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Chapter 16: Problem 68

The solubility product of \(\mathrm{PbBr}_{2}\) is \(8.9 \times 10^{-6}\). Determine the molar solubility (a) in pure water, (b) in \(0.20 M \mathrm{KBr}\) solution, (c) in \(0.20 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Short Answer

Expert verified
The molar solubility of \(PbBr_{2}\) is (a) in pure water: \(1.14 x 10^{-2}M\), (b) in \(0.20 M KBr\) solution: \(2.23 x 10^{-3}M\) and (c) in \(0.20 M Pb(NO_{3})_{2}\) solution: \(1.12 x 10^{-5}M\).

Step by step solution

01

Write the Dissociation Equation

Write the dissociation equation for \(PbBr_{2}\). \(PbBr_{2} \leftrightarrow Pb^{2+} + 2Br^-\)
02

Write the Solubility Product Expression

The solubility product expression will be \(Ksp = [Pb^{2+}][Br^-]^2\)
03

Solubility in Pure Water

In pure water, there are no common ions, so this is a straightforward application of the solubility product equation. Letting the solubility be s, we can write \(8.9 x 10^{-6} = s (2s)^2\) which simplifies to \(s = 1.14 x 10^{-2}M\)
04

Solubility in 0.20 M KBr Solution

The KBr will dissociate to give \(0.20 M Br^- ions\). Now, our \(Ksp\) equation will turn into \(8.9 x 10^{-6} = s x \(0.20\)^2\) which gives \(s = 2.23 x 10^{-3}M\)
05

Solubility in 0.20 M \(Pb(NO_{3})_{2}\) Solution

The \(Pb(NO_{3})_{2}\) will dissociate to give \(0.20 M Pb^{2+} ions\). Now the \(Ksp\) equation will turn into \(8.9 x 10^{-6} = 0.20 x (2s)^2\) which gives \(s = 1.12 x 10^{-5}M\)

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Most popular questions from this chapter

How many grams of \(\mathrm{CaCO}_{3}\) will dissolve in \(3.0 \times\) \(10^{2} \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ?\)

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