   # A solution is made by adding 50.0 mL of 0.200 M acetic acid ( K a = 1.8 × 10 −5 ) to 50.0 mL of 1.00 × 10 −3 M HCl. a. Calculate the pH of the solution. b. Calculate the acetate ion concentration. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 72E
Textbook Problem
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## A solution is made by adding 50.0 mL of 0.200 M acetic acid (Ka = 1.8 × 10−5) to 50.0 mL of 1.00 × 10−3 M HCl.a. Calculate the pH of the solution.b. Calculate the acetate ion concentration.

Interpretation Introduction

Interpretation: A solution prepared by mixing the given amount of CH3COOH and HCl is given. The pH and the acetate ion concentration of this solution is to be calculated.

Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

The number of moles of a solute is calculated by the formula,

Molesofsolute=Volume(L)×Molarity

To determine: The pH of the given solution.

### Explanation of Solution

Explanation

Given

Volume of CH3COOH is 50.0mL(0.050L).

Concentration of CH3COOH is 0.200M.

Volume of HCl is 50.0mL(0.050L).

Concentration of HCl is 1.00×103M.

The number of moles of a solute is calculated by the formula,

Molesofsolute=Volume(L)×Molarity

Substitute the value of volume and molarity of CH3COOH and HCl in the above expression.

For CH3COOH,

MolesofCH3COOH=0.050L×0.200M=0.01mol_

For HCl,

MolesofHCl=0.050L×(1.00×103)M=5.0×10-5mol_

The [H+] from HCl is 5.0×10-4M_.

The total volume of the solution is calculated by the formula,

Totalvolume=VolumeofCH3COOH+VolumeofHCl

Substitute the value of the volume of CH3COOH and the volume of HCl in the above expression.

Totalvolume=(0.050+0.050)L=0.1L

HCl is a strong acid. Therefore, it completely dissociates in water. Hence, the hydrogen ion concentration is equal to the concentration of HCl.

The concentration is calculated by the formula,

Concentration=NumberofmolesVolume(L)

Substitute the value of the number of moles of HCl and the volume in the above expression.

Concentration=5.0×105mol0.10L=5.0×10-4M_

Therefore, the [H+] from HCl is 5.0×10-4M_.

The initial [CH3COOH] is 0.1M_.

The concentration is calculated by the formula,

Concentration=NumberofmolesVolume(L)

Substitute the value of the number of moles of CH3COOH and the volume in the above expression.

Concentration=0.01mol0.10L=0.1M_

Therefore, the initial [CH3COOH] is 0.1M_

The equilibrium [CH3COO] is 1.1×10-3M_.

The change in concentration of CH3COOH is assumed to be x.

The ICE table for the stated reaction is,

CH3COOH(aq)H+(aq)+CH3COO(aq)Inititialconcentration0

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