Question

a) The solubility of calcium sulfate (CaSO4) is found to be 0.67 g/L
Calculate the values of Ksp for calcium sulfate.
b) Calculate the solubility of copper (II) hydroxide Cu(OH)2 in g/L. The Ksp if copper (II) hydroxide is 2.2×10^-20.

Answer 1

a)

Molar mass of CaSO4 = 1*MM(Ca) + 1*MM(S) + 4*MM(O)

= 1*40.08 + 1*32.07 + 4*16.0

= 136.15 g/mol

Molar mass of CaSO4= 136.15 g/mol

s = 0.67 g/L

To covert it to mol/L, divide it by molar mass

s = 0.67 g/L / 136.15 g/mol

s = 4.921*10^-3 g/mol

The salt dissolves as:

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

   s s

Ksp = [Ca2+][SO42-]

Ksp = (s)*(s)

Ksp = 1(s)^2

Ksp = 1(4.921*10^-3)^2

Ksp = 2.42*10^-5

Answer: 2.42*10^-5

b)

The salt dissolves as:

Cu(OH)2 <----> Cu2+ + 2 OH-

   s 2s

Ksp = [Cu2+][OH-]^2

2.2*10^-20=(s)*(2s)^2

2.2*10^-20= 4(s)^3

s = 1.765*10^-7 M

Molar mass of Cu(OH)2 = 1*MM(Cu) + 2*MM(O) + 2*MM(H)

= 1*63.55 + 2*16.0 + 2*1.008

= 97.566 g/mol

Molar mass of Cu(OH)2= 97.566 g/mol

s = 1.765*10^-7 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.765*10^-7 mol/L * 97.566 g/mol

s = 1.72*10^-5 g/L

Answer: 1.72*10^-5 g/L