Methylamine (CH₃NH₂) is a weakly basic compound. Calculate the K_b for methylamine if a 0.253 M solution is 4.07% ionized.

Question

Quizplus · Accepted Answer

To find the $K_b$ for methylamine, we first calculate the concentration of ionized methylamine. If the solution is 4.07% ionized, then $0.253 \times 0.0407 = 0.0103$ M of methylamine is ionized. Since methylamine is a weak base, it ionizes in water to form $CH_3NH_2 + H_2O \rightarrow CH_3NH_3^+ + OH^-$. The concentration of $OH^-$ will also be 0.0103 M because the reaction produces one hydroxide ion for each molecule of methylamine that ionizes. The $K_b$ expression is $K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$. The initial concentration of $CH_3NH_2$ is 0.253 M, but we need to subtract the ionized portion, which is negligible in this case, leaving approximately 0.253 M. Therefore, $K_b = \frac{(0.0103)(0.0103)}{0.253} = 4.19 \times 10^{-4}$.

Methylamine (CH3NH2) Is a Weakly Basic Compound ×\times× 103 B)4 ×\times× 10-4 C)5

Methylamine (CH₃NH₂) Is a Weakly Basic Compound $\times$ 10³
B)4 $\times$ 10^-4
C)5