   Chapter 16, Problem 16.120QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

a Draw a pH titration curve that represents the titration of 25.0 mL of 0.15 M propionic acid. CH3CH2COOH, by the addition of 0.15 M KOH from a buret. Label the axes and put a scale on each axis. Show where the equivalence point and the buffer region are on the titration curve. You should do calculations for the 0%, 50%, 60%, and 100% titration points. b Is the solution neutral, acidic, or basic at the equivalence point? Why?

(a)

Interpretation Introduction

Interpretation:

For titration of 25.0 mL of 0.15 M propionic acid, CH3CH2COOH by the addition of 0.15 M KOH ,

A pH titration curve showing the equivalence point and buffer region has to be drawn

1. (a) The pH of the titration points for the 0%, 50%, 60% and 100% has to be calculated
2. (b) Whether the solution at the equivalence point is neutral, acidic or basic has to be explained

Concept Introduction:

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Relationship between pH and pOH:

pH + pOH = 14

Explanation

To Draw: A pH titration curve showing the equivalence point and buffer region

The pH titration curve for the titration of 0.15 M propionic acid  with 0.15 M KOH showing the equivalence point is given in the below Figure 1,

Figure 1

(a)

To Calculate: The pH of the titration points for the 0%, 50 %, 60% and 100%

Given data:

Titration of 0.15 M propionic acid with 0.15 M KOH

pH at the 0% titration point:

Construct an equilibrium table with x as unknown concentration

Consider propionic acid as HA and propionate anion as A-

 HA     +   H2O ⇌   H3O+     +     A− Initial (M) 0.15 −x 0.15-x 0.00 0.00 Change (M) +x +x Equilibrium (M) x x

Substitute equilibrium concentrations into the equilibrium-constant equation.

The Ka for propionic acid is 1.3×105

Assume x is negligible compared to 0.15 M

Ka =[H3O+][A][HA] =(x)2(0.15x)1.3×105 (x)20.15Rearrange and solve for x      x =(0.15)(1.3×105) =1.40×103 M

Therefore, the concentration of hydronium ion [H3O+] is 1.40×103 M

In the end, pH is calculated as follows,

pH =log[H3O+] =log(1.40×103) =2.85

Therefore, the pH at the 0% titration point is 2.85

pH at the 50% titration point:

[H3O+] =[A]Therefore,Ka=[H3O+][A][HA][H3O+=1.3×105 M

The pH is calculated as follows,

pH =log[H3O+] =log(1.3×105) =4.89

Therefore, the pH at the 50% titration point is 4.89

pH at the 60% titration point:

For convenience, express the concentrations as percents.

[HA]40%[A]60%

Substitute the concentrations into the equilibrium expression.

Ka =[H3O+][A][HA] =(60%)(x)(40%)1

(b)

Interpretation Introduction

Interpretation:

For titration of 25.0 mL of 0.15 M propionic acid, CH3CH2COOH by the addition of 0.15 M KOH ,

A pH titration curve showing the equivalence point and buffer region has to be drawn

1. (a) The pH of the titration points for the 0%, 50%, 60% and 100% has to be calculated
2. (b) Whether the solution at the equivalence point is neutral, acidic or basic has to be explained

Concept Introduction:

Equivalence point:

The equivalence point in titration is the point where the amount of standard titrant solution (in moles) and the unknown concentration analyte solution (in moles) becomes equal.

In other words, the equivalence point is the point obtained in a titration once a stoichiometric amount of reactant has been added.

Relationship between pH and pOH:

pH + pOH = 14

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