Problem 35 - Chemistry: Principles and Reactions

Question

Calculate \(K_{a}\) for the weak acids that have the following \(\mathrm{pK}_{\mathrm{a}}\) values. (a) \(3.9\) (b) \(10.12\) (c) \(13.07\)

Step-by-Step Solution

Answer

Question: Calculate the Ka values for weak acids with the following pKa values: (a) 3.9, (b) 10.12, (c) 13.07. Answer: The Ka values for the weak acids with the given pKa values are: (a) 1.26 × 10^-4, (b) 7.58 × 10^-11, (c) 8.56 × 10^-14.

Step 1: Write down the given values and formula

(Write the \(\mathrm{pK}_a\) values for each part (a), (b), and (c) and the formula \(\mathrm{pK}_a = -\log_{10}(K_a)\). (a) \(\mathrm{pK}_a = 3.9\) (b) \(\mathrm{pK}_a = 10.12\) (c) \(\mathrm{pK}_a = 13.07\) Formula: \(\mathrm{pK}_a = -\log_{10}(K_a)\)

Step 2: Calculate the \(K_a\) values

(Use the formula to find the \(K_a\) values for each part (a), (b), and (c).) (a) \(K_a = 10^{-\mathrm{pK}_a} = 10^{-3.9}\) The \(K_a\) for part (a) is: \(K_{a} \approx 1.26 \times 10^{-4}\) (b) \(K_a = 10^{-\mathrm{pK}_a} = 10^{-10.12}\) The \(K_a\) for part (b) is: \(K_{a} \approx 7.58 \times 10^{-11}\) (c) \(K_a = 10^{-\mathrm{pK}_a} = 10^{-13.07}\) The \(K_a\) for part (c) is: \(K_{a} \approx 8.56 \times 10^{-14}\) The \(K_a\) values for the weak acids with the given \(\mathrm{pK}_a\) values are: (a) \(1.26 \times 10^{-4}\) (b) \(7.58 \times 10^{-11}\) (c) \(8.56 \times 10^{-14}\)

Key Concepts

Weak Acids

Weak acids are substances that partially dissociate into ions in water. This means they only lose a small number of hydrogen ions (\(H^+\)) compared to strong acids, which dissociate completely. This partial dissociation results in less acidic solutions.

When dissolved in water, a weak acid makes an equilibrium between the undissociated acid molecules and the ions formed. This equilibrium can be represented as:

• HA (aq) ⇌ H⁺ (aq) + A⁻ (aq)

Where HA is the weak acid, and H⁺ and A⁻ are the ions formed.

The dissociation constant, \(K_a\), quantifies the degree of dissociation, with smaller values indicating weaker acids. The ability to dissociate less than strong acids reflects in their lower conductance and reactive behavior in aqueous solutions.

pKa

The \(\mathrm{pK}_a\) value is a crucial concept in chemistry, particularly when dealing with weak acids. It helps to understand the strength of these acids. Essentially, \(\mathrm{pK}_a\) is a negative logarithmic scale that indicates acidity.

The formula to convert a \(K_a\) to \(\mathrm{pK}_a\) is:

• \(\mathrm{pK}_a = -\log_{10}(K_a)\)

The lower the \(\mathrm{pK}_a\) value, the stronger the acid because it indicates a higher \(K_a\), meaning better dissociation into ions.

Conversely, if the \(\mathrm{pK}_a\) is higher, the acid is weaker, showing less ion dissociation. Understanding the \(\mathrm{pK}_a\) helps to determine whether an acid will be more or less reactive in different conditions, influencing its role in chemical reactions and processes.

Logarithms

Logarithms are mathematical tools used to express large ranges of values in a compact form. In chemistry, they are often used to deal with exponential relationships, such as those found in acid dissociation contexts like \(K_a\) and \(pK_a\).

The base-10 logarithm, often written as \(\log_{10}\), is particularly useful. Its power lies in its ability to convert multiplicative relationships into additive ones. For example, the relationship:

• \(10^a \times 10^b = 10^{a+b}\)

can be simplified using logarithms:

• \(\log_{10}(10^a \times 10^b) = a + b\)

This compact form is used to simplify calculations in chemistry, especially in expressions like \(\mathrm{pK}_a = -\log_{10}(K_a)\). By understanding and using logarithms, one can easily move between pKa and Ka values to predict and explain chemical behavior, maintaining simplicity in calculations.