   # A buffer solution was prepared by adding 4.95 g of sodium acetate, NaCH 3 CO 2 , to 2.50 × 10 2 mL of 0.150 M acetic acid, CH 3 CO 2 H. (a) What is the pH of the buffer? (b) What is the pH of 1.00 × 10 2 mL of the buffer solution if you add 82 mg of NaOH to the solution? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 23PS
Textbook Problem
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## A buffer solution was prepared by adding 4.95 g of sodium acetate, NaCH3CO2, to 2.50 × 102 mL of 0.150 M acetic acid, CH3CO2H. (a) What is the pH of the buffer? (b) What is the pH of 1.00 × 102 mL of the buffer solution if you add 82 mg of NaOH to the solution?

(a)

Interpretation Introduction

Interpretation:

For the buffer solution prepared by mixing 4.95 g of sodium acetate in 2.50×102mL of 0.150 M CH3COOH . The pH of the buffer solution has to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation relates pH of a buffer with pKa of acid, concentration of conjugate base and concentration of acid. The expression is written as,

pH=pKa+log[conjugatebase][acid] (1)

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

### Explanation of Solution

The calculation of pH is done by using Henderson-Hasselbalch equation.

Given:

Refer to table no. 16.2 in the textbook for the value of Ka.

The value of Ka for acetic acid is 1.8×105.

Negative logarithm of the Ka value gives the pKa value of the acid.

pKa=log(Ka)=log(1.8×105)=4.74

Therefore, pKa value for the acetic acid is 4.74.

The concentration of acetic acid is 0.150 M.

The amount of sodium acetate added to 2.50×102 mL acetic acid is 4.95 g.

Molar mass of sodium acetate is 82.03 gmol1.

The volume of acetic acid solution is 2.50×102 mL

Conversion of 2.50×102 mL into L

(2.50×102 mL)(1 L1000 mL)=0.250 L

Therefore, volume of the solution is 0.250 L.

The concentration of sodium acetate is calculated as follows;

Molarity=Number of molesvolume of solvent (2)

Number of moles is calculated by using expression.

Number of moles=weightmolar mass

Substitute 4.95 g for weight and 82.03 gmol1 for molarmass for sodium acetate

(b)

Interpretation Introduction

Interpretation:

For the buffer solution prepared by mixing 4.95 g of sodium acetate in 2.50×102mL of 0.150 M CH3COOH . pH of the solution when 82 mg of NaOH is added to 1.0×102 mL of buffer solution has to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation relates pH of a buffer with pKa of acid, concentration of conjugate base and concentration of acid. The expression is written as,

pH=pKa+log[conjugatebase][acid] (1)

This equation shows that pH of buffer solution is controlled by two major factors. First, Strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

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