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
Calcium nitrate is limiting reactant, since it says there’s excess HCl. So you must use the mass of calcium nitrate in your calculation (see picture below):
Answer:
31.3 kJ/mol is the free energy, ΔG, for the overall reaction, A⇌D.
Explanation:
A⇌B, 
= 11.9 kJ/mol ...[1]
B⇌C, 
= -26.7 kJ/mol ...[2]
C⇌D, 
= 7.30 kJ/mol ...[3]
A⇌D, 
= ?...[4]
[4] = [1] - [2] - [3] (Using Hess's law)
31.3 kJ/mol is the free energy, ΔG, for the overall reaction, A⇌D.
