   Chapter 11, Problem 127CP

Chapter
Section
Textbook Problem

# Formic acid (HCO2H) is a monoprotic acid that ionizes only partially in aqueous solutions. A 0.10-M formic acid solution is 4.2% ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of 0.10 M formic acid.

Interpretation Introduction

Interpretation: The boiling point and freezing point has to be calculated.

Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.

The elevation in boiling point can be given by the equation,

ΔT=Kbmsolute

Where, ΔT = change in boiling point elevation

Kb = molal boiling point elevation constant

msolute = molality of solute

The depression in freezing point can be given by the equation,

ΔT=Kfmsolute

Where, ΔT =change in freezing point depression

Kf = molal freezing point depression constant

msolute = molality of solute

Explanation

Record the given data

Molarity of Formic acid = 0.10M

To calculate the molarity of ionized Formic acid solution

HCO2HH++HCO2-

Ionized rate of Formic acid = 4.2%

The amount of H+orHCO2- produced is 0.042×0.10M=0.0042M

The amount of HCO2H remaining in the solution after ionization

= 0.10M-0.0042M=0.10M

The total molarity of the species = MHCO2H+MH++MHCO2- = 0.10+0.0042+0.0042 = 0.11M

Therefore, the total molarity of the species = 0.11M

To calculate the boiling point of Formic acid

Molal boiling point elevation constant = 0.51°C

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