Question

Calculate the solubility constant, Ksp, from the following information

a) The solubility of KIO3 is 43 g/L

b) The solubility of AgI is 9.1x10^-9 M

c) The solubility of MgF2 is 2.64x10^-4 M

Answer 1

we know that

Ksp = solubility product

a) The solubility of KIO3 is 43 g/L

So molar solubility = solubility / Molecular weight = 43 / 214g/mole = 0.201 moles / L

           KIO3 ---> K+ + IO3-

molar solubility = 0.201 moles / L = [K+] = [IO3-]

Ksp = [K+] [IO3-] = 0.201 X 0.201 = 0.0404

b) The solubility of AgI is 9.1x10^-9 M

moalr solubility is given = 9.1x10^-9 M

AgI --> Ag⁺ + I-

[Ag+] = [I-] = 9.1x10^-9 M

Ksp = [Ag+] [ I-] = 9.1x10^-9 M X 9.1x10^-9 M = 8.281 X 10^-17

c) The solubility of MgF2 is 2.64x10^-4 M

Molar solubility = 2.64x10^-4

MgF2 -- > Mg⁺² + 2F-

[F-] = 2 X Mg⁺² = 2 X 2.64x10^-4

Ksp = [Mg+2] [F-]^2 = [ 2.64x10^-4] [ 2 X 2.64x10^-4]^2 = 7.36 X 10^-11