If 100.mL of 0.400 M Na₂SO₄ is added to 200.mL of 0.600 M NaCl,what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive.

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First, we need to calculate the number of moles of Na₂SO₄ and NaCl separately: n(Na₂SO₄) = V x C = 0.1 L x 0.4 mol/L = 0.04 mol n(NaCl) = V x C = 0.2 L x 0.6 mol/L = 0.12 mol Since Na⁺ ions come from Na₂SO₄, we need to calculate the number of moles of Na⁺ in 0.04 mol of Na₂SO₄: 1 mol of Na₂SO₄ produces 2 mol of Na⁺ ions 0.04 mol of Na₂SO₄ produces 0.08 mol of Na⁺ ions Next, we need to add the moles of Na⁺ to the moles of NaCl: Total moles of Na⁺ = 0.08 mol Total moles of NaCl = 0.12 mol To get the concentration of Na⁺, we divide the total moles of Na⁺ by the total volume of the solution: Total volume = 0.1 L + 0.2 L = 0.3 L C(Na⁺) = total moles of Na⁺ / total volume of the solution = 0.08 mol / 0.3 L C(Na⁺) = 0.267 M However, we need to convert this to the concentration of Na⁺ ions: 1 mol of Na⁺ produces 2 mol of Na+ and 1 mol of SO42- Thus, the concentration of Na⁺ ions is: C(Na⁺ ions) = 2 x C(Na⁺) = 2 x 0.267 M = 0.534 M Therefore, the answer is (A) 0.534 M.

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