General Chemistry

Question

What is the final sodium ion concentration when 35.0 ml of 0.150 M NaCl is mixed with 75.0 of 0.050 M Na2SO4 and 20.0 ml of 0.100 M KCI?

Answer 1

Remember mole concentration = number of moles /number of liters of solution.

35.0ml of 0.150M NaCl contains 0.035 times 0.150 = 0.0525 mol NaCl thus 0.00525 mol of Na⁺ ions

75.0m of 0.050M Na₂SO₄ contains 0.075 times 0.050 = 0.00375 mol Na₂SO₄ thus 0.0075 mol of Na⁺ ions

Total moles of Na⁺ = 0.00525 +0.00750 = 0.01275 mol

Total volume of solution = 35.0ml + 75.0ml = 120 ml = 0.120 l

Thus mole concentration of mixed solution = 0.01275 mol/0.120 l = 0.10625 M = 0.106M (in 3s.f)

If you need the concentration in grams per liter of solution, just multiply by molar mass of sodium.