   Chapter 16, Problem 41E

Chapter
Section
Textbook Problem

# Calculate the solubility of solid Ca3(PO4)2 (Ksp = 1.3 × 10−32) in a 0.20-M Na3PO4 solution.

Interpretation Introduction

Interpretation: The solubility product of Ca3(PO4)2 is given. By using this value, the solubility of Ca3(PO4)2 is to be calculated in a given concentration of Na3PO4 solution.

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 Ca3(PO4)2 in 0.20M Na3PO4 .

The solubility of Ca3(PO4)2 in 0.20M Na3PO4 is 2.3×1011mol/L_ .

Given

Solubility product of Ca3(PO4)2 is 1.3×1032 .

Concentration of Na3PO4 is 0.20M .

The dissociation reaction of Ca3(PO4)2 is,

Ca3(PO4)2(s)3Ca2+(aq)+2PO43(aq)

The ratio of moles between ions is 3:2 .

Since, Na3PO4 is soluble in aqueous solution. The dissociation reaction of Na3PO4 is,

Na3PO4(aq)3Na+(aq)+PO43(aq)

The concentration of PO4 is,

[PO43]=0.20M

The solubility of Ca3(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/LCa3(PO4)23smol/LCa2++2smol/LPO43

Make the ICE table for the dissociation reaction of Ca3(PO4)2

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