Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 16, Problem 16.111AP

(a)

Interpretation Introduction

Interpretation: The two equilibrium constant for the given equilibrium reactions are given. The species that is more likely present in the drinking water, the equilibrium constant of the overall reaction of the two reactions, the pH and the equilibrium concentration of HF2 is to be calculated.

Concept introduction: The solution is supersaturated if the ionic product is more than the solubility product for a salt.

The overall equilibrium constant of the two equilibrium reaction is calculated as,

Ka(netreaction)=Ka1×Ka2

The pH of the solution is calculated by the formula,

pH=log[H3O+]

To determine: The fluorine containing species that is more likely to be present in drinking water.

(a)

Expert Solution
Check Mark

Answer to Problem 16.111AP

Solution:

The fluorine containing species that is more likely to be present in drinking water is HF2

Explanation of Solution

Given

The equilibrium of hydrofluoric acid in water is given as,

HF(aq)+H2O(l)H3O+(aq)+F(aq)F(aq)+HF(aq)HF2(aq)

The equilibrium of the first equilibrium reaction, Ka1 is 1.1×103 .

The equilibrium of the first equilibrium reaction, Ka2 is 2.6×101 .

The equilibrium constant of first reaction is less than that of the second reaction, therefore, the amount of the reactant of first equilibrium is present in large amount and the amount of product of second equilibrium is present in large amount.

Thus, undissociated HF is present in large amount and this is utilized in second equilibrium to form HF2 . Therefore, the concentration of HF2 is maximum in the drinking water.

(b)

Interpretation Introduction

To determine: The equilibrium constant of the given reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 16.111AP

Solution:

The equilibrium constant of the given reaction is 2.86×10-4_ .

Explanation of Solution

Given

The equilibrium of hydrofluoric acid in water is given as,

HF(aq)+H2O(l)H3O+(aq)+F(aq)F(aq)+HF(aq)HF2(aq)

The equilibrium of the first equilibrium reaction, Ka1 is 1.1×103 .

The equilibrium of the first equilibrium reaction, Ka2 is 2.6×101 .

The overall reaction of the above two equilibriums is,

2HF(aq)+H2O(l)H3O+(aq)+HF2(aq)

The equilibrium constant of the overall reaction is the product of the equilibrium constant of the individual reaction.

Therefore, the overall equilibrium constant of the two equilibrium reaction is calculated as,

Ka(netreaction)=Ka1×Ka2

Substitute the value of the equilibrium constant of the individual reactions in the above formula,

Ka(netreaction)=1.1×103×2.6×101=2.86×104

Therefore, the equilibrium constant of the given reaction is 2.86×10-4_ .

(c)

Interpretation Introduction

To determine: The pH of the solution of HF and the equilibrium concentration of HF2

(c)

Expert Solution
Check Mark

Answer to Problem 16.111AP

Solution:

The pH of the solution of HF and the equilibrium concentration of HF2 is 2.6_ and 2.54×10-3M_ , respectively.

Explanation of Solution

Given

The concentration of HF is 0.150M .

The equilibrium reaction of HF is,

2HF(aq)+H2O(l)H3O+(aq)+HF2(aq)

The equilibrium constant, Ka , of the above reaction is 2.86×104 .

The concentration of hydronium ion and HF2 is assumed to be x .

The initial and the equilibrium concentrations of the species involved in the above equilibrium reaction is given as,

2HF(aq)+H2O(l)H3O+(aq)+HF2(aq)Initial(M):0.15000Change(M):x+x+xEquilibrium(M):0.150xxx

The equilibrium constant is given by the formula,

Ka=[H3O+]×[HF2][HF]2

Substitute the value of Ka , [H3O+] , [HF2] and [HF] in the above equation.

2.86×104=x×x(0.150x)2

Since, the value of Ka is very small, the value of x is neglected compared to 0.150 .

2.86×104=x×x(0.150)2x2=2.86×104×2.25×102=6.432×106x=2.54×103M

Therefore, the concentration of hydronium ion and HF2 at equilibrium is 2.54×10-3M_ .

The pH of the solution is calculated by the formula,

pH=log[H3O+]

Substitute the concentration of hydronium ion in the above equation.

pH=log(2.54×103M)=2.6

Thus, the pH of the solution of HF is 2.6_ .

