   Chapter 17, Problem 80GQ

Chapter
Section
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 1.8×105.

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 0.15 M.

The volume of acetic acid solution is 20 mL

Conversion of 20 mL into L

(20 mL)(1 L1000 mL)=0.020 L

Therefore, volume of the solution is 0.020 L.

The concentration of NaOH is 0.17 M.

The volume of NaOH solution is 5 mL.

Conversion of 5 mL into L

(5 mL)(1 L1000 mL)=0.005 L

Therefore, volume of the solution is 0.005 L.

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 and affects the equilibrium conditions.

CH3COOH(aq)+ OH(aq)H2O(l)+ CH3COO(aq)

Table gives the relationship of the concentration for the reaction between OH, CH3COOH and CH3COO.

OH(from added NaOH)  CH3COOH CH3COO(produced)initial(mol)0.000850.0030.0change(mol)0.000850.00085+0.00085after reaction(mol)00.002150.00085

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

Molarity=Number of molesvolume of solvent (2)

The total volume of the solution is calculated as follows;

total volume of solution =  volume of acetic acid + volume of NaOH added=0.020 L+0.005 L= 0.025 L

The total volume of the resultant solution is 0.025 L.

Substitute, 0.00215 mol for Number of moles and 0.025 L for  volume of solvent to calculate the concentration of acetic acid in equation (2).

Molarity=0.00215 mol0.0250 L=0.086 molL1

Therefore, concentration of acetic acid left after reaction is 0.086 molL1

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