Problem 40 - Chemistry The Science in Context

Question

On the basis of the following degree-of-ionization data for\(0.100 M\) solutions, select which acid has t he largest \(K_{\mathrm{a}} .\) $$\begin{array}{cc}\text { Acid } & \text { Degree of lonization }(\%) \\\\\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} & 2.5 \\\\\hline \mathrm{HF} & 8.5 \\\\\hline \mathrm{HN}_{3} & 1.4 \\\\\hline \mathrm{CH}_{3} \mathrm{COOH} & 1.3 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

The acid with the largest Ka is HF, as it has the highest ion concentration (0.0085 M) among the given acids.

Step 1: Recall the relationship between degree of ionization, molarity, and ion concentration

The degree of ionization represents the percentage of molecules that have ionized in solution. So, to calculate the concentration of ions (\([\text {ions}]\)) formed due to ionization, we can use the following relation: $$[\text{ions}] = \text{Degree of Ionization} (\%) \cdot \text{Molarity}$$

Step 2: Calculate ion concentration for each acid

Using the given degree of ionization for each acid and the initial molarity, we can calculate the concentration of ions formed in each solution. C\(_6\)H\(_5\)COOH: \([\text{C}_6\text{H}_5\text{COO}^-] = 0.025 \cdot 0.100~\text{M} = 0.0025~\text{M}\) HF: \([\text{F}^-] = 0.085 \cdot 0.100~\text{M} = 0.0085~\text{M}\) HN\(_3\): \([\text{N}_3^-] = 0.014 \cdot 0.100~\text{M} = 0.0014~\text{M}\) CH\(_3\)COOH: \([\text{CH}_3\text{COO}^{-}] = 0.013 \cdot 0.100~\text{M} = 0.0013~\text{M}\)

Step 3: Use ion concentration to find \(K_a\) for each acid

Since all the acids are weak acids, we can write the expression for the dissociation of each acid and use the ion concentrations we just calculated to determine the \(K_a\). The general form of the dissociation of a weak acid is: $$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^−$$ And the expression for \(K_a\) is given by: $$K_a = \frac{[\text{H}^+][\text{A}^−]}{[\text{HA}]}$$ For all these solutions, we can consider the concentrations of H\(^+\) and A\(^-\) to be equal to the concentration of ions formed (as given in Step 2). Since the initial molarity of the acids is the same, we can compare these calculated values directly.

Step 4: Compare calculated K\(_a\) values to find the largest K\(_a\)

Using the expressions from Step 2, we can now compare the ion concentrations to see which is the largest, indicating which of the acids has the largest K\(_a\). C\(_6\)H\(_5\)COOH: \(0.0025~\text{M}\) HF: \(0.0085~\text{M}\) HN\(_3\): \(0.0014~\text{M}\) CH\(_3\)COOH: \(0.0013~\text{M}\) Since the HF solution has the highest ion concentration (\(0.0085\)), it has the largest \(K_a\). Therefore, the acid with the largest \(K_a\) is HF.

Key Concepts

Degree of Ionization

The degree of ionization is a crucial concept in understanding how acids behave in solutions. It refers to the percentage of acid molecules that dissociate into ions.
For example, in a solution of an acid like HF, the degree of ionization tells us what fraction of HF molecules split into ions. If the degree of ionization is 8.5%, this means that 8.5% of the total molecules contributed to ions in the solution.
By knowing the degree of ionization, we can predict the strength and behavior of an acid. Strong acids ionize almost completely, while weak acids ionize only partially. The given percentages help us to estimate the acid's ability to ionize and release hydrogen ions. Some key points:

• A high degree of ionization indicates a stronger acid.
• Low ionization suggests a weaker acid.
• It directly affects the calculation of ion concentration in solutions.

Ion Concentration Calculation

Calculating ion concentration is an essential step in understanding the acidity of a solution. When you know the degree of ionization and the initial molarity, it's straightforward to find how many ions are present in solution.
To calculate ion concentration, use the formula:\[[\text{ions}] = \text{Degree of Ionization} (\%) \cdot \text{Molarity}\]Let's see this with our given acids:

• For HF with 8.5% ionization and 0.100 M concentration: \([\text{F}^-] = 0.085 \times 0.100~\text{M} = 0.0085~\text{M}\)
• For C\(_6\)H\(_5\)COOH with 2.5% ionization: \([\text{C}_6\text{H}_5\text{COO}^-] = 0.025 \times 0.100~\text{M} = 0.0025~\text{M}\)

These calculations allow us to see which acid produces more ions, thereby acting as a stronger acid. Understanding ion concentrations helps in further calculations and comparisons, as seen with the dissociation constant.

Acid Dissociation Constant (Ka)

The acid dissociation constant, \(K_a\), is a vital parameter that measures the strength of an acid in a solution. It represents how well an acid ionizes in water. The greater the \(K_a\), the stronger the acid.
For weak acids, \(K_a\) can be calculated using the concentrations of the ionized species:\[K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]Here, \([\text{H}^+]=[\text{A}^-]\), reflecting the ions calculated from the degree of ionization.With our acids:

• Higher ion concentrations mean larger \(K_a\).
• The HF with concentration \(0.0085\,\text{M}\) leads to the highest \(K_a\), indicating stronger ionizing ability compared to the others.

Understanding \(K_a\) values helps in selecting the best acid for your specific needs, be it in industrial applications or chemical experiments. It gives you a quantifiable measure of acidity and strength.