Question

Acetic acid is a weak acid that ionizes in solution as follows: $$ \mathrm{CH}_{3} \mathrm{COOH}(a q) \rightleftarrows \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}^{+}(a q) $$ 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

The percent of acetic acid ionized is approximately 3%.

Step-by-step solution

Identify the initial freezing point depression

Determine the normal freezing point of water, which is 0°C. Calculate the freezing point depression, which is the difference between the normal freezing point and the observed freezing point of the solution:\[\Delta T_f = 0°C - (-0.203°C) = 0.203°C\]

Calculate Molality Using Freezing Point Depression

Use the formula for freezing point depression:\[\Delta T_f = i \,K_f \, m\]where \(K_f\) for water is 1.86°C/m and \(m\) is the molality. We will solve for the van 't Hoff factor \(i\), which represents the total moles of particles produced per mole of solute.

Solve for van 't Hoff factor (i)

Rearranging the formula from Step 2:\[i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.203}{1.86 \times 0.106} \approx 1.03\]The van 't Hoff factor \(i\) indicates the degree of ionization of the acetic acid in solution.

Determine the percent ionization

For acetic acid, complete ionization gives two ions per molecule hence theoretically \(i = 2\). The ionization percentage is calculated by the formula:\[\text{Percent ionization} = \left(\frac{i - 1}{2 - 1}\right) \times 100\%\]\[\text{Percent ionization} = (1.03 - 1) \times 100\% = 3\%\]This indicates that 3% of the original acetic acid undergoes ionization in the solution.

Key concepts

Understanding Weak Acids

A weak acid is a type of acid that does not completely ionize in a solution. This means not all acid molecules dissociate into ions. Is only partially ionized when dissolved in water. This property of weak acids is crucial in determining their behavior in solutions.
Acetic acid is a classic example of a weak acid. It partially dissociates into acetate ions (\( ext{CH}_3 ext{COO}^-\)) and hydrogen ions (\( ext{H}^+\)) in water, as shown in the balanced chemical equation:

• The equation: \[\text{CH}_3 ext{COOH} ightleftarrows ext{CH}_3 ext{COO}^- + ext{H}^+\]
• This equilibrium signifies the partial ionization, which is characteristic of weak acids.

Understanding this equilibrium concept helps analyze the degree of ionization and affects calculations in various chemical properties, like freezing point depression.

Percent Ionization of Weak Acids

Percent ionization reflects how much of a weak acid ionizes compared to the original amount of the acid dissolved. It is an important measure when analyzing weak acids, as it tells us the extent to which the acid has dissociated in solution.
The formula to calculate percent ionization is:

• \[\text{Percent Ionization} = \left(\frac{i - 1}{n - 1}\right) \times 100\%\]
• For acetic acid, the complete ionization implies \(i = n = 2\), where each molecule should produce two ions.

In our example, we discovered that only about 3% of acetic acid is ionized, which aligns with expected weak acid behavior, as they don't fully dissociate.

Explaining the Van 't Hoff Factor

The van 't Hoff factor (\(i\)) is used to indicate the number of particles in a solution after solute dissociation. It plays a crucial role in calculating colligative properties such as freezing point depression.
In simple terms, it tells us how effectively the solute impacts the solution properties:

• A solute that doesn’t dissociate has \(i = 1\)
• A solute that fully dissociates should have \(i\) equal to the number of particles formed.

For weak acids like acetic acid, the observed van 't Hoff factor is slightly above 1 (\(i \approx 1.03\)). This reflects slight ionization into two distinct ions: acetate and hydrogen. It's important for understanding why the freezing point depression was calculated as lower than expected for complete dissociation.