Glencoe Chemistry: Matter and Change, Student Edition
Glencoe Chemistry: Matter and Change, Student Edition
1st Edition
ISBN: 9780076774609
Author: McGraw-Hill Education
Publisher: MCGRAW-HILL HIGHER EDUCATION
Question
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Chapter 17.3, Problem 20PP

(a)

Interpretation Introduction

Interpretation:

When solubility product is given then the solubility in mol/L is to be calculated.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

(a)

Expert Solution
Check Mark

Answer to Problem 20PP

The solubility of the compound PbCrO4 mol/L is 4.795 × 10-7 mol/L.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

PbCrO4Pb2++ CrO42-

I =             0         0          (initial concentration)

C =          +x     +x         (change in concentration or change in amount from initial state to equilibrium)

E =             x        x          (equilibrium concentration)

The solubility product is given as

             Ksp =    [Pb2+] [CrO42-]                      =     x . x                      =     x2          Ksp  =     x

Or, x = 23 × 10-14

= 4.795 × 10-7 mol/L

(b)

Interpretation Introduction

When solubility product is given then the solubility in mol/L is to be calculated

Interpretation:

The positron should be defined and described.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

(b)

Expert Solution
Check Mark

Answer to Problem 20PP

The solubility of the compound AgCl in mol/L is 1.342 × 10-5 mol/L.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

          AgClAg1++ Cl1-

I =             0         0          (initial concentration)

C =          +x     +x         (change in concentration or change in amount from initial state to equilibrium)

E =             x        x          (equilibrium concentration)

The solubility product is given as

              Ksp =    [Ag+] [Cl1-]                      =     x . x                      =     x2

Ksp =     x

Or, x = 1.8× 10-10

= 1.342 × 10-5 mol/L

(c)

Interpretation Introduction

Interpretation:

When solubility product is given then the solubility in mol/L is to be calculated

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

(c)

Expert Solution
Check Mark

Answer to Problem 20PP

The solubility of the compound CaCO3 in mol/L is 5.831 × 10-5 mol/L.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

CaCO3Ca2++CO3-2

I =             0         0          (initial concentration)

C =          +x     +x         (change in concentration or change in amount from initial state to equilibrium)

E =             x        x          (equilibrium concentration)

The solubility product is given as

             Ksp =    [Ca2+] [CO32-]                      =     x . x                      =     x2

 Ksp =     x

Or, x = 3.4× 10-9

                         = 5.831 × 10-5 mol/L

Chapter 17 Solutions

Glencoe Chemistry: Matter and Change, Student Edition

Ch. 17.1 - Prob. 11SSCCh. 17.1 - Prob. 12SSCCh. 17.2 - Prob. 13SSCCh. 17.2 - Prob. 14SSCCh. 17.2 - Prob. 15SSCCh. 17.2 - Prob. 16SSCCh. 17.2 - Prob. 17SSCCh. 17.3 - Prob. 18PPCh. 17.3 - Prob. 19PPCh. 17.3 - Prob. 20PPCh. 17.3 - Prob. 21PPCh. 17.3 - Prob. 22PPCh. 17.3 - Prob. 23PPCh. 17.3 - Prob. 24PPCh. 17.3 - Prob. 25PPCh. 17.3 - Prob. 26PPCh. 17.3 - Prob. 27SSCCh. 17.3 - Prob. 28SSCCh. 17.3 - Prob. 29SSCCh. 17.3 - Prob. 30SSCCh. 17.3 - Prob. 31SSCCh. 17.3 - Prob. 32SSCCh. 17 - Prob. 33ACh. 17 - Prob. 34ACh. 17 - Prob. 35ACh. 17 - Prob. 36ACh. 17 - Prob. 37ACh. 17 - Prob. 38ACh. 17 - Prob. 39ACh. 17 - Prob. 40ACh. 17 - Prob. 41ACh. 17 - Prob. 42ACh. 17 - Prob. 43ACh. 17 - Prob. 44ACh. 17 - Prob. 45ACh. 17 - Prob. 46ACh. 17 - Prob. 47ACh. 17 - Prob. 48ACh. 17 - Prob. 49ACh. 17 - Prob. 50ACh. 17 - Prob. 51ACh. 17 - Prob. 52ACh. 17 - Prob. 53ACh. 17 - Prob. 54ACh. 17 - Prob. 55ACh. 17 - Prob. 56ACh. 17 - Prob. 57ACh. 17 - Prob. 58ACh. 17 - Prob. 59ACh. 17 - Prob. 60ACh. 17 - Prob. 61ACh. 17 - Prob. 62ACh. 17 - Prob. 63ACh. 17 - Prob. 64ACh. 17 - Why are compounds such as sodium chloride usually...Ch. 17 - Prob. 66ACh. 17 - Prob. 67ACh. 17 - Prob. 68ACh. 17 - Prob. 69ACh. 17 - Prob. 70ACh. 17 - Prob. 71ACh. 17 - Prob. 72ACh. 17 - Prob. 73ACh. 17 - Prob. 74ACh. 17 - Prob. 75ACh. 17 - Prob. 76ACh. 17 - Prob. 77ACh. 17 - Prob. 78ACh. 17 - Evaluate this statement: A low value for Keq means...Ch. 17 - Prob. 80ACh. 17 - Prob. 81ACh. 17 - Prob. 82ACh. 17 - Prob. 83ACh. 17 - Prob. 84ACh. 17 - Prob. 85ACh. 17 - Prob. 86ACh. 17 - Prob. 87ACh. 17 - Prob. 88ACh. 17 - Prob. 89ACh. 17 - Prob. 90ACh. 17 - Prob. 91ACh. 17 - Prob. 92ACh. 17 - Prob. 93ACh. 17 - Prob. 94ACh. 17 - Prob. 95ACh. 17 - Prob. 96ACh. 17 - Prob. 97ACh. 17 - Prob. 98ACh. 17 - Prob. 99ACh. 17 - Prob. 100ACh. 17 - Prob. 101ACh. 17 - Prob. 102ACh. 17 - Prob. 103ACh. 17 - Prob. 104ACh. 17 - Prob. 105ACh. 17 - Prob. 1STPCh. 17 - Prob. 2STPCh. 17 - Prob. 3STPCh. 17 - Prob. 4STPCh. 17 - Prob. 5STPCh. 17 - Prob. 6STPCh. 17 - Prob. 7STPCh. 17 - Prob. 8STPCh. 17 - Prob. 9STPCh. 17 - Prob. 10STPCh. 17 - Prob. 11STPCh. 17 - Prob. 12STPCh. 17 - Prob. 13STPCh. 17 - Prob. 14STPCh. 17 - Prob. 15STPCh. 17 - Prob. 16STPCh. 17 - Prob. 17STPCh. 17 - Prob. 18STP
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