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 15, Problem 15.31QP

(a)

Interpretation Introduction

Interpretation: The values of pH and pOH with the given [H3O+] or [OH] values are to be calculated. Also, whether the solutions are acidic, basic or neutral is to be indicated.

Concept introduction: The potential of H+ ions is called as pH and the potential of OH ions is called as pOH . The value of pH is less than 7 for acidic solutions and greater than 7 for basic solutions. For neutral solution the value of pH is 7 .

To determine: The values of pH and pOH with the given [H3O+] value and if the solution is acidic, basic or neutral.

(a)

Expert Solution
Check Mark

Answer to Problem 15.31QP

Solution

The pH of the solution is 2.275_ .

The pOH of the solution is 11.725_ .

The given solution is acidic.

Explanation of Solution

Explanation

Given

The value of [H3O+] is 5.3×103M .

The auto-dissociation of pure water is shown by the reaction,

2H2OH3O++OH

Concentrations of [H3O+] and [OH] in the solution are related according to the following relation,

[H3O+]×[OH]=Kw=1014

The value of pH is calculated by using the expression,

pH=log[H3O+]

The value of pOH is calculated by using the expression,

pOH=log[OH]

Hence, pH and pOH of the solution are related with each other as,

[H3O+]×[OH]=1014log([H3O+]×[OH])=log(1014)(log[H3O+])+(log[OH])=14pH+pOH=14

Now, the pH is calculated by using the expression,

pH=log[H3O+]

Substitute the value of [H3O+] in the above expression to calculate pH of the solution.

pH=log(5.3×103)=2.275_

Hence, pH of the solution is 2.275_ .

The value of pOH is calculated by using the expression,

pOH=14pH

Substitute the value of pH to calculate pOH of the solution in the above expression.

pOH=142.275=11.725_

Hence, pOH of the solution is 11.725_ .

The pH of the solution is less than 7 , thus it is an acidic solution.

(b)

Interpretation Introduction

To determine: The values of pH and pOH with the given [H3O+] value and if the solution is acidic, basic or neutral.

(b)

Expert Solution
Check Mark

Answer to Problem 15.31QP

Solution

The pH of the solution is 8.42_ .

The pOH of the solution is 5.58_ .

The given solution is acidic.

Explanation of Solution

Explanation

Given

The value of [H3O+] is 3.8×109M .

The pH is calculated by using the expression,

pH=log[H3O+]

Substitute the value of [H3O+] in the above expression to calculate pH of the solution.

pH=log(3.8×109)=8.42_

Hence, pH of the solution is 8.42_ .

The value of pOH is calculated by using the expression,

pOH=14pH

Substitute the value of pH to calculate pOH of the solution in the above expression.

pOH=148.42=5.58_

Hence, pOH of the solution is 5.58_ .

The pH of the solution is less than 7 , thus it is an acidic solution.

(c)

Interpretation Introduction

To determine: The values of pH and pOH with the given [H3O+] value and if the solution is acidic, basic or neutral.

(c)

Expert Solution
Check Mark

Answer to Problem 15.31QP

Solution

The pH of the solution is 5.14_ .

The pOH of the solution is 8.86_ .

The given solution is basic.

Explanation of Solution

Explanation

Given

The value of [H3O+] is 7.2×106M .

The pH is calculated by using the expression,

pH=log[H3O+]

Substitute the value of [H3O+] in the above expression to calculate pH of the solution.

pH=log(7.2×106)=5.14_

Hence, pH of the solution is 5.14_ .

The value of pOH is calculated by using the expression,

pOH=14pH

Substitute the value of pH to calculate pOH of the solution in the above expression.

pOH=145.14=8.86_

Hence, pOH of the solution is 8.86_ .

The pH of the solution is greater than 7 , thus it is a basic solution.

(d)

Interpretation Introduction

To determine: The values of pH and pOH with the given [OH] value and if the solution is acidic, basic or neutral.

(d)

Expert Solution
Check Mark

Answer to Problem 15.31QP

Solution

The pH of the solution is 0_ .

The pOH of the solution is 14_ .

The given solution is acidic.

Explanation of Solution

Explanation

Given

The value of [OH] is 1.0×1014M .

The pOH is calculated by using the expression,

pOH=log[OH]

Substitute the value of [OH] in the above expression to calculate pOH of the solution.

pOH=log(1×1014)=14_

Hence, pOH of the solution is 14_ .

The value of pH is calculated by using the expression,

pH=14pOH

Substitute the value of pOH to calculate pH of the solution in the above expression.

pH=1414=0_

Hence, pH of the solution is 0_ .

The pH of the solution is less than 7 , thus it is an acidic solution.

