Need help with a molar solubility question [Solved]

The Ksp of PbBr2 is 6.60 X 10^-6.

The Ksp of PbBr2 is 6.60 X 10^-6.
The balanced equation for the dissolution of lead (II) bromide in water is
PbBr2(s) Pb^2 (aq) 2 Br^-(aq)
Ksp = [Pb^2 ][Br^-]^2

First question: What is the molar solubility of PbBr2 in pure water
From the above equation we see that molar concentration of the lead (II) ions equals the molar concentration of lead (II) bromide (see same coefficients). On the other hand, the molar concentration of bromide ions is twice the molar concentration of lead (II) ions.
Let x equal the molar concentration of lead (II) ions, and let 2x equal the molar concentration of bromide ions. Substituting in the above equation for the solubility product constant, we ge:
6.60 10-6 = [x][2x]^2
6.60 10-6 = [x][4x^2]
6.60 10-6 = [4x^3]
1.65 10-6 = [x^3]
1.18166575 10-2 = x
[Pb^2 ] = 1.18 10-2 mol/L = [PbBr2]
Answer: The molar solubility of PbBr2 in pure water is approximately 1.18 10-2 mol/L.

Second question: What is the molar solubility of PbBr2 in 0.500 M KBr solution
Judging from the value of Ksp constant, the molar concentrations of lead (II) ions and molar concentrations of bromide ions are very small, and seem insignificant compared with the molar concentration of bromide ions in 0.500 M solution of potassium bromide. We can calculate the lead (II) ions concentration in this solution by substituting 0.500 M for the bromide ions concentration in the equation for Ksp constant of lead (II) bromide:
Ksp = [Pb^2 ][2 Br^-]^2
6.60 10-6 = [Pb^2 ][(2)(0.500)]^2
6.60 10-6 = [Pb^2 ][(1.00)]^2
6.60 10-6 = [Pb^2 ][1.00]
6.60 10-6 = [Pb^2 ]
6.60 10-6 mol/L = [Pb^2 ] = [PbBr2]
Answer: The molar solubility of PbBr2 in 0.500 M KBr solution is approximately 6.60 10-6 mol/L.

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Third question: What is the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution
Judging from the value of Ksp constant, the molar concentrations of lead (II) ions and molar concentrations of bromide ions in a saturated solution of lead (II) bromide are very small, and seem insignificant compared with the molar concentration of lead (II) ions in 0.500 M solution of lead (II) nitrate. We can calculate the bromide ions molar concentration in this solution by substituting 0.500 M for the lead (II) ions concentration in the equation for Ksp constant of lead (II) bromide:
Ksp = [Pb^2 ][(2)(Br^-)]^2
6.60 10^-6 = [0.500][(2)(Br^-)]^2
6.60 10^-6 = [0.500](4)[Br^-]^2
13.2 10^-6 = (4)[Br^-]^2
3.30 10^-6 = [Br^-]^2
1.8165902 10^-6 = [Br^-]
Since the concentration of lead (II) ions from lead (II) bromide dissolved in this solution will be half the concentration of the bromide ions, then
[Pb^2 ] = (0.5)(1.8165902 10^-6) = 0.908296 10^-6 mol/L or
[PbBr2] = 9.08 10^-7 mol/L
Answer: The molar solubility of PbBr2 in 0.500 M Pb(NO3)2 solution is approximately 9.08 10^-7 mol/L.

Source(s): personal knowledge

This post is last updated on hrtanswers.com at Date : 1st of September – 2022