   Chapter 16, Problem 16.87QP 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 1.24-g sample of benzoic acid was dissolved in water to give 50.0 mL or solution. This solution was titrated with 0.180 M NaOH. What was the pH of the solution when the equivalence point was reached?

Interpretation Introduction

Interpretation:

The pH at the equivalence point when the given amount of benzoic acid is titrated with 0.180 M NaOH has to be calculated.

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Explanation

Given data:

A1.24 g of benzoic acid is dissolved in water to get 50.0 mL of solution.

The concentration of NaOH = 0.180 M

Explanation:

Let us represent benzoic acid as HBen and benzoate anion as Ben-

The moles of benzoic acid is calculated as follows,

moles =massmolar massmol HBen =12.4 g HBen122.1 g HBen/mol HBen =0.01015 mol HBen

At the equivalence point, equal molar amounts of benzoic acid and NaOH reacts to give the solution.

volume of NaOH  =0.01015 mol NaOH0.180 mol NaOH/L =0.05641 L

Hence, the total volume is as follows,

Total volume = 0.05641 L + 0.0500 L HBen soln =0.10641 L

The concentration of benzoate ion [Ben-] at the equivalence point is,

[Ben-] =0.01015 mol Ben- from HBen0.10641 L =0.09538 M

Now, the hydrolysis of Ben- is considered.

The base ionization constant of Ben- can be calculated as follows,

The Ka of HBen = 6.3×105 ( HBen is the conjugate acid of Ben- )

Kb =KwKa =1.0×10146.3×105 =1

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