the concentration of Na+ ions in the final solution, assuming the final volume is the sum of the volumes of the 2 solutions?

I know that the answer is 0.167M but i don't know why.

回答 (1)

戇戇居士

2016-11-02 5:19 am

✔ 最佳答案

No. of moles of Na₂SO₄ = (0.100 mol) × (100.0/1000) = 0.0100 mol
Each moles of Na₂SO₄ contains 2 moles of Na⁺ ions.
No. of moles of Na⁺ ions in Na₂SO₄ = (0.0100 mol) × 2 = 0.0200 mol

No. of moles of NaCl = (0.150 mol) × (200.0/1000) = 0.0300 mol
Each moles of NaCl contains 1 moles of Na⁺ ions.
No. of moles of Na⁺ ions in NaCl = 0.0300 mol

Total no. of moles of Na⁺ ions in the final solution = (0.0200 + 0.0300) mol = 0.0500 mol
Volume of the final solution = (100.0 + 200.0) mL = 300.0 mL = 0.3000 L
Concentration of Na⁺ ions in the final solution = (0.0500 mol) / (0.3000 L) = 0.167 M

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