CHEM 212:CHEMISTSRY V 2
CHEM 212:CHEMISTSRY V 2
8th Edition
ISBN: 9781260304503
Author: SILBERBERG
Publisher: MCGRAW-HILL CUSTOM PUBLISHING
Question
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Chapter 22, Problem 22.80P

(a)

Interpretation Introduction

Interpretation:

The affect of acid rain on the leaching of phosphate into ground water from terrestrial phosphate rock has to be given.  The solubility of Ca3(PO4)2 in pure water with pH 7.0 has to be calculated.

(a)

Expert Solution
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Explanation of Solution

The solubility of Ca3(PO4)2 can be calculated from its Ksp.  It is 1.2×10-29.

The phosphate ion is derived from a weak acid.  the pH of the solution impacts the acid-base equilibrium.

The solubility of Ca3(PO4)2 in pure water can be calculated by using the equilibrium expression and concentration of H+.

   Ca3(PO4)2(s)Ca2+(aq)+ 2PO43-(aq)Initial - 0 0Change -x +3x +2x_Equilibrium - 3x 2x

The equilibrium constant can be calculated as

  Ksp= [Ca2+]3[PO43-]21.2×10-29 = (3x)3(2x)2 = 108x5x = 6.4439×10-7M

The solubility of Ca3(PO4)2 in pure water is 6.4×10-7M.

(b)

Interpretation Introduction

Interpretation:

The affect of acid rain on the leaching of phosphate into ground water from terrestrial phosphate rock has to be given.  The solubility of Ca3(PO4)2 in moderately acidic rain water with pH 4.5 has to be calculated.

(b)

Expert Solution
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Explanation of Solution

The concentration of H3O+ at pH 4.5 is

  [H3O+] = 10-4.5 = 3.162×10-5M

The phosphate ion can gain one H+ and forms HPO42-. On further gain of one more H+, it forms H2PO4-, on addition of last H+, it forms H3PO4.  The Ka values for phosphate ions are Ka1=7.2×10-3,Ka2=6.3×10-8andKa3=4.2×10-13.

  H3PO4(aq)+ H2O(l)H2PO4-(aq)+ H3O+

The equilibrium constant can be calculated as

  Ka1[H2PO4-][H3O+][H3PO4][H2PO4-][H3PO4]=Ka1[H3O+]7.2×10-33.162×10-5=227.70

  H2PO4-(aq)+ H2O(l)HPO42-(aq)+ H3O+(aq)

The equilibrium constant can be calculated as

  Ka2[HPO42-][H3O+][H2PO4-][HPO42-][H2PO4-]=Ka2[H3O+]6.3×10-83.162×10-5=1.9924×10-3

  HPO42-(aq)+ H2O(l)PO43-(aq)+ H3O+(aq)

The equilibrium constant can be calculated as

  Ka3[PO43-][H3O+][HPO42-][PO43-][HPO42-]=Ka3[H3O+]4.2×10-133.162×10-5=1.32827×10-8

From the above calculations, the concentration of H2PO4- is at least 1000 times more than the concentration of any other species.  It is the dominant species and from this the value of [PO43-] is calculated.

  [PO43-][HPO42-] = 1.32827×10-8 and [HPO42-][H2PO4-] = 1.9924×10-3[PO43-] = (1.32827×10-8)[HPO42-] and [HPO42-] = (1.9924×10-3)[H2PO4-][PO43-] = (1.32827×10-8)[HPO42-] = (1.32827×10-8)(1.9924×10-3)[H2PO4-][PO43-] = (2.6464×10-11)[H2PO4-]

Substitute this in the Ksp expression,

  Ksp = [Ca2+]3[PO43-]2 = 1.2×10-29Ksp = [Ca2+]3{(2.6464×10-11)[H2PO4-]}2

Rearrange,

  Ksp/(2.6464×10-11)2 = [Ca2+]3[H2PO4-]2

The concentration of calcium ions is represented as 3x and the concentration of dihydrogen phosphate ion as 2x.

  Ksp/(2.6464×10-11)2 = [3x]3[2x]2(1.2×10-29)(2.6464×10-11)2 = 108x5x = 1.0967×10-2 = 1.1×10-2M

Acid rains increases the leaching of phosphate into ground water from terrestrial phosphate rock due to the protonation of PO43- to HPO42-andH2PO4-.

From the calculations of (a) and (b), the solubility increases from pure water 6.4×10-7M to acidic rain water 1.1×10-2M.

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Chapter 22 Solutions

CHEM 212:CHEMISTSRY V 2

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