Final answer:
The hydroxide ion concentration ([OH-]) in a 0.33 M methylamine solution is 0.0387 M and the pH is 12.59, calculated using the base dissociation constant (Kb) and standard equilibrium equations for weak bases.
Step-by-step explanation:
The question asks to find the hydroxide ion concentration ([OH⁻¹]) and pH of a 0.33 M Methyl Amine solution given its base dissociation constant (Kb = 4.4 x 10⁻⁴). Methyl amine is a weak base, and its reaction with water can be represented as:
CH3NH2 + H2O ⇌ CH3NH3+ + OH⁻¹
Assuming the degree of ionization is small, we can use the approximation that the concentration of OH⁻¹ is much smaller than the initial concentration of methylamine, so the equilibrium concentrations can be described as follows:
[CH3NH2] ≈ 0.33 M - x
[OH⁻¹] = x
Where x is the change in concentration due to the ionization. The equilibrium expression for the base can be written as:
Kb = \frac{[CH3NH3+][OH⁻¹]}{[CH3NH2]} \approx \frac{x2}{0.33}
Solving for x:
x = \sqrt{Kb × [CH3NH2]} = \sqrt{4.4 x 10⁻⁴ × 0.33} ≈ 0.0387 M
The hydroxide ion concentration is equal to x, which is 0.0387 M. To find the pH, we first calculate the pOH:
pOH = -log [OH⁻¹] = -log (0.0387) ≈ 1.41
To convert pOH to pH, we use the relationship pH = 14 - pOH:
pH = 14 - 1.41 = 12.59
Thus, the hydroxide ion concentration in a 0.33 M methylamine solution is 0.0387 M and the pH is 12.59.