General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 16, Problem 16.17QP

(a)

Interpretation Introduction

Interpretation:

The pH of the given solutions has to be calculated

Concept Information:

pH definition:

The concentration of hydrogen ion is measured using pH scale. The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H+]

Relationship between pH and pOH

pOH is similar to pH.  The only difference is that in pOH the concentration of hydroxide ion is used as a scale while in pH, the concentration of hydronium ion is used.

The equation of equilibrium for autoionization of water is,

Kw=[H+][OH-]

The equilibrium expression for water at 25oC is,

[H+][OH-]= 1×10-14

Taking negative logarithm on both sides, we get

log([H+][OH-])= -log(1×10-14)(log[H+])+(-log[OH-])= 14)

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

pH+pOH=14,at25oC

As pOH and pH are opposite scale, the total of both has to be equal to 14.

The pH of the given solutions

(a)

Expert Solution
Check Mark

Answer to Problem 16.17QP

The pH of the given solution (a) is 3.00

Explanation of Solution

Calculation of hydrogen ion concentration:

HCl is a strong acid, so the concentration of hydrogen ion is also 0.0010M

Calculation of pH

The pH is calculated as follows,

pH=-log[H+]=-log(0.0010)=3.00

(b)

Interpretation Introduction

Interpretation:

The pH of the given solutions has to be calculated

Concept Information:

pH definition:

The concentration of hydrogen ion is measured using pH scale. The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

pH=-log[H+]

Relationship between pH and pOH

pOH is similar to pH.  The only difference is that in pOH the concentration of hydroxide ion is used as a scale while in pH, the concentration of hydronium ion is used.

The equation of equilibrium for autoionization of water is,

Kw=[H+][OH-]

The equilibrium expression for water at 25oC is,

[H+][OH-]= 1×10-14

Taking negative logarithm on both sides, we get

log([H+][OH-])= -log(1×10-14)(log[H+])+(-log[OH-])= 14)

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

pH+pOH=14,at25oC

As pOH and pH are opposite scale, the total of both has to be equal to 14.

(b)

Expert Solution
Check Mark

Answer to Problem 16.17QP

The pH of the given solution (b) is 13.89

Explanation of Solution

Calculation of hydrogen ion concentration:

KOH is an ionic compound and gets fully ionized in water.

The hydrogen ion concentration can be found as follows,

[pOH]=log[OH-]=log[0.76][pOH]=0.119

Calculation of pH

The pH is calculated as follows,

pH  +pOH = 14.00pH  =14.00 - pOH =14.00  -0.119  pH=13.88=13.89

(c)

Interpretation Introduction

Interpretation:

The pH of the given solutions has to be calculated

Concept Information:

pH definition:

The concentration of hydrogen ion is measured using pH scale. The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

pH=-log[H+]

Relationship between pH and pOH

pOH is similar to pH.  The only difference is that in pOH the concentration of hydroxide ion is used as a scale while in pH, the concentration of hydronium ion is used.

The equation of equilibrium for autoionization of water is,

Kw=[H+][OH-]

The equilibrium expression for water at 25oC is,

[H+][OH-]= 1×10-14

Taking negative logarithm on both sides, we get

log([H+][OH-])= -log(1×10-14)(log[H+])+(-log[OH-])= 14)

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

pH+pOH=14,at25oC

As pOH and pH are opposite scale, the total of both has to be equal to 14.

The pH of the given solutions

(c)

Expert Solution
Check Mark

Answer to Problem 16.17QP

The pH of the given solution (c) is 10.75

Explanation of Solution

Calculation of hydrogen ion concentration:

Ba(OH)2 is ionic and gets fully ionized in water.  The concentration of the hydroxide ion is 2.8×104

The hydrogen ion concentration can be found as follows,

[pOH]=log[OH-]=log[2×2.8×10-4]=log[5.6×10-4]=3.251[pOH]=3.251

Calculation of pH

The pH is calculated as follows,

pH  +pOH = 14.00pH  =14.00 - pOH =14.00  -3.25  pH=10.75

(d)

Interpretation Introduction

Interpretation:

The pH of the given solutions has to be calculated

Concept Information:

pH definition:

The concentration of hydrogen ion is measured using pH scale. The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

pH=-log[H+]

Relationship between pH and pOH

pOH is similar to pH.  The only difference is that in pOH the concentration of hydroxide ion is used as a scale while in pH, the concentration of hydronium ion is used.

