   Chapter 16, Problem 42E

Chapter
Section
Textbook Problem

# Calculate the solubility of solid Pb3(P04)2 (Ksp = 1 × 10−54) in a 0.10-M Pb(NO3)2 solution.

Interpretation Introduction

Interpretation: The solubility product of Pb3(PO4)2 is given. The solubility of Pb3(PO4)2 is to be calculated in a given concentration of Pb(NO3)2 .

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

Where,

• x is coefficient of concentration of A .
• y is coefficient of concentration of B .
Explanation

Explanation

To determine: The solubility of Pb3(PO4)2 in 0.10M Pb(NO3)2 .

The solubility of Pb3(PO4)2 in 0.10M Pb(NO3)2 is 1.6×1026mol/L_ .

Given

Solubility product of Pb3(PO4)2 is 1.0×1054 .

Concentration of Pb(NO3)2 is 0.10M .

The dissociation reaction of Pb3(PO4)2 is,

Pb3(PO4)2(s)3Pb2+(aq)+2PO43(aq)

The ratio of moles between ions is 3:2 .

Since, Pb(NO3)2 is soluble in aqueous solution. The dissociation reaction of Pb(NO3)2 is,

Pb(NO3)2(aq)Pb2+(aq)+2NO3(aq)

The concentration of Pb2+ is,

[Pb2+]=0.10M

The solubility of Pb3(PO4)2 can be calculated from the concentration of ions at equilibrium.

It is assumed that smol/L of solid is dissolved to reach the equilibrium. The meaning of 3:2 stoichiometry of salt is,

smol/LPb3(PO4)23smol/LPb2++2smol/LPO43

Make the ICE table for the dissociation reaction of Pb3(PO4)2

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