Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 15, Problem 15.61QE

(a)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 0.20 MC6H5COOH solution has to be determined.

Concept Introduction:

A weak acid in water produces a hydrogen ion and conjugate base. When weak acid dissolves in water, some acid molecules transfer proton to water.

In solution of weak acid, the actual concentration of the acid molecules becomes less because partial dissociation of acid has occurred and lost protons to form hydrogen ions.

The reaction is as follows:

  HA(aq)+H2O(l)H3O+(aq)+A(aq)

The reaction is as follows:

  HA(aq)+H2O(l)H3O+(aq)+A(aq)

The expression for Ka is as follows:

  Ka=[H3O+][A][HA]

For value of Ka of different acids. (Refer to 15.6 in the book).

(a)

Expert Solution
Check Mark

Answer to Problem 15.61QE

The ICE table for 0.20 MC6H5COOH is as follows:

  EquationC6H5COOH+H2OH3O++C6H5COOInitial(M)0.20000Change(M)x +x+xEquilibrium(M)0.20xxx

The equation needed to solve for the concentration of the hydrogen ion for 0.20 MC6H5COOH solution is as follows:

  6.3×105=(x)(x)0.20x

Explanation of Solution

The chemical equation for ionization of C6H5COOH is as follows:

    C6H5COOH+H2OH3O++C6H5COO

The concentration of C6H5COOH is 0.20 M.

Also, C6H5COOH is ionized into H3O+ and C6H5COO. Therefore, concentration of H3O+ is equal to C6H5COO.

Consider the concentration of H3O+ and C6H5COO be x.

The ICE table for the above reaction is as follows:

  EquationC6H5COOH+H2OH3O++C6H5COOInitial(M)0.20000Change(M)x +x+xEquilibrium(M)0.20xxx

The expression for Ka for ionization of C6H5COOH  is as follows:

  Ka=[H3O+][C6H5COO][C6H5COOH]        (1)

Substitute 0.20 M for [C6H5COOH], x for [H3O+], x for [C6H5COO] and 6.3×105 for Ka in equation (1), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  6.3×105=(x)(x)0.20x

(b)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 1.50 MHCOOH solution has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 15.61QE

The ICE table for 1.50 MHCOOH is as follows:

  EquationHCOOH+H2OH3O++HCOOInitial(M)1.50000Change(M)x +x+xEquilibrium(M)1.50xxx

The equation needed to solve for the concentration of the hydrogen ion for 1.50 MHCOOH solution is as follows:

  1.8×104=(x)(x)1.50x

Explanation of Solution

The chemical equation for ionization of HCOOH is as follows:

    HCOOH+H2OH3O++HCOO

The concentration of HCOOH is 1.50 M.

Also, HCOOH is ionized into H3O+ and HCOO. Therefore, concentration of H3O+ is equal to HCOO.

Consider the concentration of H3O+ and HCOO be x.

The ICE table for the above reaction is as follows:

  EquationHCOOH+H2OH3O++HCOOInitial(M)1.50000Change(M)x +x+xEquilibrium(M)1.50xxx

The expression for Ka for ionization of HCOOH is as follows:

  Ka=[H3O+][HCOO][HCOOH]        (2)

Substitute 1.50 M for [HCOOH], x for [H3O+], x for [HCOO] and 1.8×104 for Ka in equation (2), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  1.8×104=(x)(x)1.50x

(c)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 0.0055 MHCN solution has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 15.61QE

The ICE table for 0.0055 MHCN is as follows:

  EquationHCN+H2OH3O++CNInitial(M)0.0055000Change(M)x +x+xEquilibrium(M)0.055xxx

The equation needed to solve for the concentration of the hydrogen ion for 0.0055 MHCN solution is as follows:

  6.2×1010=(x)(x)0.0055x

Explanation of Solution

The chemical equation for ionization of HCN is as follows:

    HCN+H2OH3O++CN

The concentration of HCN is 0.0055 M.

Also, HCN is ionized into H3O+ and CN. Therefore, concentration of H3O+ is equal to CN.

Consider the concentration of H3O+ and CN be x.

The ICE table for the above reaction is as follows:

  EquationHCN+H2OH3O++CNInitial(M)0.0055000Change(M)x +x+xEquilibrium(M)0.055xxx

The expression for Ka for ionization of HCN is as follows:

  Ka=[H3O+][CN][HCN]        (3)

Substitute 0.0055 M for [HCN], x for [H3O+], x for and CN and 6.2×105 for Ka in equation (3), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  6.2×1010=(x)(x)0.0055x

(d)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 0.075 MHNO2 solution has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 15.61QE

The ICE table for 0.075 MHNO2 is as follows:

  EquationHNO2+H2OH3O++NO2Initial(M)0.075000Change(M)x +x+xEquilibrium(M)0.075xxx

The equation needed to solve for the concentration of the hydrogen ion for 0.075 MHNO2 solution is as follows:

  5.6×104=(x)(x)0.075x

Explanation of Solution

The chemical equation for ionization of HNO2 is as follows:

    HNO2+H2OH3O++NO2

The concentration of HNO2 is 0.075 M.

Also, HNO2 is ionized into H3O+ and NO2. Therefore, concentration of H3O+ is equal to NO2.

Consider the concentration of H3O+ and NO2 be x.

The ICE table for the above reaction is as follows:

  EquationHNO2+H2OH3O++NO2Initial(M)0.075000Change(M)x +x+xEquilibrium(M)0.075xxx

The expression for Ka for ionization of HNO2 is as follows:

  Ka=[H3O+][NO2][HNO2]        (4)

Substitute 0.075 M for [HNO2], x for [H3O+], x for [NO2] and 5.6×104 for Ka in equation (4), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  5.6×104=(x)(x)0.075x

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Chapter 15 Solutions

Chemistry: Principles and Practice

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