Question

If 130.0 mL of 0.0030 MNa2SO4(aq) is saturated with CaSO4, how many grams of CaSO4 would be present in the solution?

Answer 1

Solution :-

Na2SO4 ---- > 2 Na^+ + SO4^2-

[0.0030 M Na2SO4*1 M SO4^2- / 1 Na2SO4] = 0.0030 M SO4^2-

Using the concentration of the SO4^2- ions we need to find the concentration of the Ca^2+

CaSO4 ---- > Ca^2+ + SO4^2-

                           X                0.0030+x

Ksp = 4.93*10^-5 for the CaSO4

Ksp = [Ca^2+] [SO4^2-]

4.93*10^-5 = [x][0.0030+x ]

4.93*10^-5 = 0.003x + x^2

-x^2-0.003x +4.93*10^-5 = 0

Solving for using the quadratic formula we get

X = 0.00568 M= [Ca^2+]

Using the molarity of the Ca^2+ we can find the mass of CaSO4 in the 130 ml solution

(0.00568 mol per Ca^2+ * 0.130 L)*(1 mol CaSO4 / 1 mol Ca^2+)*(136.14 g / 1 mol CaSO4) = 0.101 g CaSO4

Therefore the mass of CaSO4 present in the solution is 0.101 g