Question

Given the solubility, calculate the solubility product constant (ksp) of each salt at 25°c: (a) pbcro4, s = 4.0 × 10−5 g/l; (b) bac2o4, s = 0.29 g/l; (c) mnco3, s = 4.2 × 10−6 g/l.

Answer 1

a) PbCrO4:

according to the equation:

PbCrO4(s)  →  Pb2+(aq)   +  CrO42-(aq)

so, Ksp = [Pb2+][CrO42-]

by assuming [Pb2+] = [CrO42-] = X

and when S (the solubility) = 4 x 10^-5 g/L

we have first to convert solubility from g/L to mol/L by getting the molar mass of the salt 

solubility mol/L =  solubility g/L / molar mass of salt

                           = 4 x 10^-5g/L / 323.2 g/mol

                            = 1.24 x 10^-7 mol / L

by substitution in Ksp formula:

∴Ksp = X* X

         = (1.24 x 10^-7)^2

         = 1.54 x 10^-14

b) BaC2O4:

according to this equation:

BaC2O4(s)→  Ba 2+(aq)  +  C2O4 2- (aq)  

So Ksp = [Ba2+][C2O42-]

Assume that [Ba2+] = [C2O42-] = X

when the solubility S =  0.29 g/L = X  , so we need to convert S from g/L to
mol / L 

solubility mol / L= solubility g/L / molar mass of salt

                           = 0.29 g/L / 225.34 g/mol

                           = 0.0013 mol/L

by substitution in ksp formula:

∴Ksp = X^2 

         = (0.0013)^2 

         = 1.69 x 10^-6 

C)  MnCO3: 

according to this equation :

MnCO3(s)→ Mn2+(aq)   +  CO3 2-(aq)

so, Ksp = [Mn2+][CO32-]

assume [Mn2+] = [CO32-] = X

when the solubility s = 4.2 x 10^-6 g/L so we need to convert S from g/L to mol/L by dividing on molar mass.

the solubility mol/L = solubility g/L / molar mass g/mol

                                 =.4.2 x 10^-6 /114.9 

                                 = 3.7 x 10^-8 mol/L

by substitution on ksp formula:

∴ Ksp = X*X

           = (3.7 x 10^-8)^2

           = 1.369 x 10^-15