   Chapter 14, Problem 72E

Chapter
Section
Textbook Problem

# A solution is made by adding 50.0 mL of 0.200 M acetic acid (K4 = 1.8 × 10–3) to 50.0 mL of 1.00 × 10–3 M HCl  a. Calculate the pH of the solution  b. Calculate the acetate ion concentration.

(b)

Interpretation Introduction

Interpretation:

A solution is made by adding 50.0 mL of 0.200M Acetic acid to 50mL of 1×10-3M HCl .

1. (a) pH of the solution is to be calculated.
2. (b) Acetate ion concentration 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

Explanation

To determine: Equilibrium concentration of Acetate ion

The moles of CH3COOH are 0.01mol_ and that of HCl are 5.0×10-5mol_ .

Given data

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.

Calculation of moles of Acetic acid and Hydrochloric acid

For CH3COOH ,

MolesofCH3COOH=0.050L×0.200M=0.01mol_

For HCl ,

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

Calculation of hydrogen ion forms HCl

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_ .

Calculation of initial concentration of Acetic acid

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_

(a)

Interpretation Introduction

Interpretation:

A solution is made by adding 50.0 mL of 0.200M Acetic acid to 50mL of 1×10-3M HCl .

1. (a) pH of the solution is to be calculated.
2. (b) Acetate ion concentration 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

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