   # A 0.100-g sample of the weak acid HA (molar mass = 100.0 g/mol) is dissolved in 500.0 g water. The freezing point of the resulting solution is −0.0056°C. Calculate the value of K a for this acid. Assume molality equals molarity in this 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 185CP
Textbook Problem
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## A 0.100-g sample of the weak acid HA (molar mass = 100.0 g/mol) is dissolved in 500.0 g water. The freezing point of the resulting solution is −0.0056°C. Calculate the value of Ka for this acid. Assume molality equals molarity in this solution.

Interpretation Introduction

Interpretation: The value of Ka for a 0.100g weak acid dissolved in 500.0g water is to be calculated.

Concept introduction: Molality is a measure of concentration of solute and is defined as the number of moles of solute per kg of solvent. Van’t Hoff factor is used to calculate the extent of association and dissociation of solute added in a solution. Depression in freezing point is the colligative property observed when a solute is added to a solution and the freezing point of the solution becomes less than the freezing point of the solvent.

### Explanation of Solution

Explanation

The molality is 0.002m_ .

Given

Mass of the given weak acid is 0.100g .

Mass of the water is 500g .

The molar mass of the weak acid is 100g/mol .

The molality of a solution is calculated by the formula,

Molality=Numberofmolesofsolute×1000Massofsolvent(g)Molality=Givenmassofsolute×1000Molarmassofsolute×Massofsolvent(g)

Substitute the values of masses of solute and solvent and the molar mass of the solute, that is the weak acid) in the above equation.

Molality=0.100g×1000100g/mol×500gMolality=0.002m_

The value of Van’t Hoff factor (i) is 1.5053_ .

Given

The depression in freezing point is 0.0056°C .

The molality of given solution is 0.002m_ .

The value of Van’t Hoff factor (i) is calculated by the formula.

ΔTf=i×msolute×Kf

Where,

• ΔTf is change in freezing point of the solution.
• i is Van’t Hoff factor.
• msolute is the molality of the solute.
• Kf is the cryoscopic constant. For water the value of Kf is 1

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