   # You mix 30 0 mL of 0.15 M NaOH with 30.0 mL of 0.15 M acetic acid. What molecules and ions exist in this solution? List them in order of decreasing concentration. ### 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 16, Problem 108GQ
Textbook Problem
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## You mix 30 0 mL of 0.15 M NaOH with 30.0 mL of 0.15 M acetic acid. What molecules and ions exist in this solution? List them in order of decreasing concentration.

Interpretation Introduction

Interpretation:

Molecules and ion present in equimolar concentrations of sodium hydroxide and acetic acid are to be determined. and concentrations of all ions are arranged in decreasing order.

Concept introduction:

Moles of reactants are calculated from given concentration and volume Since equal moles of each reactant combine in the reaction, therefore neither acid nor base is in excess on the product side. From Henderson Hasselbalch equation, the value of pH of solution is obtained Concentration of hydrogen ion and hydroxide ion is obtained from pH and pOH values.

### Explanation of Solution

Amount of CH3COOH consumed in the reaction is

= 0.03 mole×0.15 mole/L= 0.0045 mole CH3COOH

Amount of NaOH consumed in the reaction is

= 0.03 mole×0.15 mole/L= 0.0045 mole NaOH

The reaction of CH3COOH and NaOH is given below:

CH3COOH(aq)+NaOH(aq)CH3COONa+(aq)+H2O(l)

Consider the stoichiometry of the reaction;

One mole of acetic acid gives one mole of sodium acetate.

0.0045 mole CH3COOH(1mole CH3COONa+1mole CH3COOH )=0.0045 mole CH3COONa+

From Henderson Hasselbalch equation, the value of pH of solution is given as:

pH=pKa+log(CH3COOCH3COOH)

Since the concentration of sodium acetate and acetic acid is equal, therefore pH=pKa for the solution.

The value of Ka is 1.8×105.

From the value of acid dissociation constant,pKa is calculated as given below:

pKa=log(Ka)=log(1.8×105)=4.75

Since,pH=pKa

Therefore, pH=4.75

pH=log[H+]

Therefore, Concentration of hydronium ion is given as:

[H+]=1

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