Question

Indicate the concentration of each ion present in the solution formed by mixing the the following.

44.0 mL of 0.100 M Na2SO4 and 25.0 mL of .200 M KCl. Assume that the volumes are additive.

Use three significant figures for all answers.

Answer 1

Molarity of Na₂SO₄ (M₁)= 0.100M = 0.100 mol/L

Volume of Na₂SO₄ (V₁)= 44.0 mL

Volume of KCl added = 25 mL

Molarity of KCl = 0.200 M

Total volume of solution(V₂) = 44 + 25 = 69.0 mL

Let molarity of Na₂SO₄ after mixing be M₂

From molarity equation we have

M₁V₁ =M₂V₂

0.100M*44 mL = M₂ * 69 mL

M₂ = 0.0638 M

Na₂SO₄ dissociates as :

Na₂SO₄ 2Na⁺ + SO₄^2-

Thus, [Na⁺] = 2*[ Na₂SO₄ ] = 2* 0.0638M = 0.128M

[SO₄^2-] =[ Na₂SO₄ ] = 0.0638M

Similarly KCl concentration is

0.200M * 25 mL = M₂ * 69 mL

M₂ = 0.0725 M

KCl dissociates as:

KCl K⁺ + Cl^-

[KCl] = [K⁺ ] = [Cl^-] = 0.0725 M

Hence, [Na⁺] =  0.128M ; [SO₄^2-] = 0.0638M ; [Cl^-] = 0.0725 M ; [K⁺ ] = 0.0725 M