Question

Determine the pH of a 0.35 M aqueous solution of CH₃NH₂ (methylamine). The K_b of methylamine is 4.4 × 10⁻⁴.     

can you please show your work and explain the steps on how you get 12.09 as the answer

Answer 1

The dissociation reaction with ICE TABLE

CH3NH2 + H2O = CH3NH3+ + OH-

I 0.35

C - x +x +x

E 0.35-x x x

Equilibrium constant expression of the reaction

Kb = [CH3NH3+][OH-]/[CH3NH2]

4.4*10^-4 = (x)(x)/(0.35 - x)

x^2 = 1.54*10^-4 - 4.4*10^-4x

Since Kb is small

0.35 >> x

x^2 = 1.54*10^-4

x = 1.24*10^-2

[OH-] = x = 1.24*10^-2 M

pOH = - log [OH-] = - log (1.24*10^-2)

pOH = 1.9

pH = 14 - pOH = 14 - 1.91

pH = 12.09