Question

A 45-mL sample of 0.015 M calcium chloride, CaCl₂, is added to 55 mL of 0.010 M sodium sulfate, Na₂SO₄. Is a precipitate expected? Explain.

Hint -- You must calculate the [Ca²⁺] and [SO₄^2-] of the total solution before calculating Q_c and comparing to K_sp. K_sp of calcium sulfate is 2.4 x 10^-5.

Answer 1

Lets find the concentration after mixing for CaCl2

Concentration after mixing = mol of component / (total volume)

M(CaCl2) after mixing = M(CaCl2)*V(CaCl2)/(total volume)

M(CaCl2) after mixing = 0.015 M*45.0 mL/(45.0+55.0)mL

M(CaCl2) after mixing = 6.75*10^-3 M

Lets find the concentration after mixing for Na2SO4

Concentration after mixing = mol of component / (total volume)

M(Na2SO4) after mixing = M(Na2SO4)*V(Na2SO4)/(total volume)

M(Na2SO4) after mixing = 0.01 M*55.0 mL/(55.0+45.0)mL

M(Na2SO4) after mixing = 5.5*10^-3 M

So, we have now

[Ca2+] = 6.75*10^-3 M

[SO42-] = 5.5*10^-3 M

At equilibrium:

CaSO4 <----> Ca2+ + SO42-

Qsp = [Ca2+][SO42-]

Qsp = (6.75*10^-3)*(5.5*10^-3)

Qsp = 3.713*10^-5

we have,

Ksp = 2.4*10^-5

Since Qsp is greater than ksp, precipitate will form