Conclusion:

  1. a. The fluorine containing species that is more likely to be present in drinking water is HF2 .
  2. b. The equilibrium constant of the given reaction is 2.86×10-4_ .
  3. c. The pH of the solution of HF and the equilibrium concentration of HF2 is 2.6_ and 2.54×10-3M_ , respectively

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Chapter 16 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 16.4 - Prob. 11PECh. 16.4 - Prob. 12PECh. 16.5 - Prob. 13PECh. 16.6 - Prob. 14PECh. 16.8 - Prob. 15PECh. 16.8 - Prob. 16PECh. 16.8 - Prob. 17PECh. 16.8 - Prob. 18PECh. 16.8 - Prob. 19PECh. 16 - Prob. 16.1VPCh. 16 - Prob. 16.2VPCh. 16 - Prob. 16.3VPCh. 16 - Prob. 16.4VPCh. 16 - Prob. 16.5VPCh. 16 - Prob. 16.6VPCh. 16 - Prob. 16.7VPCh. 16 - Prob. 16.8VPCh. 16 - Prob. 16.9VPCh. 16 - Prob. 16.10VPCh. 16 - Prob. 16.11QPCh. 16 - Prob. 16.12QPCh. 16 - Prob. 16.13QPCh. 16 - Prob. 16.14QPCh. 16 - Prob. 16.15QPCh. 16 - Prob. 16.16QPCh. 16 - Prob. 16.17QPCh. 16 - Prob. 16.18QPCh. 16 - Prob. 16.19QPCh. 16 - Prob. 16.20QPCh. 16 - Prob. 16.21QPCh. 16 - Prob. 16.22QPCh. 16 - Prob. 16.23QPCh. 16 - Prob. 16.24QPCh. 16 - Prob. 16.25QPCh. 16 - Prob. 16.26QPCh. 16 - Prob. 16.27QPCh. 16 - Prob. 16.28QPCh. 16 - Prob. 16.29QPCh. 16 - Prob. 16.30QPCh. 16 - Prob. 16.31QPCh. 16 - Prob. 16.32QPCh. 16 - Prob. 16.33QPCh. 16 - Prob. 16.34QPCh. 16 - Prob. 16.35QPCh. 16 - Prob. 16.36QPCh. 16 - Prob. 16.37QPCh. 16 - Prob. 16.38QPCh. 16 - Prob. 16.39QPCh. 16 - Prob. 16.40QPCh. 16 - Prob. 16.41QPCh. 16 - Prob. 16.42QPCh. 16 - Prob. 16.43QPCh. 16 - Prob. 16.44QPCh. 16 - Prob. 16.45QPCh. 16 - Prob. 16.46QPCh. 16 - Prob. 16.47QPCh. 16 - Prob. 16.48QPCh. 16 - Prob. 16.49QPCh. 16 - Prob. 16.50QPCh. 16 - Prob. 16.51QPCh. 16 - Prob. 16.52QPCh. 16 - Prob. 16.53QPCh. 16 - Prob. 16.54QPCh. 16 - Prob. 16.55QPCh. 16 - Prob. 16.56QPCh. 16 - Prob. 16.57QPCh. 16 - Prob. 16.58QPCh. 16 - Prob. 16.59QPCh. 16 - Prob. 16.60QPCh. 16 - Prob. 16.61QPCh. 16 - Prob. 16.62QPCh. 16 - Prob. 16.63QPCh. 16 - Prob. 16.64QPCh. 16 - Prob. 16.65QPCh. 16 - Prob. 16.66QPCh. 16 - Prob. 16.67QPCh. 16 - Prob. 16.68QPCh. 16 - Prob. 16.69QPCh. 16 - Prob. 16.70QPCh. 16 - Prob. 16.71QPCh. 16 - Prob. 16.72QPCh. 16 - Prob. 16.73QPCh. 16 - Prob. 16.74QPCh. 16 - Prob. 16.75QPCh. 16 - Prob. 16.76QPCh. 16 - Prob. 16.77QPCh. 16 - Prob. 16.78QPCh. 16 - Prob. 16.79QPCh. 16 - Prob. 16.80QPCh. 16 - Prob. 16.81QPCh. 16 - Prob. 16.82QPCh. 16 - Prob. 16.83QPCh. 16 - Prob. 16.84QPCh. 16 - Prob. 16.85QPCh. 16 - Prob. 16.86QPCh. 16 - Prob. 16.87QPCh. 16 - Prob. 16.88QPCh. 16 - Prob. 16.89QPCh. 16 - Prob. 16.90QPCh. 16 - Prob. 16.91QPCh. 16 - Prob. 16.92QPCh. 16 - Prob. 16.93QPCh. 16 - Prob. 16.94QPCh. 16 - Prob. 16.95QPCh. 16 - Prob. 16.96QPCh. 16 - Prob. 16.97QPCh. 16 - Prob. 16.98QPCh. 16 - Prob. 16.99QPCh. 16 - Prob. 16.100QPCh. 16 - Prob. 16.101QPCh. 16 - Prob. 16.102QPCh. 16 - Prob. 16.103QPCh. 16 - Prob. 16.104QPCh. 16 - Prob. 16.105QPCh. 16 - Prob. 16.106QPCh. 16 - Prob. 16.107QPCh. 16 - Prob. 16.108QPCh. 16 - Prob. 16.109QPCh. 16 - Prob. 16.110QPCh. 16 - Prob. 16.111APCh. 16 - Prob. 16.112APCh. 16 - Prob. 16.113APCh. 16 - Prob. 16.114APCh. 16 - Prob. 16.115APCh. 16 - Prob. 16.116APCh. 16 - Prob. 16.117APCh. 16 - Prob. 16.118AP
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