   # You mix 0.40 g of NaOH with 100 mL of 0.10 M acetic acid. What is the pH of the resulting solution? (a) less than 7 (b) equal to 7 (c) greater than 7 ### 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 16.7, Problem 4RC
Textbook Problem
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## You mix 0.40 g of NaOH with 100 mL of 0.10 M acetic acid. What is the pH of the resulting solution? (a) less than 7 (b) equal to 7 (c) greater than 7

Interpretation Introduction

Interpretation:

The pH range should be identified given the statement of equilibrium reaction.

Concept introduction:

pH concept of hydronium ions: A strong acid completely dissociates into its constituent ions in aqueous solution, as a result, the concentration of its ion is same as the initial concentration of that strong acid.

For example, strong acid dissociates as follows in water,

HA(aq)+H2O(l)H3O+(aq)+A(aq)

[H3O+]=[A]=[HA]

The pH of a solution is basically the measure of the molar concentration of the H+ or H3O+ ion in the solution. More the concentration of H+ or H3O+ ion in the solution, lesser will be the pH value and more acidic will be the solution.

The expression for pH is given as,

pH=log[H3O+]

The concentration of OH ions is calculated by using the ionic product of water.

Kw=[H3O+][OH]

The value of Kw is 1.0×1014 .

If pH<7 then, the solution is acidic in nature.

If pH<7 then, the solution is basic in nature.

If pH<7 then, the solution is neutral in nature.

### Explanation of Solution

The value of pH for sodium hydroxide and acetic acid solution and the concentration is calculated below.

Given

The initial concentration of NaOHandCH3COOH solution is 0.10 M .

NaOH has strong base and acetic acid is week acid it is equlibrium reaction showes below.

NaOH(aq)+CH3COOH(l)CH3COO-Na+(aq)+H2O(aq)

The ICE table is as follows,

EquilibriumNaOH(aq)+CH3COOH(l)CH3COO-Na+(aq)+H2O(aq)Initial(M)0.1000Change(M)x+x+xEquilibrium(M)(0.10x)XX

TheequilibriumexpressionshownbelowKa=[CH3COO-Na+][H2O][NaOH][CH3COOH]TheKbvalueofCH3COO-Na+=5

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