Problem 83 - Chemistry The Science in Context

Question

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

Step-by-Step Solution

Answer

Answer: The \(pK_b\) of saccharin is approximately \(3.32\).

Step 1: Find the relationship between \(K_a\), \(K_b\), and \(KW\)

To begin, we need to find the relationship between the ion product of water (\(KW\)), the acidity constant (\(K_a\)), and the basicity constant (\(K_b\)). The relationship between them is given by the formula: \(KW = K_a \times K_b\)

Step 2: Calculate \(K_b\) using the given \(K_a\)

We're given the acidity constant (\(K_a\)) of the conjugate acid of saccharin as \(2.1 \times 10^{-11}\). The ion product of water (\(KW\)) is a constant value at \(25^{\circ}\mathrm{C}\), which is approximately \(1.0 \times 10^{-14}\). Now we can find the basicity constant (\(K_b\)) using the relationship found in Step 1: \(K_b = \frac{KW}{K_a} = \frac{1.0 \times 10^{-14}}{2.1 \times 10^{-11}}\)

Step 3: Calculate the \(\mathrm{p} K_{\mathrm{b}}\)

Now that we have \(K_b\), we can find the \(pK_b\) by using the definition of \(pK_b\): \(pK_b = -\log_{10}(K_b)\) First, calculate \(K_b\) from Step 2: \(K_b = \frac{1.0 \times 10^{-14}}{2.1 \times 10^{-11}} \approx 4.76 \times 10^{-4}\) Now, find \(pK_b\) by taking the negative logarithm: \(pK_b = -\log_{10}(4.76 \times 10^{-4}) \approx 3.32\) So, the \(pK_b\) for saccharin is approximately \(3.32\).

Key Concepts

pKb Calculation

Understanding how to calculate the base dissociation constant's negative logarithm, known as the \(pK_b\), is essential in acid-base chemistry. When you're presented with the acidity constant, \(K_a\), for a conjugate acid, as in the saccharin example, you can delve into the process of calculating \(pK_b\). After determining the basicity constant, \(K_b\), using the provided \(K_a\) and the ion product of water, \(K_W\), the next step is simple. You take the negative logarithm (base 10) of \(K_b\) to find \(pK_b\). This conversion is vital as it transforms the often cumbersome exponential notation into a more easily comparable number.

For instance, in the saccharin case, we found that the \(K_b\) was approximately \(4.76 \times 10^{-4}\). By taking its negative logarithm, we calculated the \(pK_b\) to be about 3.32, which enables us to discuss and compare the strength of bases more effectively.

Ka and Kb Relationship

In acid-base equilibrium, the key relationship between the acidity constant (\(K_a\)) and the basicity constant (\(K_b\)) is their connection through the ion product of water (\(K_W\)). The equation \(K_W = K_a \times K_b\) signifies that the product of the acidity and basicity constants at a given temperature always equals \(K_W\), which is constant at a given temperature, typically \(1.0 \times 10^{-14}\) at 25°C. This relationship is fundamental because it allows us to switch perspectives between the acidity of a conjugate acid and the basicity of the corresponding base of an acid-base pair.

When dealing with the saccharin example, by knowing the \(K_a\) of saccharin’s conjugate acid, we efficiently compute \(K_b\) for saccharin itself, allowing for a full characterization of saccharin’s chemical behavior in aqueous solutions.

Ion Product of Water

The ion product of water (\(K_W\)) is an essential constant in acid-base chemistry, standing at approximately \(1.0 \times 10^{-14}\) at 25°C. This constant reflects the equilibrium concentration of hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)) in pure water, resulting from the self-ionization of water molecules. The value of \(K_W\) gives us the foundation for the relationship between \(K_a\) and \(K_b\), as it underpins the equilibrium state of water, a substance commonly found in acid-base reactions. Understanding \(K_W\) is critical when solving problems related to acidity and basicity, as it is used to bridge the gap between the strengths of acids and their conjugate bases.

Conjugate Acid-Base Pair

Acids and bases exist in pairs, known as conjugate acid-base pairs. When an acid donates a proton, it becomes its conjugate base, whereas when a base accepts a proton, it forms its conjugate acid. These pairs are intimately related through their \(K_a\) and \(K_b\) values as previously discussed. For instance, saccharin's conjugate acid has a specific \(K_a\), which tells us how readily the acid gives up its proton in water. Once it does so, saccharin (as a base) is ready to accept a proton, and this propensity is quantified by its \(K_b\).

Grasping the nature of these pairs is crucial in acid-base chemistry because it allows us to predict the outcomes of chemical reactions involving acids and bases, to calculate pH, and to understand buffer solutions' behavior.