Answer to Question #96202 in General Chemistry for Truong

Question

Some Ca(NO3)2 is dissolved in 100mL of water. This solution is mixed with 300mL of 0.010M Na2SO4. A very faint precipitate of calcium sulfate is formed. If the Ksp value of calcium sulfate is 2.4 x 10-5, how much calcium nitrate was dissolved to make the initial solution?

Answer 1

The dissolution of calcium sulfate forms equal amounts of calcium ions and sulfate ions.

Then we can solve for the molarity of calcium sulfate that would be soluble.

2.4 x 10^5- = (x)*(x)

Solving for x we get [Ca²⁺] = [SO₄^2-] = 4.90 x 10^-3 M

Since the equation above shows a 1:1 mole ratio of calcium sulfate to Ca²⁺ ions, we can assume that 4.90 x 10-3 moles of CaSO₄ will dissolve.

Na₂SO₄ + Ca(NO₃)₂ = CaSO₄ + 2NaNO₃

n(Na₂SO₄) = C*V = 0.01mol/L*0.3L =0.003 mol

According to the equation 0.003 mol of Na₂SO_{4 }reacts with 0.003 mol of Ca(NO₃)_2, and this amount corresponds to that specified in K_{sp .}

n(Ca(NO₃)₂) = 0.003 mol

m(Ca(NO₃)₂) = n*M = 0.003mol*164g/mol = 0.492 g