Question

The \(K_{\mathrm{b}}\) values for ammonia and methylamine are \(1.8 \times 10^{-5}\) and \(4.4 \times 10^{-4}\), respectively. Which is the stronger acid, \(\mathrm{NH}_{4}^{+}\) or \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ?\)

Short answer

The stronger acid is NH4+ because its Ka value (\(5.5 \times 10^{-10}\)) is higher than the Ka value for CH3NH3+ (\(2.3 \times 10^{-11}\)).

Step-by-step solution

Find the Ka values of NH4+ and CH3NH3+

We will find the Ka values of NH4+ and CH3NH3+ using the Kb values given and the Ka-Kb relationship. For NH4+, we are given the Kb value of its conjugate base, ammonia (NH3), which is 1.8 × 10^{-5}. Using the relation Ka x Kb = Kw, \(K_{a} \times(1.8 \times 10^{-5})=1 \times 10^{-14}\) Solving for Ka: \(K_{a} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\) For CH3NH3+, we are given the Kb value of its conjugate base, methylamine (CH3NH2), which is 4.4 × 10^{-4}. Using the relation Ka x Kb = Kw, \(K_{a} \times(4.4 \times 10^{-4})=1 \times 10^{-14}\) Solving for Ka: \(K_{a} = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}}\)

Compare the Ka values

Now we will compare the Ka values of NH4+ and CH3NH3+ to determine which is the stronger acid. Ka for NH4+: \(K_{a} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\) \[K_{a} \approx 5.5 \times 10^{-10}\] Ka for CH3NH3+: \(K_{a} = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}}\) \[K_{a} \approx 2.3 \times 10^{-11}\] Because the Ka value for NH4+ (5.5 × 10^{-10}) is higher than the Ka value for CH3NH3+ (2.3 × 10^{-11}), NH4+ is the stronger acid.

Key concepts

Ka and Kb relationship

Understanding the relationship between the acid dissociation constant \( K_a \) and the base dissociation constant \( K_b \) is key in analyzing acid-base equilibria. These constants reveal the strength of acids and bases in solution and tell us how well they dissociate into their ions. The relationship between \( K_a \) and \( K_b \) is expressed by the equation:

• \( K_a \times K_b = K_w \)

where \( K_w \) is the ion product of water, equal to \( 1 \times 10^{-14} \) at 25°C.
The formula comes from the concept of conjugate acids and bases. A strong acid will have a weak conjugate base, and vice versa. For example, let’s consider the conjugate acid-base pairs in this exercise:

• Ammonium ion \( \text{NH}_4^+ \) is the conjugate acid of ammonia.
• Methylammonium ion \( \text{CH}_3\text{NH}_3^+ \) is the conjugate acid of methylamine.

Given their \( K_b \) values, we can calculate their \( K_a \) values by rearranging the equation to:\[ K_a = \frac{K_w}{K_b} \]
This allows us to switch between the acidic and basic strengths of these species.

Acid strength comparison

Comparing acid strength involves looking at their \( K_a \) values, which indicates how completely an acid ionizes in solution. A larger \( K_a \) value means a stronger acid, as it shows a greater tendency to donate protons. In our exercise, we calculated:

• For \( \text{NH}_4^+ \), \( K_a \approx 5.5 \times 10^{-10} \).
• For \( \text{CH}_3\text{NH}_3^+ \), \( K_a \approx 2.3 \times 10^{-11} \).

This shows \( \text{NH}_4^+ \) is a stronger acid than \( \text{CH}_3\text{NH}_3^+ \) because its \( K_a \) value is higher. A stronger acid donates protons more readily, leading to a greater concentration of \( \text{H}^+ \) ions in solution.
As a result, you can determine relative acid strengths by comparing \( K_a \) values directly. Remember, a stronger acid has weaker conjugate base, reflected inversely in their \( K_b \) values.

Conjugate acid-base pairs

Every acid has a conjugate base, and every base has a conjugate acid. This relationship is crucial in understanding how acids and bases behave during chemical reactions. When an acid donates a proton, the remaining part becomes its conjugate base. Similarly, when a base accepts a proton, it becomes its conjugate acid.
In our problem, we looked at the conjugate pairs:

• \( \text{NH}_3 \) (ammonia) and \( \text{NH}_4^+ \) (ammonium ion)
• \( \text{CH}_3\text{NH}_2 \) (methylamine) and \( \text{CH}_3\text{NH}_3^+ \) (methylammonium ion)

This dynamic equilibrium allows us to focus on their behavior not just as standalone substances, but in relation to their mutual conversions.
The strength of a conjugate acid-base pair is inversely related. This is why a strong acid's conjugate base is weak and a strong base's conjugate acid is weak. The inverse \( K_a \) and \( K_b \) values we calculated demonstrate this principle, helping us predict the resulting pH changes in a solution based on which form predominates.