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 15.62QE

(a)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 1.25 MHOCl 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 15.62QE

The ICE table for 1.25 MHOCl is as follows:

  EquationHOCl+H2OH3O++OClInitial(M)1.25000Change(M)x +x+xEquilibrium(M)1.25xxx

The equation needed to solve for the concentration of the hydrogen ion for 1.25 MHOCl solution is as follows:

  4.0×108=(x)(x)1.25x

Explanation of Solution

The chemical equation for ionization of HOCl is as follows:

    HOCl+H2OH3O++OCl

The concentration of HOCl is 1.25 M.

Also, HOCl is ionized into H3O+ and OCl. Therefore concentration of H3O+ is equal to OCl.

Let us assume the concentration of H3O+ and OCl be x.

The ICE table for the above reaction is as follows:

  EquationHOCl+H2OH3O++OClInitial(M)1.25000Change(M)x +x+xEquilibrium(M)1.25xxx

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

  Ka=[H3O+][OCl][HOCl]        (1)

Substitute 1.25 M for [HOCl], x for [H3O+], x for [OCl] and 4.0×108 for Ka in equation (1), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  4.0×108=(x)(x)1.25x

(b)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 0.80 MHF solution has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 15.62QE

The ICE table for 0.80 MHF is as follows:

  EquationHF+H2OH3O++FInitial(M)0.80000Change(M)x +x+xEquilibrium(M)0.80xxx

The equation needed to solve for the concentration of the hydrogen ion for 0.80 MHF solution is as follows:

  6.3×104=(x)(x)0.80x

Explanation of Solution

The chemical equation for the ionization of HF is as follows:

    HF+H2OH3O++F

The concentration of HF is 0.80 M.

Also, HF is ionized into H3O+ and F. Therefore concentration of H3O+ is equal to F.

Let us assume the concentration of H3O+ and F be x.

The ICE table for the above reaction is as follows:

  EquationHF+H2OH3O++FInitial(M)0.80000Change(M)x +x+xEquilibrium(M)0.80xxx

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

  Ka=[H3O+][F][HF]        (2)

Substitute 0.80 M for [HF], x for [H3O+], x for [F] and 6.3×104 for Ka in equation (2), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  6.3×104=(x)(x)0.80x

(c)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 0.14 MCH3COOH solution has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 15.62QE

The ICE table for 0.14 MCH3COOH is as follows:

  EquationCH3COOH+H2OH3O++CH3COOInitial(M)0.14000Change(M)x +x+xEquilibrium(M)0.14xxx

The equation needed to solve for the concentration of the hydrogen ion for 0.14 MCH3COOH solution is as follows:

  1.8×105=(x)(x)0.14x

Explanation of Solution

The chemical equation for ionization of CH3COOH is as follows:

    CH3COOH+H2OH3O++CH3COO

The concentration of CH3COOH is 0.14 M.

Also, CH3COOH is ionized into H3O+ and CH3COO. Therefore, concentration of H3O+ is equal to CH3COO.

Let us assume the concentration of H3O+ and CH3COO be x.

The ICE table for the above reaction is as follows:

  EquationCH3COOH+H2OH3O++CH3COOInitial(M)0.14000Change(M)x +x+xEquilibrium(M)0.14xxx

The expression for Ka for ionization of CH3COOH is as follows:

  Ka=[H3O+][CH3COO][CH3COOH]        (3)

Substitute 0.14 M for [CH3COOH], x for [H3O+], x for [CH3COO] and 1.8×105 for Ka in equation (3), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  1.8×105=(x)(x)0.14x

(d)

Interpretation Introduction

Interpretation:

The ICE table and the equation needed to solve for the concentration of the hydrogen ion for 0.25 MHCOOH solution has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 15.62QE

The ICE table for 0.25 MHCOOH is as follows:

  EquationHCOOH+H2OH3O++HCOOInitial(M)0.25000Change(M)x +x+xEquilibrium(M)0.25xxx

The equation needed to solve for the concentration of the hydrogen ion for 0.25 MHCOOH solution is as follows:

  1.8×104=(x)(x)0.25x

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.

Let us assume the concentration of H3O+ and HCOO be x.

The ICE table for the above reaction is as follows:

  EquationHCOOH+H2OH3O++HCOOInitial(M)0.25000Change(M)x +x+xEquilibrium(M)0.25xxx

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

  Ka=[H3O+][HCOO][HCOOH]        (4)

Substitute 0.25 M for [HCOOH], x for [H3O+], x for [HCOO] and 1.8×104 for Ka in equation (4), the equation needed to solve for the concentration of the hydrogen ion is as follows:

  1.8×104=(x)(x)0.25x

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Chapter 15 Solutions

Chemistry: Principles and Practice

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