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Chapter 12: Problem 120

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
2.9% of the Acetic acid has undergone ionization.

Step by step solution

01

Calculate the theoretical molality

First of all, let's rearrange the freezing point depression formula to calculate the theoretical molality (m) of the solution.\nΔΤf = Kf * m => m = ΔΤf / Kf\nHere, ΔΤf = - (-0.203°C) = 0.203°C is the change in freezing point, and Kf = 1.86°C/m is the cryoscopic constant for water.\nSo, m = 0.203°C / 1.86°C/m = 0.1092 mol/kg.
02

Compare the theoretical molality with the experimentally measured molality

The experimentally measured molality was 0.106 mol/kg. This is less than the theoretical molality calculated above. The reason for this difference is because some of the Acetic acid has undergone ionization, thereby leading to an increase in the number of particles in solution, and hence an increase in the molality. Therefore, the percentage of Acetic acid that has undergone ionization can be calculated as follows:\nPercentage ionization = ((theoretical molality - experimental molality) / theoretical molality) * 100\nTherefore Percentage ionization = ((0.1092 mol/kg - 0.106 mol/kg) / 0.1092 mol/kg) * 100 = 2.9%
03

Interpret the Result

The percent ionization tells us the fraction of the original acid that has ionized. In this case, 2.9% of the Acetic acid has ionized, indicating that only a small fraction of the original acid ionizes in solution. This is expected, as Acetic acid is a weak acid.

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