Mastering Chemistry With Pearson Etext -- Standalone Access Card -- For General Chemistry: Principles And Modern Applications (11th Edition)
Mastering Chemistry With Pearson Etext -- Standalone Access Card -- For General Chemistry: Principles And Modern Applications (11th Edition)
11th Edition
ISBN: 9780133387803
Author: Ralph H. Petrucci; F. Geoffrey Herring; Jeffry D. Madura; Carey Bissonnette
Publisher: PEARSON
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Textbook Question
Chapter 17, Problem 1E

For a solution that e 0.275M CH 2 CH 2 COOH (propionic acid, K 8 = 1.3 × 10 9 ) and 0.0892 M HI, calculate (a) [HO-]; (b) [OH-] (c) CH2CH2COO; (d) [1-].

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

A solution consists propionic acid, CH3CH2COOH of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of [H3O+] is to be calculated. The value of equilibrium constant of acid is Ka= 1.3×105.

Concept introduction:

The concentration of any solution is calculated by the molarity of the solution. It is defined as the number of moles of solute in 1 L of the solution.

M=nV(L)

Here, n is number of moles of solute and V is volume of solution in L.

The dissociation constant of the weak acid is calculated by the following formula-

HAH++A

The expression can be written as follows:

Ka=[H+][A][HA]

Answer to Problem 1E

The concentration of [H3O+] = 0.0892 M.

Explanation of Solution

The complete dissociation of strong acid occurs in the solution, whereas weak acid is not able to dissociate completely.

From the given information,

Propionic acid, CH3CH2COOH or C3H6O2 is a weak acid.

The molarity of propionic acid [CH3CH2COOH] is 0.275 M.

Hydroiodic acid, HI = strong acid

The molarity of [HI] or [H3O+] ion = 0.0892 M

The chemical equation of this reaction is given below:

HC3H5O2(aq)+H2O(l)C3H5O2(aq)+H3O+(aq)

Now, the concentration will be calculated by using the following ICE table:

HC3H5O2 H2O C3H5O2 H3O+
Initial (I) 0.275 - 0 0.0892
Change (C) - x - + x + x
Equilibrium (E) 0.275- x - x 0.0892+ x

Given that-

Ka = 1.3×10-5

C3H5O2 = x

HC3H5O2 = 0.275-x

H3O+ = 0.0892+x

The equation for the Ka is as follows:

Ka=[C3H5O2-][H3O+][C3H6O2] …………… (1)

Put the given values in equation (1).

1.3×105=x(0.0892+x)0.275x

On calculation −

x=4.0×105M

Now, the concentration of H3O+

= 0.0892+x=0.0892+4.0×105

H3O+0.0892 M

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

A solution consists propionic acid, CH2CH2COOH of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of [OH] is to be calculated. The value of equilibrium constant of acid is Ka= 1.3×103.

Concept introduction:

The concentration of any solution is calculated by the molarity of the solution. It is defined as the number of moles of solute in 1 L of the solution.

M=nV(L)

Here, n is number of moles of solute and V is volume of solution in L.

The dissociation constant of the weak acid is calculated by the following formula:

HAH++A

The expression can be written as follows:

Ka=[H+][A][HA]

The relation between the dissociation constant of acid and the base is given by the following equation-

Ka=KwKb

Here, Kw is ionic product of water with value 1014.

Answer to Problem 1E

The concentration of [OH] = 1.1×1013M.

Explanation of Solution

The chemical equation of this reaction is given below:

HC3H5O2(aq)+H2O(l)C3H5O2-(aq)+H3O+(aq)

The concentration of [OH] is calculated by the following equation:

[OH]=Kw[H3O+] ……………… (1)

Given that-

[H3O+]=0.0892M

And,

Kw=1×1014

Put the above values in equation (1)

[OH]=1×10140.0892 M

[OH]=1.1×1013 M

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

A solution consists propionic acid, CH2CH2COOH of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of [CH2CH2COO] is to be calculated. The value of equilibrium constant of acid is Ka= 1.3×103.

Concept introduction:

The concentration of any solution is calculated by the molarity of the solution. It is defined as the number of moles of solute in 1 L of the solution.

M=nV(L)

Here, n is number of moles of solute and V is volume of solution in L.

The dissociation constant of the weak acid is calculated by the following formula:

HAH++A

The expression can be written as follows:

Ka=[H+][A][HA]

Answer to Problem 1E

The concentration of [CH2CH2COO] = 4.0×10-5 M.

Explanation of Solution

The chemical equation of this reaction is given below:

HC3H5O2(aq)+H2O(l)C3H5O2-(aq)+H3O+(aq)

Now, the concentration will be calculated by using the following ICE table:

HC3H5O2 H2O C3H5O2 H3O+
Initial (I) 0.275 - 0 0.0892
Change (C) -x - +x +x
Equilibrium (E) 0.275-x - x 0.0892+x

From the ICE table:

C3H5O2- or CH3CH2O2=x

Ka = 1.3×10-5

HC3H5O2 = 0.275-x

H3O+ = 0.0892+x

The equation for the Ka is given as

Ka=[C3H5O2-][H3O+][C3H6O2] …………… (1)

Put the given values in equation (1).

1.3×105=x(0.0892+x)0.275x

On calculation −

x=4.0×105M

Therefore, C3H5O2 or CH3CH2O2=x=4.0×105M

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

A solution consists propionic acid, CH2CH2COOH of molarity 0.275 M and HI of molarity 0.0892 M. The concentration of [I] is to be calculated. The value of equilibrium constant of acid is Ka= 1.3×103.

Concept introduction:

The dissociation constant of the weak acid is calculated by the following formula:

HAH++A

The expression can be written as follows:

Ka=[H+][A][HA]

The strong acid dissociates completely in comparison to the weak acid.

Answer to Problem 1E

The concentration of [HI] = 0.0892 M.

Explanation of Solution

The complete dissociation of strong acid occurs in the solution, whereas weak acid is not able to dissociate completely.

From the given question −

Hydroiodic acid, HI = strong acid

The molarity of [HI] or [H3O+] ion = 0.0892 M

As hydroiodic acid is strong acid it will dissociate completely in the solution. Hence, the initial concentration of [HI] = 0.0892 M.

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Chapter 17 Solutions

Mastering Chemistry With Pearson Etext -- Standalone Access Card -- For General Chemistry: Principles And Modern Applications (11th Edition)

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