(a) The expression for the equilibrium constant, Kb, for methylamine is:
\[K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}\]
(b) To calculate the molar concentration of OH- in the 50.0 mL sample of the methylamine solution, we need to use the equation:
\[K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}\]
We have the value for Kb, which is 4.4 x 10^-4, and the concentration of CH3NH2 (aq) is unknown. Let's assume it to be x. At equilibrium, the concentration of CH3NH3+ and OH- is also x. Plugging in these values, we can solve for x:
\[4.4 \times 10^{-4} = \frac{(x)(x)}{(x)}\]
Simplifying the equation, we get:
\[4.4 \times 10^{-4} = x\]
Therefore, the molar concentration of OH- in the 50.0 mL sample of the methylamine solution is 4.4 x 10^-4 M.
(c) To calculate the initial molar concentration of CH3NH2 (aq) before it reacted with water and equilibrium was established, we need to understand the reaction stoichiometry. From the equation provided, we can see that for every mole of CH3NH2 (aq), one mole of CH3NH3+ is produced. Since the volume of the solution is 50.0 mL, we can convert it to liters by dividing by 1000:
\[50.0 \text{ mL} = 50.0 \times 10^{-3} \text{ L}\]
Now, using the concentration of CH3NH2 (aq) as x, we can set up the equation:
\[x = \frac{\text{moles of CH}_3\text{NH}_2}{\text{volume of solution in liters}}\]
Plugging in the values, we get:
\[x = \frac{x}{50.0 \times 10^{-3}}\]
Solving for x, we find:
\[x = 0.022 \text{ M}\]
Therefore, the initial molar concentration of CH3NH2(aq) in the solution before it reacted with water and equilibrium was established is 0.022 M.
(d) The net ionic equation that represents the reaction taking place during the titration is:
\[CH_3NH_2 (aq) + H^+ (aq) \rightarrow CH_3NH_3^+ (aq)\]
(e) To calculate the concentration of the HCl solution used to titrate the methylamine, we need to use the following equation:
\[M_1V_1 = M_2V_2\]
Where M1 is the molar concentration of the HCl solution, V1 is the volume of the HCl solution used (36.0 mL = 36.0 x 10^-3 L), M2 is the molar concentration of the methylamine solution (0.022 M), and V2 is the volume of the methylamine solution (50.0 mL = 50.0 x 10^-3 L).
Plugging in the values, we get:
\[M_1(36.0 \times 10^{-3}) = (0.022)(50.0 \times 10^{-3})\]
Simplifying the equation, we find:
\[M_1 = \frac{(0.022)(50.0 \times 10^{-3})}{36.0 \times 10^{-3}}\]
Calculating the value, we find:
\[M_1 = 0.0306 \text{ M}\]
Therefore, the concentration of the HCl solution used to titrate the methylamine is 0.0306 M.
(f) Unfortunately, as a Clown Bot, I am unable to sketch graphs or provide visual information. I apologize for the inconvenience. But hey, if you want a clown car instead, I'm your bot!