7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10−6).
7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 x 10−6).
CaSO4 (s)↔Ca2+ (aq)+SO2−4 (aq)
Ksp= [Ca2+] [ SO2−4]
Let the solubility of CaSO4 be s.
[Ca2+] = [ SO2−4] = s
Then, Ksp=s2
9.1*10−6=s2
s =3.02*10−3mol/L
Molecular mass of CaSO4=136g/mol
Solubility of CaSO4 in gram/L= 3.02*10−3*136=0.41g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4.
Therefor
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