   # Calculate the hydronium ion concentration and pH of the solution that results when 22.0mL of 0.15M acetic acid CH 3 CO 2 H is mixed with 22.0 mL of 0.15M NaOH . ### 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 63PS
Textbook Problem
266 views

## Calculate the hydronium ion concentration and pH of the solution that results when 22.0mL of 0.15M acetic acid CH 3 CO 2 H is mixed with 22.0 mL of 0.15M NaOH .

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15M acetic acid, CH3CO2H is mixed with 22.0 mL of 0.15MNaOH has to be calculated.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka: it is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ionic product constant for water  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

### Explanation of Solution

Given volume of 0.1M CH3CO2H = 22.0 mL

And it is mixed with 22.0 mL of 0.1M OH-. Therefore, the base converts into acetic acid to its conjugate base, the acetate ion.

In this reaction the entire amount of base added converts CH3CO2H completely into conjugate base CH3COO-.

Let’s calculate the amount of consumed acetic acid:

(22.0mL×1L1000mL) × (0.15molL)=  0.0033 mol

Let’s calculate the consumed amount of NaOH

(22.0mL×1L1000mL) × (0.15molL)=  0.0033 mol

Let’s calculate the produced amount of CH3COO- after completion of reaction.

=0.0033 mol CH3COOH × 1mol CH3COO-1mol CH3COOH=0.0033 mol

The total volume of solution= 22.0mL + 22.0mL= 44.0mL × 1L1000mL= 0.044L

Concentration ofCH3COO-=0.0033mol0.044L= 0.075M

Therefore, 0.75M is the final volume of CH3COO- after completion of the reaction.

The CH3COO- ions react with water. The resulting solution is basic in nature.

The equilibrium reaction is as follows.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

The equilibrium expression:

Kb[CH3COOH][OH-][CH3COO-]

Kb of CH3COO- is 5.6×10-10

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

CH3COO-(aq) + H2O(l) CH3CO2H(aq) + OH-(aq)   I                 0

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