Question

A) The solubility of BaF₂ is measured and found to be 1.29 g/L. Use this information to calculate a K_sp value for barium fluoride.

K_sp = ___

B) The solubility of Ag₂CrO₄ is measured and found to be 4.25×10^-2 g/L. Use this information to calculate a K_sp value for silver chromate.

K_sp = ___

C)

The solubility of CrPO₄ is measured and found to be 6.95×10^-10 g/L. Use this information to calculate a K_sp value for chromium(III) phosphate.

K_sp = ___

Answer 1

A)

Molar mass of BaF2,

MM = 1*MM(Ba) + 2*MM(F)

= 1*137.3 + 2*19.0

= 175.3 g/mol

Molar mass of BaF2= 175.3 g/mol

s = 1.29 g/L

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

s = 1.29 g/L / 175.3 g/mol

s = 7.359*10^-3 mol/L

At equilibrium:

BaF2 <----> Ba2+ + 2 F-

   s 2s

Ksp = [Ba2+][F-]^2

Ksp = (s)*(2s)^2

Ksp = 4(s)^3

Ksp = 4(7.359*10^-3)^3

Ksp = 1.594*10^-6

Answer: 1.59*10^-6

B)

Molar mass of Ag2CrO4,

MM = 2*MM(Ag) + 1*MM(Cr) + 4*MM(O)

= 2*107.9 + 1*52.0 + 4*16.0

= 331.8 g/mol

Molar mass of Ag2CrO4= 331.8 g/mol

s = 4.25*10^-2 g/L

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

s = 4.25*10^-2 g/L / 331.8 g/mol

s = 1.281*10^-4 mol/L

At equilibrium:

Ag2CrO4 <----> 2 Ag+ + CrO42-

   2s s

Ksp = [Ag+]^2[CrO42-]

Ksp = (2s)^2*(s)

Ksp = 4(s)^3

Ksp = 4(1.281*10^-4)^3

Ksp = 8.406*10^-12

Answer: 8.41*10^-12

C)

Molar mass of CrPO4,

MM = 1*MM(Cr) + 1*MM(P) + 4*MM(O)

= 1*52.0 + 1*30.97 + 4*16.0

= 146.97 g/mol

Molar mass of CrPO4= 146.97 g/mol

s = 6.95*10^-10 g/L

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

s = 6.95*10^-10 g/L / 146.97 g/mol

s = 4.729*10^-12 mol/L

At equilibrium:

CrPO4 <----> Cr3+ + PO43-

   s s

Ksp = [Cr3+][PO43-]

Ksp = (s)*(s)

Ksp = 1(s)^2

Ksp = 1(4.729*10^-12)^2

Ksp = 2.236*10^-23

Answer: 2.24*10^-23