   # Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00- M HC 2 H 3 O 2 to double the pH of the solution (assume that the added NaOH does not change the volume of the solution). ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 180CP
Textbook Problem
48 views

## Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00-M HC2H3O2 to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

Interpretation Introduction

Interpretation: The mass of NaOH required to be added to 1.00L of 1.00M acetic acid (CH3COOH) in order to double the pH of the solution is to be calculated.

Concept introduction: Weak acid dissociates up to a very small extent in an aqueous solution and on addition of a base, neutralization reaction occurs and formation of a salt takes place. This salt acts as a buffer solution.

To determine: The mass of NaOH required to be added to 1.00L of 1.00M acetic acid (CH3COOH) in order to double the pH of the solution.

### Explanation of Solution

Explanation

The pH of given solution of acetic acid is 2.38_.

A weak acid like acetic acid dissociates partially in aqueous solution and ICE (Initial Change Equilibrium) is used to calculate the concentration of products at equilibrium.

CH3COOH+H2OH3O++CH3COOInitial(M):1M00Change(M):xxxEquilibrium(M):1xxx

The acid dissociation constant formula for the above reaction is.

Ka=[H3O+][CH3COO][CH3COOOH]

Substitute the values of concentration and standard value of Ka in above equation.

1.8×105=x×x1.00xx2+1.8×1051.8×105=0x=0.0042M

Therefore, the [H+] is 0.0042M.

The pH is calculated by the formula.

pH=log[H+]

Substitute the value of [H+] in the above expression.

pH=log(0.0042)pH=2.38_

The mass of NaOH added is 20g_.

The addition of NaOH causes the neutralization reaction and the resulting product acts as a buffer solution. The amount of NaOH added is assumed to be xM then the amount of salt formed is xM after reacting with xM of acid

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