Question

Calculate the solubility (in g/L) of Caso (s) in 0.350 M Na,so, (aq) at 25°C. The K. of Caso, is 4.93 x 10- solubility:

Answer 1

Equation for ionisation of CaSO4,

CaSO4 + Ca²+ + S02-

Solubility product,

Kyp = [Ca2+][80-1

The concentration of Sodium sulfate (Na2SO4) = 0.350M

It ionised as ​​​​​​

Na2SO4 + 2Na+

That is concentration of Na2SO4 = concentration of SO42-

[Na2SO4] = [SO-

[SO2-) = 0.350M

Therefore the concentration of SO42- total in solution is 0.350M ( we can ignore the concentration of SO42- produced by the dissociation of CaSO4 because it is too small).

Ksp = 4.93 * 10-5

4.93 * 10-5 = [Ca2+] * 0.350

4.93 + 10-5 10.350

​​​​​​

[Ca²+1 = 14.08 * 10-5M

Ignore the of infinitely small value of [SO42-] in CaSO4.

Therefore solubility = 14.08*10^{-5} mol/L

Molar mass of CaSO4 = 136 g/mol

Solubility in (g/L) of CaSO4 =

14.08 * 10-mol/L x 136g/mol = 0.0191g/L

=  0.0191 g/L