General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 16, Problem 16.55QP

(a)

Interpretation Introduction

Interpretation:

The pH for the given solutions has to be calculated

Concept Information:

Strong base and weak base:

Strong base dissociates into its constituent ions fully. It produces more of hydroxide ions while dissolved in water.  Weak bases partially dissociates into its constituent ions.

According to Bronsted-Lowry, strong base is a good proton acceptor whereas weak base is a poor proton acceptor

Since, the ionization of a weak base is incomplete; it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base B is given by the below equation.

  B(aq)+H2O(l)HB+(aq)+OH-(aq)

The equilibrium expression for the ionization of weak base B will be,

  Kb=[HB+][OH-][B]

Where,

  Kb is base ionization constant,

  [OH] is concentration of hydroxide ion

  [HB+]  is concentration of conjugate acid

  [B] is concentration of the base

pOHdefinition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH-] concentration. pOH scale is analogous to pH scale.

  pOH=-log[OH-]

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 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.

To Calculate: The pH of the given solutions

To calculate the pH of 0.10 M NH3

(a)

Expert Solution
Check Mark

Answer to Problem 16.55QP

The pH of the given solution (a) is 11.11

Explanation of Solution

Record the given data

A 0.10-M solution of ammonia at 25oC.

From the concentration and Kb of the given ammonia, the hydroxide ion concentration can be found.  From hydroxide ion concentration, the pOH and finally pH can be calculated

Calculation of Kb

The equilibrium table for ammonia can be constructed as follows,

 NH3(aq)+H2O(l)NH4+(aq)+OH-(aq)
Initial (M)

0.10

x

0.10x

0.000.00
Change (M)+x+x
Equilibrium (M)xx

  Kb=[NH4+][OH-][NH3]The base ionization constant for ammonia is 1.8×10-51.8×10-5=x2(0.10x)Assuming (0.10-x)0.10, we have:1.8×10-5=x2(0.10)x=1.3×103M

Therefore, the concentration of hydroxide ion is 1.3×103M

Calculation of pOH:

The pOH can be calculated as follows,

  pOH=-log[OH-]=log(1.3×103)=2.89

Calculation of pH

The pH can be calculated using the following formula as follows,

  pH+pOH=14pH=14.00-2.89=11.11

Therefore, the pH of 0.10 M ammonia solution is 11.11

(b)

Interpretation Introduction

Interpretation:

The pH for the given solutions has to be calculated

Concept Information:

Strong base and weak base:

Strong base dissociates into its constituent ions fully. It produces more of hydroxide ions while dissolved in water.  Weak bases partially dissociates into its constituent ions.

According to Bronsted-Lowry, strong base is a good proton acceptor whereas weak base is a poor proton acceptor

Since, the ionization of a weak base is incomplete; it is treated in the same way as the ionization of a weak acid.

The ionization of a weak base B is given by the below equation.

  B(aq)+H2O(l)HB+(aq)+OH-(aq)

The equilibrium expression for the ionization of weak base B will be,

  Kb=[HB+][OH-][B]

Where,

  Kb is base ionization constant,

  [OH] is concentration of hydroxide ion

  [HB+]  is concentration of conjugate acid

  [B] is concentration of the base

pOHdefinition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH-] concentration. pOH scale is analogous to pH scale.

  pOH=-log[OH-]

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 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.

To Calculate: The pH of the given solutions

To calculate the pH of 0.050 M pyridine

(b)

Expert Solution
Check Mark

Answer to Problem 16.55QP

The pH of the given solution (b) is 8.96

Explanation of Solution

Record the given datas

A 0.050 M pyridine at 25oC.

From the concentration and Kb of the given pyridine, the hydroxide ion concentration can be found.  From hydroxide ion concentration, the pOH and finally pH can be calculated

Calculation of Kb

The equilibrium table for ammonia can be constructed as follows,

 C6H5N(aq)+H2O(l)C6H5NH+(aq)+OH-(aq)
Initial (M)

0.050

x

0.050x

0.000.00
Change (M)+x+x
Equilibrium (M)xx

  Kb=[C6H5NH+][OH-][C6H5N]The base ionization constant for ammonia is 1.7×10-91.7×10-9=x2(0.050x)Assuming (0.050-x)0.050, we have:1.8×10-5=x2(0.050)x=3×104M

Therefore, the concentration of hydroxide ion is 3×104M

Calculation of pOH:

The pOH can be calculated as follows,

  pOH=-log[OH-]=log(3×104)=3.52

Calculation of pH

The pH can be calculated using the following formula as follows,

  pH+pOH=14pH=14.00-3.52=10.48

Therefore, the pH of 0.050 M pyridine solution is 10.48

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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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