Answer:
[CH₃NH₃⁺] = 0.00355 M
[CH₃NH₂] = 0.028 M
pH = 11.55
Explanation:
The reaction is the following:
CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻ (1)
0.0317 - x x x
The equilibrium constant of the reaction (1) is:
[tex] K_{b} = \frac{[CH_{3}NH_{3}^{+}][OH^{-}]}{[CH_{3}NH_{2}]} [/tex]
[tex] 4.47\cdot 10^{-4} = \frac{x \cdot x}{0.0317 - x} [/tex]
[tex] x^{2} - 4.47 \cdot 10^{-4}(0.0317 - x) = 0 [/tex] (2)
By solving equation (2) for x, we have:
x₁ = -0.00399
x₂ = 0.00355
Taking the positive value, we have that:
x = [CH₃NH₃⁺] = [OH⁻] = 0.00355 M
[CH₃NH₂] = 0.0317 - x = (0.0317 - 0.00355)M = 0.028 M
Therefore, the concentrations of CH₃NH₂ and CH₃NH₃⁺ are 0.028 M and 0.00355 M, respectively.
The pH of the solution is:
[tex] pOH = -log [OH^{-}] = -log (0.00355) = 2.45 [/tex]
[tex] pH = 14 - pOH = 14 - 2.45 = 11.55 [/tex]
Hence, the pH of the solution is 11.55.
I hope it helps you!
