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Chapter 13: Problem 97

Acetic acid is a weak acid that ionizes in solution as follows: $$ \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q) $$ If the freezing point of a \(0.106 \mathrm{~m} \mathrm{CH}_{3} \mathrm{COOH}\) solution is \(-0.203^{\circ} \mathrm{C},\) calculate the percent of the acid that has undergone ionization.

Short Answer

Expert verified
The percentage of acetic acid that has undergone ionization in the given solution is approximately 2.83%.

Step by step solution

01

Determine solution's molality using freezing point depression.

The relationship between freezing point depression and molality can be given as \( ΔT = k_f x m \) where \( ΔT \) is the freezing point depression, \( k_f \) is the freezing point depression constant, and \( m \) is the molality. You can rearrange the equation for molality: \( m = ΔT / k_f \). The freezing point depression of water is \( 0.203^{\circ} C \) (from 0 to -0.203), and the freezing-point-depression constant for water (\( k_f \)) is \( 1.86 \(\frac{C}{m}\) \). So, \( m = 0.203 / 1.86 = 0.109 mol/kg \).
02

Calculate the theoretical molality.

The theoretical molality (if no ions were produced) is equal to the given molality of acetic acid, which is 0.106 mol/kg.
03

Calculate the van't Hoff factor.

The van't Hoff factor \( i \) is the ratio of the observed molality to the theoretical molality. It signifies the number of particles into which a compound breaks up in solution. So, \( i = observed\_molality / theoretical\_molality = 0.109 / 0.106 = 1.028 \). This factor is higher than 1, indicating that some ionization has occurred and the solution contains more than just non-ionized molecules of acetic acid.
04

Calculate the percent ionization.

The percent ionization of the acid is the percent difference in \( i \) from 1, since 1 would indicate no ionization. It can be calculated as follows: \( percent\_ionization = [(i - 1) / 1] x 100 = [(1.028 - 1) / 1] x 100 = 2.83\% \). This means about 2.83% of the acetic acid has undergone ionization in solution.

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Most popular questions from this chapter

The density of an aqueous solution containing 10.0 percent ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) by mass is \(0.984 \mathrm{~g} / \mathrm{mL}\). (a) Calculate the molality of this solution. (b) Calculate its molarity. (c) What volume of the solution would contain 0.125 mole of ethanol?

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