Conclusion

  1. a. The pH of the solution is 2.275_ and the pOH of the solution is 11.725_ . The given solution is acidic.
  2. b. The pH of the solution is 8.42_ and the pOH of the solution is 5.58_ . The given solution is acidic.
  3. c. The pH of the solution is 5.14_ and the pOH of the solution is 8.86_ . The given solution is basic.
  4. d. The pH of the solution is 0_ and the pOH of the solution is 14_ . The given solution is acidic.

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Chapter 15 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 15.5 - Prob. 11PECh. 15.5 - Prob. 12PECh. 15.6 - Prob. 13PECh. 15.6 - Prob. 14PECh. 15.7 - Prob. 15PECh. 15.8 - Prob. 16PECh. 15.8 - Prob. 17PECh. 15.8 - Prob. 18PECh. 15 - Prob. 15.1VPCh. 15 - Prob. 15.2VPCh. 15 - Prob. 15.3VPCh. 15 - Prob. 15.4VPCh. 15 - Prob. 15.5VPCh. 15 - Prob. 15.6VPCh. 15 - Prob. 15.7VPCh. 15 - Prob. 15.8VPCh. 15 - Prob. 15.9VPCh. 15 - Prob. 15.10VPCh. 15 - Prob. 15.11QPCh. 15 - Prob. 15.12QPCh. 15 - Prob. 15.13QPCh. 15 - Prob. 15.14QPCh. 15 - Prob. 15.15QPCh. 15 - Prob. 15.16QPCh. 15 - Prob. 15.17QPCh. 15 - Prob. 15.18QPCh. 15 - Prob. 15.19QPCh. 15 - Prob. 15.20QPCh. 15 - Prob. 15.21QPCh. 15 - Prob. 15.22QPCh. 15 - Prob. 15.23QPCh. 15 - Prob. 15.24QPCh. 15 - Prob. 15.25QPCh. 15 - Prob. 15.26QPCh. 15 - Prob. 15.27QPCh. 15 - Prob. 15.28QPCh. 15 - Prob. 15.29QPCh. 15 - Prob. 15.30QPCh. 15 - Prob. 15.31QPCh. 15 - Prob. 15.32QPCh. 15 - Prob. 15.33QPCh. 15 - Prob. 15.34QPCh. 15 - Prob. 15.35QPCh. 15 - Prob. 15.36QPCh. 15 - Prob. 15.37QPCh. 15 - Prob. 15.38QPCh. 15 - Prob. 15.39QPCh. 15 - Prob. 15.40QPCh. 15 - Prob. 15.41QPCh. 15 - Prob. 15.42QPCh. 15 - Prob. 15.43QPCh. 15 - Prob. 15.44QPCh. 15 - Prob. 15.45QPCh. 15 - Prob. 15.46QPCh. 15 - Prob. 15.47QPCh. 15 - Prob. 15.48QPCh. 15 - Prob. 15.49QPCh. 15 - Prob. 15.50QPCh. 15 - Prob. 15.51QPCh. 15 - Prob. 15.52QPCh. 15 - Prob. 15.53QPCh. 15 - Prob. 15.54QPCh. 15 - Prob. 15.55QPCh. 15 - Prob. 15.56QPCh. 15 - Prob. 15.57QPCh. 15 - Prob. 15.58QPCh. 15 - Prob. 15.59QPCh. 15 - Prob. 15.60QPCh. 15 - Prob. 15.61QPCh. 15 - Prob. 15.62QPCh. 15 - Prob. 15.63QPCh. 15 - Prob. 15.64QPCh. 15 - Prob. 15.65QPCh. 15 - Prob. 15.66QPCh. 15 - Prob. 15.67QPCh. 15 - Prob. 15.68QPCh. 15 - Prob. 15.69QPCh. 15 - Prob. 15.70QPCh. 15 - Prob. 15.71QPCh. 15 - Prob. 15.72QPCh. 15 - Prob. 15.73QPCh. 15 - Prob. 15.74QPCh. 15 - Prob. 15.75QPCh. 15 - Prob. 15.76QPCh. 15 - Prob. 15.77QPCh. 15 - Prob. 15.78QPCh. 15 - Prob. 15.79QPCh. 15 - Prob. 15.80QPCh. 15 - Prob. 15.81QPCh. 15 - Prob. 15.82QPCh. 15 - Prob. 15.83QPCh. 15 - Prob. 15.84QPCh. 15 - Prob. 15.85QPCh. 15 - Prob. 15.86QPCh. 15 - Prob. 15.87APCh. 15 - Prob. 15.88APCh. 15 - Prob. 15.89APCh. 15 - Prob. 15.90APCh. 15 - Prob. 15.91APCh. 15 - Prob. 15.92APCh. 15 - Prob. 15.93APCh. 15 - Prob. 15.94APCh. 15 - Prob. 15.95APCh. 15 - Prob. 15.96APCh. 15 - Prob. 15.97APCh. 15 - Prob. 15.98APCh. 15 - Prob. 15.99APCh. 15 - Prob. 15.100AP
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