 Chemistry
1 February, 10:57

# 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.

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1. 1 February, 11:28
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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