To find the acid dissociation constant (Ka) for the conjugate acid of methylamine (CH₃NH₂), we can use the relationship between the base dissociation constant (Kb) and the ion product of water (Kw). The ion product of water (Kw) at 25°C is a constant value of [tex]\(1.0 \times 10^{-14}\)[/tex].
The relationship between Ka and Kb is given by the equation:
[tex]\[ K_a \cdot K_b = K_w \][/tex]
Given:
- [tex]\( K_b \)[/tex] for [tex]\( CH_3NH_2 \)[/tex] is [tex]\( 4.4 \times 10^{-4} \)[/tex]
- [tex]\( K_w \)[/tex] is [tex]\( 1.0 \times 10^{-14} \)[/tex]
We need to find [tex]\( K_a \)[/tex].
Step-by-Step Solution:
1. Write down the equation that relates Ka, Kb, and Kw:
[tex]\[ K_a \cdot K_b = K_w \][/tex]
2. Substitute the given values into the equation:
[tex]\[ K_a \cdot 4.4 \times 10^{-4} = 1.0 \times 10^{-14} \][/tex]
3. Solve for Ka by isolating it on one side of the equation. You do this by dividing both sides of the equation by [tex]\( 4.4 \times 10^{-4} \)[/tex]:
[tex]\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}} \][/tex]
4. Perform the division:
[tex]\[ K_a = \frac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}} \approx 2.2727272727272724 \times 10^{-11} \][/tex]
Therefore, the acid dissociation constant ([tex]\( K_a \)[/tex]) for the conjugate acid of [tex]\( CH_3NH_2 \)[/tex] is approximately [tex]\( 2.27 \times 10^{-11} \)[/tex].
