Calculate the pH of a 0.100 M aqueous solution of ethylamine CH₃CH₂NH₂.

Data: K_b [CH₃CH₂NH₂] = 4.30 x 10^-4

Question

Calculate the pH of a 0.100 M aqueous solution of ethylamine CH₃CH₂NH₂.

Data: K_b [CH₃CH₂NH₂] = 4.30 x 10^-4

| General Chemistry 3

Accepted Answer

Acido-basic reaction:

CH₃CH₂NH₂ (aq) + H₂O (l) $⇌$ CH₃CH₂NH₃⁺ (aq) + HO^- (aq) [K_b]

A t = 0:

[CH₃CH₂NH₂] = 0.100 M and [CH₃CH₂NH₃⁺] = [HO^-] = 0 M

At the equilibrium:

[CH₃CH₂NH₂] = 0.100 - X M and [CH₃CH₂NH₃⁺] = [HO^-] = X M

K_b = $\frac{[{CH}_{3} {CH}_{2} {NH}_{3}^{+}] [{HO}^{-}]}{[{CH}_{3} {CH}_{2} {NH}_{2}]}$ = $\frac{X^{2}}{0.100 - X}$

Let’s suppose 0.100 >> X

K_b = 4.30 x 10^-4 = $\frac{X^{2}}{0.100}$

⇒ X = 6.56 x 10^-3 (X cannot be < 0)

⇒ the hypothesis 0.100 >> X is verified

pOH = - log [HO^-] = 2.18

pH = 14 – pOH = 11.8

Midterm 1 - Question 3 | General Chemistry 3