Chapter 17, Problem 80GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Calculate the hydronium ion concentration and the pH of the solution that results when 20.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 5.0 mL of 0.17 M NaOH.

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH for the resultant solution has to be calculated when 20 mL of 0.15 M CH3COOH is mixed with 5 mL of 0.17 M NaOH.

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.

As, NaOH is a strong base it ionize completely into its ions in aqueous solution.

NaOH(aq)Na+(aq)+ OH(aq)

The hydroxide ions will react with acetic acid.

CH3COOH(aq)+ OH(aq)H2O(l)+ CH3COO(aq)

The acetic acid left after the reaction and acetate ion produced will form a buffer solution.

Explanation

The calculation of pH is done by using Henderson-Hasselbalch equation. The value of hydronium ion concentration is calculated by using the pH value of the resultant solution.

Given:

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

The value of Ka for acetic acid is 1.8Ã—10âˆ’5.

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

pKa=âˆ’log(Ka)=âˆ’log(1.8Ã—10âˆ’5)=4.74

Therefore, pKa value for the acetic acid is 4.74.

The concentration of acetic acid is 0.15Â M.

The volume of acetic acid solution is 20Â mL

Conversion of 20Â mL into L

(20Â mL)(1Â L1000Â mL)=0.020Â L

Therefore, volume of the solution is 0.020Â L.

The concentration of NaOH is 0.17Â M.

The volume of NaOH solution is 5Â mL.

Conversion of 5Â mL into L

(5Â mL)(1Â L1000Â mL)=0.005Â L

Therefore, volume of the solution is 0.005Â L.

As, NaOH is a strong base it ionize completely into its ions in aqueous solution.

NaOH(aq)â†’Na+(aq)+Â OHâˆ’(aq)

The hydroxide ions will react with acetic acid and affects the equilibrium conditions.

CH3COOH(aq)+Â OHâˆ’(aq)â†’H2O(l)â€‰+Â CH3COOâˆ’(aq)

Table gives the relationship of the concentration for the reaction between OHâˆ’, CH3COOH and CH3COOâˆ’.

Â Â Â Â Â Â Â Â Â Â Â OHâˆ’(fromÂ addedÂ NaOH)Â Â CH3COOHÂ CH3COOâˆ’(produced)initial(mol)0.000850.0030.0change(mol)âˆ’0.00085âˆ’0.00085+0.00085afterÂ reaction(mol)00.002150.00085

The concentration of acetic acid and acetate ion after the reaction is calculated by using relation.

Molarity=NumberÂ ofÂ molesvolumeÂ ofÂ solvent (2)

The total volume of the solution is calculated as follows;

totalÂ volumeÂ ofÂ solutionÂ =Â Â volumeÂ ofÂ aceticÂ acidÂ +Â volumeÂ ofÂ NaOHÂ added=0.020â€‹Â L+0.005Â L=Â 0.025Â L

The total volume of the resultant solution is 0.025Â L.

Substitute, 0.00215Â mol for NumberÂ ofÂ moles and 0.025Â L for Â volumeÂ ofÂ solvent to calculate the concentration of acetic acid in equation (2).

Molarity=0.00215Â mol0.0250Â L=0.086Â molâ‹…Lâˆ’1

Therefore, concentration of acetic acid left after reaction is 0.086Â molâ‹…Lâˆ’1

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