http://www.ck12.org Chapter 19. Equilibrium
Sample Problem 19.2: Calculating Ks pfrom Solubility
The solubility of lead(II) fluoride is found experimentally to be 0.533 g/L. Calculate the Ks pfor lead(II) fluoride.
Step 1: List the known quantities and plan the problem.
Known
- solubility of PbF 2 = 0.533 g/L
- molar mass of PbF 2 = 245.20 g/mol
Unknown
- Ks pof PbF 2 =?
The dissociation equation for PbF 2 and the corresponding Ks pexpression can be constructed as follows:
PbF 2 (s)⇀↽Pb^2 +(aq)+2F−(aq) Ksp= [Pb^2 +][F−]^2The steps above will be followed to calculate Ks pfor PbF 2.
Step 2: Solve.
molar solubility:- (^533) g
1 L
×
1 mol
245. 20 g= 2. 17 × 10 −^3 M
The dissociation equation shows that, for every mole of PbF 2 that dissociates, 1 mol of Pb^2 +and 2 mol of F−are
produced. Therefore, at equilibrium, the concentrations of the ions are:
Pb^2 + = 2.17× 10 −^3 M and [F−] = 2×2.17× 10 −^3 = 4.35× 10 −^3 MSubstitute into the equilibrium expression, and solve for Ks p.
Ks p= (2.17× 10 −^3 )(4.35× 10 −^3 )^2 = 4.11× 10 −^8Step 3: Think about your result.
The solubility product constant is significantly less than 1 for a nearly insoluble compound such as PbF 2.
Practice Problem- From the given solubility data, calculate Ks pfor each of the following compounds.
a. copper(II) iodide, CuI = 4.30× 10 −^4 g/L
b. silver sulfide, Ag 2 S = 2.84× 10 −^15 g/L
The known Ks pvalues from the table above (Table19.5) can be used to calculate the solubility of a given compound
by following the steps listed below.
- Set up an ICE problem (Initial, Change, Equilibrium) in order to use the Ks pvalue to calculate the concentra-
tion of each of the ions. Assume that no ions are initially present in the solution. - The concentrations of the ions can be used to calculate the molar solubility of the compound.