Problem 83 - Chemistry: An Atoms-Focused Approach

Question

The \(K_{\mathrm{a}}\) of the conjugate acid of the artificial sweetener saccharin is \(2.1 \times 10^{-11} .\) What is the p \(K_{h}\) for saccharin?

Step-by-Step Solution

Answer

Answer: The pKh value of saccharin is approximately 10.68.

Step 1: Find the Kb value of saccharin

Since we are given the Ka value of the conjugate acid of saccharin, we can use the relationship between Ka and Kb to find the Kb value of saccharin itself. The relationship is as follows: \(K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}\) Where \(K_\mathrm{w}\) is the ion product constant of water, which has a value of \(1.0 \times 10^{-14}\) at 25°C. We can solve for Kb using the given Ka value: \(K_\mathrm{b} = \cfrac{K_\mathrm{w}}{K_\mathrm{a}} \)

Step 2: Calculate the Kb value of saccharin

Now, we can substitute the given Ka value and the Kw value into the equation: \(K_\mathrm{b} = \cfrac{1.0 \times 10^{-14}}{2.1 \times 10^{-11}}\) Calculate the Kb value: \(K_\mathrm{b} = 4.76 \times 10^{-4}\)

Step 3: Calculate the pKb value of saccharin

Now that we have the Kb value, we can calculate the pKb value using the following formula: \(pK_\mathrm{b} = -\log {K_\mathrm{b}}\) Substitute the Kb value into the equation: \(pK_\mathrm{b} = -\log {(4.76 \times 10^{-4})}\) Calculate the pKb value: \(pK_\mathrm{b} ≈ 3.32\)

Step 4: Find the pKh value of saccharin

Finally, we can find the pKh value of saccharin using the relationship between pKa and pKb: \(pK_\mathrm{a} + pK_\mathrm{b} = 14\) Solve for pKh: \(pK_\mathrm{h} = 14 - pK_\mathrm{b}\) Substitute the pKb value obtained in Step 3: \(pK_\mathrm{h} = 14 - 3.32\) Calculate the pKh value: \(pK_\mathrm{h} ≈ 10.68\) So, the pKh value for saccharin is approximately 10.68.

Key Concepts

Conjugate Acids and Bases

In the world of chemistry, understanding conjugate acids and bases is crucial. When an acid donates a proton (an H⁺ ion), it becomes a conjugate base. Conversely, when a base accepts a proton, it becomes a conjugate acid. This concept is essential in acid-base reactions and helps in predicting the behavior of substances in different chemical environments.

Consider saccharin: when it donates a proton, the leftover is its conjugate base. The behavior of saccharin in reactions is governed by these shifts between its protonated and deprotonated forms. By knowing the properties of saccharin's conjugate acid and base, chemists can predict how it will behave in solutions, including its acidity or basicity.

To navigate these reactions, we look at constants such as the acid dissociation constant \((K_a)\) and the base dissociation constant \((K_b)\), together helping us understand the strength of the acid or base in question.

Ion Product Constant of Water (Kw)

The ion product constant of water, \(K_w\), is a fundamental value in chemistry, especially in acid-base equilibria. It represents the extent to which water can dissociate into its ions: hydronium \((H_3O^+)\) and hydroxide \((OH^-)\). At 25°C, \(K_w\) is consistently at \(1.0 \times 10^{-14}\).

This constant forms the basis of the relationship between \(K_a\) and \(K_b\):

• \(K_a \times K_b = K_w\)

By utilizing this relationship, as in the case of saccharin, we can determine either \(K_a\) or \(K_b\) when the other is known. It helps in exploring the nature of a substance as an acid or base in water. Understanding \(K_w\) is pivotal as it also aids in calculating \(pH\) and \(pOH\), the measures of acidity and basicity in a solution.

pKa and pKb Calculations

Calculations of \(pK_a\) and \(pK_b\) provide insight into the strengths of acids and bases. Using these calculations, we can derive how easily an acid donates a proton or how readily a base accepts one.

To calculate \(pK_b\), we take the negative logarithm of \(K_b\):

• \(pK_b = -\log{K_b}\)

In the case of saccharin with \(K_b = 4.76 \times 10^{-4}\), the computation yields a \(pK_b\) of approximately 3.32.

Additionally, the relationship \(pK_a + pK_b = 14\) helps find one value if the other is known, due to the consistent nature of the ion product constant of water. Once we have the \(pK_b\) for saccharin, calculating its \(pK_h\) becomes straightforward:

• \(pK_h = 14 - pK_b\)

Here, the result is a \(pK_h\) of about 10.68, indicating the overall stability of saccharin in solution.