The equation of equilibrium for autoionization of water is,

Kw=[H+][OH-]

The equilibrium expression for water at 25oC is,

[H+][OH-]= 1×10-14

Taking negative logarithm on both sides, we get

log([H+][OH-])= -log(1×10-14)(log[H+])+(-log[OH-])= 14)

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

pH+pOH=14,at25oC

As pOH and pH are opposite scale, the total of both has to be equal to 14.

The pH of the given solutions

(d)

Expert Solution
Check Mark

Answer to Problem 16.17QP

The pH of the given solution (d) is 3.28

Explanation of Solution

Calculation of hydrogen ion concentration:

Nitric acid is a strong acid, so the concentration of hydrogen ion is also 5.2×104M

Calculation of pH

The pH is calculated as follows,

pH=-log[H+]=-log(5.2×10-4)=3.28

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Chapter 16 Solutions

General Chemistry

Ch. 16.4 - Prob. 1RCCh. 16.5 - Prob. 1PECh. 16.5 - Prob. 2PECh. 16.5 - Prob. 1RCCh. 16.5 - Prob. 3PECh. 16.5 - Prob. 2RCCh. 16.6 - Prob. 1PECh. 16.6 - Prob. 1RCCh. 16.7 - Prob. 1RCCh. 16.8 - Prob. 1PECh. 16.8 - Rank the following acids from strongest to...Ch. 16.9 - Prob. 1PECh. 16.9 - Practice Exercise Predict whether the following...Ch. 16.9 - Prob. 1RCCh. 16.10 - Prob. 1RCCh. 16.11 - Prob. 1PECh. 16.11 - Prob. 1RCCh. 16 - Prob. 16.1QPCh. 16 - Prob. 16.2QPCh. 16 - Prob. 16.3QPCh. 16 - Prob. 16.4QPCh. 16 - Prob. 16.5QPCh. 16 - Prob. 16.6QPCh. 16 - Prob. 16.7QPCh. 16 - Prob. 16.8QPCh. 16 - Prob. 16.9QPCh. 16 - Prob. 16.10QPCh. 16 - Prob. 16.12QPCh. 16 - Prob. 16.13QPCh. 16 - Prob. 16.14QPCh. 16 - 16.15 Calculate the hydrogen ion concentration for...Ch. 16 - 16.16 Calculate the hydrogen ion concentration in...Ch. 16 - Prob. 16.17QPCh. 16 - Prob. 16.18QPCh. 16 - 16.19 Complete this table for a...Ch. 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 - 16.40 Which of the following solutions has the...Ch. 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 - 16.47 A 0.040 M solution of a monoprotic acid is...Ch. 16 - Prob. 16.48QPCh. 16 - Prob. 16.49QPCh. 16 - 16.50 Write all the species (except water) that...Ch. 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 - 16.57 What is the original molarity of a solution...Ch. 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 - 16.93 Most of the hydrides of Group 1A and Group...Ch. 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 - 16.100 Hydrocyanic acid (HCN) is a weak acid and a...Ch. 16 - Prob. 16.101QPCh. 16 - Prob. 16.102QPCh. 16 - Prob. 16.103QPCh. 16 - Prob. 16.104QPCh. 16 - 16.105 You are given two beakers containing...Ch. 16 - Prob. 16.106QPCh. 16 - Prob. 16.107QPCh. 16 - Prob. 16.108QPCh. 16 - Prob. 16.109QPCh. 16 - Prob. 16.110QPCh. 16 - Prob. 16.111QPCh. 16 - Prob. 16.112QPCh. 16 - Prob. 16.113QPCh. 16 - Prob. 16.114QPCh. 16 - Prob. 16.115QPCh. 16 - Prob. 16.116QPCh. 16 - Prob. 16.117QPCh. 16 - Prob. 16.118QPCh. 16 - Prob. 16.119QPCh. 16 - Prob. 16.120QPCh. 16 - Prob. 16.121SPCh. 16 - Prob. 16.122SPCh. 16 - Prob. 16.123SPCh. 16 - Prob. 16.124SPCh. 16 - Prob. 16.125SPCh. 16 - Prob. 16.126SPCh. 16 - Prob. 16.127SPCh. 16 - Prob. 16.128SPCh. 16 - Prob. 16.129SPCh. 16 - 16.130 Use the data in Appendix 2 to calculate the...
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