2.00mol of the weak base methylamine, NH2CH3, is added to water to make up 1.500L of solution. The Kb of hydroxylamine is 4.4×10−4. What is the pH of the solution?

Question

2.00mol of the weak base methylamine, NH2CH3, is added to water to make up 1.500L of solution. The Kb of hydroxylamine is 4.4×10−4. What is the pH of the solution?

Answer 1

CH₃NH₂ + H₂O ⇔ CH₃NH₃⁺ + OH^-

Molarity of CH₃NH₂ = 2.00 mol / 1.500 L = 1.33 M

K_b = [CH₃NH₃⁺][OH^-] / [CH₃NH₂] = 4.4 x 10^-4

Setup an ICE table. At equilibrium,

[CH₃NH₂] = 1.33 M - x

[CH₃NH₃⁺] = x

[OH^-] = x

Substitute these values into the K_b expression, you get

(x) (x) / (1.33 - x) = 4.4 x 10^-4

assuming x is significantly smaller than 1.33, the above expression simplifies to

x² / 1.33 = 4.4 x 10^-4

x² = 4.4 x 10^-4 x 1.33

x² = 5.9 x 10^-4

x = √5.9 x 10^-4 = 2.4 x 10^-2

x = [OH^-] = 2.4 x 10^-2

pOH = -log [OH^-] = -log 2.4 x 10^-2 = 1.62

pH = 14.00 - 1.62 = 12.38