General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 16, Problem 16.94QP
Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Information:

Acid ionization constant Ka:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid HA can be given as follows,

  HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

  Ka=[H+][A-][HA]

Where,

  Ka is acid ionization constant,

  [H+]  is concentration of hydrogen ion

  [A-]  is concentration of acid anion

  [HA] is concentration of the acid

Autoionization of water:

The equation of equilibrium for autoionization of water is,

  H2OH++OH-

  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.

Therefore,

  Kw=[H+][OH-]=1×10-14

To Calculate: The equilibrium constant for the given reaction

Expert Solution & Answer
Check Mark

Answer to Problem 16.94QP

  • The equilibrium constant for the given reaction is 1.7×1010

Explanation of Solution

Record the given datas

  CH3COOH(aq)+NO2-    CH3COO-(aq)+HNO2(aq)

From the given data, we know that the equilibrium reaction between acetic acid and NO2- ion in aqueous medium gives CH3COO- ions and nitrous acid as products.

Splitting of the given equilibrium reaction

Write the given equilibrium reaction into two equilibria as follows,

  CH3COOH(aq)     CH3COO-(aq)+H+(aq)H+(aq)+ NO2-(aq)    HNO2(aq)

Calculation of Ka for CH3COOH(aq)   CH3COO-(aq)+H+(aq)

The acid ionization constant Ka for CH3COOH(aq)   CH3COO-(aq)+H+(aq) can be calculated as follows,

  Ka=[H+][CH3COO-][CH3COOH]

The acid ionization constant for formic acid is 1.8×105

Therefore,

  Ka=[H+][CH3COO-][CH3COOH]1.8×105=[H+][CH3COO-][CH3COOH]

Calculation of K'a for H+(aq)+ NO2-(aq)   HNO2(aq)

The acid ionization constant K'a for H+(aq)+ NO2-(aq)   HNO2(aq) can be calculated as follows,

  K'a=[HNO2][H+][NO2-]

The acid ionization constant for nitrous acid is 4.5×104

Therefore,

  K'a=[HNO2][H+][NO2-]K'a=1Ka(HNO2)=14.5×104=2.2×103

Equilibrium constant for the given reaction:

Adding up of two equilibria reaction in step 2, will result in the given equilibrium reaction.

  CH3COOH(aq)    CH3COO-(aq)+H+(aq)H+(aq)+NO2-(aq)   HNO2(aq)On adding up the above two reactions, we getCH3COOH(aq)+NO2-  CH3COO-(aq)+HNO2(aq)

The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions,

  K=[CH3COO-][HNO2][CH3COOH][NO2-]=Ka×K'a=(1.8×105)(2.2×103)=4.0×102

Therefore, the equilibrium constant for the given reaction is 4.0×102

Conclusion

The equilibrium constant for the given reaction was calculated.

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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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