Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 18, Problem 73E
Interpretation Introduction

Interpretation:

The pH of 0.12MHC2H3O2 is to be calculated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer
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Answer to Problem 73E

The pH of 0.12MHC2H3O2 is 5.43.

Explanation of Solution

The formula to calculate the number of moles of NaC2H3O2 is given below.

NumberofmolesofNaC2H3O2=MassofNaC2H3O2MolarmassofNaC2H3O2…(1)

The mass of NaC2H3O2 is 24.0g.

The molar mass of NaC2H3O2 is 82.03gmol1.

Substitute the mass and molar mass of NaC2H3O2 in equation (1).

NumberofmolesofNaC2H3O2=24.0g82.03gmol1=0.293mol

The volume of the solution is 5.00×102mL.

The relation between L and mL is given below.

1L=1000mL

The probable conversion factors are given below.

1L1000mLand1000mL1L

The conversion factor to determine L from mL is given below.

1L1000mL

Therefore, the volume in liters is calculated below.

Volume=5.00×102mL×1L1000mL=0.50L

The molarity of a solution is calculated by the formula given below.

Molarity=NumberofmolesVolumeofsolutioninliters…(2)

The number of moles of NaC2H3O2 is 0.293mol.

The volume of solution is 0.50L.

Substitute the number of moles and volume of solution in equation (2).

Molarity=0.293mol0.50L=0.586mol/L

The relation between M and mol/L as shown below.

1M=1mol/L

The probable conversion factors are given below.

1M1mol/Land1mol/L1M

The conversion factor to determine M from mol/L is given below.

1M1mol/L

So, 0.586mol/L can be written as shown below.

Molarity=0.586mol/L×1M1mol/L=0.586M

The equation for the dissociation of HC2H3O2 is shown below.

HC2H3O2(aq)C2H3O2(aq)+H+(aq)

The dissociation constant, Ka for the above reaction is given below.

Ka=[C2H3O2][H+][HC2H3O2]…(3)

The concentration of HC2H3O2, [HC2H3O2] is 0.12.

The concentration of C2H3O2, [C2H3O2] is 0.586M.

The value of Ka is 1.8×105.

Substitute the value of Ka, [HC2H3O2] and [C2H3O2] in equation (3).

1.8×105=0.586M×[H+]0.12M[H+]=0.12M×1.8×105M0.586M=3.7×106M

The formula to the pH is given below.

pH=log[H+]…(4)

Substitute the value of [H+] in equation (4).

pH=log[H+]pH=log(3.7×106M)=5.43

Therefore, the pH of 0.12MHC2H3O2 is 5.43.

Conclusion

The pH of 0.12MHC2H3O2 is 5.43.

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Chapter 18 Solutions

Introductory Chemistry: An Active Learning Approach

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