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.

Na₂SO₄ → 2Na⁺ + SO₄²⁻

c₁ (Na⁺) = 2c (Na₂SO₄)

n₁ (Na⁺) = c₁ (Na⁺) v₁=2c (Na₂SO₄) v₁

NaCl → Na⁺ + Cl⁻

c₂ (Na⁺) = c (NaCl)

n₂ (Na⁺) = c₂ (Na⁺) v₂=c (NaCl) v₂

n (Na⁺) = n₁ (Na⁺) + n₂ (Na⁺) = 2c (Na₂SO₄) v₁+c (NaCl) v₂

v=v₁+v₂

c (Na⁺) = n (Na⁺) / v={2c (Na₂SO₄) v₁+c (NaCl) v₂} / (v₁+v₂)

c (Na⁺) = {2*0.400*0.1+0.600*0.2} / (0.1+0.2) = 0.667 mol/L

c (Na⁺) = 0.667M