   # You have 75.0 mL of 0.10 M HA. After adding 30.0 mL 0.10 M NaOH, the pH is 5.50. What is the K a value of HA? ### 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 14, Problem 67E
Textbook Problem
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## You have 75.0 mL of 0.10 M HA. After adding 30.0 mL 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA?

Interpretation Introduction

Interpretation:

A 0.10M acid HA is given. The value of Ka for the given volume of this acid is to be calculated.

Concept introduction:

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

To determine: The value of Ka for the given volume of the acid.

### Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, Ka=[H+][A][HA]

Given

The volume of HA is 75.0mL .

Molarity of HA is 0.10M .

The volume of NaOH is 30.0mL .

Molarity of NaOH is 0.10M .

The value of pH is 5.50 .

In the given case, the strong base OH reacts with the donor species that is HA .

The stoichiometry of the reaction is represented as,

HA(aq)+OH(aq)A(aq)Beforereaction75.0mL×0.10M=7.5mmol30.0mL×0.10M=3.0mmol0Afterreaction7.53.0=4.5mmol3.03.0=03.0mmol

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[H+][A][HA] (1)

The initial concentrations are calculated as,

[HA]=4.5mmol(75.0+30.0)mL=0.043M

[A]=3.0mmol(75.0+30.0)mL=0.029M

The ICE table for the stated reaction is,

HA(aq)H+(aq)+A(aq)Inititialconcentration0

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