   # A buffer is made using 45.0 mL of 0.750 M HC 3 H 5 O 2 (K a = 1.3 × 10 −5 ) and 55.0 mL of 0.700 M NaC 3 H 5 O 2 . What volume of 0.10 M NaOH must be added to change the pH of the original buffer solution by 2.5%? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 109CP
Textbook Problem
1 views

## A buffer is made using 45.0 mL of 0.750 M HC3H5O2 (Ka = 1.3 × 10−5) and 55.0 mL of 0.700 M NaC3H5O2. What volume of 0.10 M NaOH must be added to change the pH of the original buffer solution by 2.5%?

Interpretation Introduction

Interpretation: The addition of NaOH to a buffer solution of HC3H5O2 and NaC3H5O2 and the change in pH is given. The volume of NaOH added to change the pH of original buffer solution by 2.5% is to be calculated.

Concept introduction: The solution of weak acid and their conjugate base which upon addition of an acid or base resists the change in pH is called buffer solution. The pH of the buffer solution remains constant.

### Explanation of Solution

Explanation

Given

The concentration of HC3H5O2 is 0.750M .

The concentration of NaC3H5O2 is 0.700M .

The volume of HC3H5O2 is 45.0ml .

The volume of NaC3H5O2 is 55.0ml .

The value of Ka for HC3H5O2 is 1.3×105 .

The conversion of ml into l is done as,

1ml=0.001l

Hence, the conversion of 45.0ml into l is done as,

45.0ml=45.0×0.001l=0.045l

Similarly, the conversion of 55.0ml into l is done as,

55.0ml=55.0×0.001l=0.055l

The pKa value of the given weak acid HA is calculated as,

pKa=logKa

Where,

• Ka is the dissociation constant of weak acid.
• pKa is the negative logarithm of dissociation constant of the acid.

Substitute the value of Ka in the above equation as,

pKa=logKa=log(1.3×105)=4.9

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres

Substitute the value of Concentration and volume of solution of CH3CH2COONa in the above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.70M×0.055l=0.0385moles

Similarly substitute the value of Concentration and volume of solution of CH3CH2COOH in the above equation as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.750M×0.045l=0.0338moles

When the ratio of moles of salt and acid is almost similar to ratio of their molarities, the concentrations are replaced with number of moles.

The Henderson-Hasselbalch equation is represented as shown below.

pH=pKa+log[A][HA]

Where,

• pH is the negative logarithm of H+ ions concentration in the solution.
• pKa is the negative logarithm of dissociation constant of the acid.
• [A] is the concentration of conjugate base of the given weak acid.
• [HA] is the concentration of given weak acid.

Substitute the value of amount of salt and acid in the above equation as,

pH=pKa+log[A][HA]=4

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