Consider the reaction

CaSO4 (s) ⇌ C a2 + (aq) + S O2-4 (aq)

At 25 ∘C the equilibrium constant is Kc=2.4 * 10-5 for this reaction.

- If excess CaSO4 (s) is mixed with water at 25 ∘C to produce a saturated solution of CaSO4, what is the equilibrium concentration of SO2-4?

- If the resulting solution has a volume of 1.2 L, what is the minimum mass of CaSO4 (s) needed to achieve equilibrium?

a) According to the reaction equation:

CaSO4 (s) ↔ Ca2 + (aq) + SO4 - (aq)

when Kc = [Ca2+][SO4-]

when [Ca2+] = [SO4-] = X

and we have Kc = 2.4 x 10^-5 so, by substitution we can get [Ca2+]&[SO4-]

2.4 x 10^-5 = X^2

∴X = 0.0049

∴[SO4-] = [Ca2+] = 0.0049 M

b) when we have [Ca2+] = 0.0049 M so we can get the no. of moles of CaSO4:

moles of CaSO4 = molarity * volume

= 0.0049 M * 1.2 L

= 0.00588 moles

when we know the molar mass of CaSO4 = 136.14 g / mol, So we can get the mass:

∴mass of CaSO4 = moles of CaSO4 * molar mass of CaSO4

= 0.00588 moles * 136.14 g/mol

= 0.8 g