Question

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.

Answer 1

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