Question

calculate the molar solubility of the following salts under specific conditions:

Calcium carbonate(CaCO3) in water [Ksp= 4.5*10^-9]

Calcium fluoride (CaF2) in water [Ksp= 3.2*10^-11]

Zinc (11) hydroxide (Zn(OH)2) in a solution of pH= 7.50 [Ksp=3.0*10^-16]

Answer 1

1

Calcium carbonate(CaCO3) in water [Ksp= 4.5*10^-9]

Ca+2 and CO3-2

then

Ksp = S*S

S= sqrt(ksp) = sqrt(4.5*10^-9) = 0.00006708203 M = 6.7*10^-5 M

2

Calcium fluoride (CaF2) in water [Ksp= 3.2*10^-11]

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

Ksp = (S)(2S)^2

Ksp = 4S^3

S= (Ksp/4)^(1/3)

S = ((3.2*10^-11)/4)^1/3

S = 2.666*10^-12 M

3

Zinc (11) hydroxide (Zn(OH)2) in a solution of pH= 7.50 [Ksp=3.0*10^-16]

Ksp = Zn(OH)^2

Ksp = S*(OH-)^2

pOH = 14-pH = 14-7.5 = 6.5

OH = 10^-pOH = 10^-6.5 = 3.1622*10^-7

Ksp = S*(OH-)^2

3*10^-16 = S*(3.1622*10^-7)^2

S = (3*10^-16)/((3.1622*10^-7)^2) = 0.00300014735 M