   # A buffer solution is prepared by dissolving 1.50 g each of benzoic acid, C 6 H 5 CO 2 H, and sodium benzoate, NaC 6 H 5 CO 2 , in 150.0 mL of solution. (a) What is the pH of this buffer solution? (b) Which buffer component must be added, and in what quantity, to change the pH to 4.00? (c) What quantity of 2.0 M NaOH or 2.0 M HCl must be added to the buffer to change the pH to 4.00? ### 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 89GQ
Textbook Problem
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## A buffer solution is prepared by dissolving 1.50 g each of benzoic acid, C6H5CO2H, and sodium benzoate, NaC6H5CO2, in 150.0 mL of solution.(a) What is the pH of this buffer solution?(b) Which buffer component must be added, and in what quantity, to change the pH to 4.00?(c) What quantity of 2.0 M NaOH or 2.0 M HCl must be added to the buffer to change the pH to 4.00?

(a)

Interpretation Introduction

Interpretation:

A buffer solution contains 1.5 g of each t and C6H5COONa in 150 mL solution.

pH of the 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 is strength of the acid which 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 benzoic acid is 6.3×105.

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

pKa=log(Ka)=log(6.3×105)=4.20

Therefore, pKa value for the benzoic acid is 4.20.

The 1.5g of benzoic acid (C6H5COOH) is dissolved in 150 mL  of H2O.

Molar mass of benzoic acid is 122.1gmol1.

The 1.5g of sodium benzoate (C6H5COONa) is dissolved in 150 mL of H2O.

Molar mass of benzoic acid is 144.1gmol1.

The concentration of benzoic acid is calculated as follows;

Molarity=Numberofmoles1L of solvent (2)

Number of moles is calculated by using expression.

Numberofmoles=weightmolarmass

Substitute 1.5g for weight and 122.1gmol1 for molarmass for benzoic acid.

Numberofmoles=1.5g122.1gmol1=0.01228 mol

Similarly for sodium benzoate, substitute 1.5g for weight and 144.1gmol1 for molarmass

(b)

Interpretation Introduction

Interpretation:

A buffer solution contains 1.5 g of each t and C6H5COONa in 150 mL solution. Find the component and its amount which is added to get the pH of the buffer solution.

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 is strength of the acid which can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

(c)

Interpretation Introduction

Interpretation:

A buffer solution contains 1.5 g of each t and C6H5COONa in 150 mL solution. The quantity of 2.0 M NaOH or 2.0 M HCl is to be calculated on which addition pH value change to 4.00.

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 is strength of the acid which can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

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