Question

A 125.0 mg sample of methylamine (CH3NH2 MW = 31.06) was dissolved in 25.00 mL of carbonate-free water. {The Ka for CH3NH3+ is 2.3 x 10-11.} Determine the pH of the resulting solution after the addition of each of the following solutions.

A) 25.00 mL of pure carbonate-free water

B) 25.00 mL of 0.1609 M aqueous methylamine hydrochloride solution

C) 25.00 mL of 0.0512 M aqueous HCl solution

Answer 1

a) in pure water, this is a weak base in water

This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibirum Kb:

Kb = [HB+][OH-]/[B]

let "x" be OH- in solution

in equilibrium due to sotichiometry:

[HB+]= x= [OH-]

Account for the dissolved base in solution vs. not in solution:

moles = mass/MW = (123*10^-3)/(31.06) = 0.003960 mol

Vtotal = 25 + 25 = 50 mL

[B]initially = mol/V = (0.003960)/(50*10^-3) = 0.0792 M

then, in equilibrium:

[B] = 0.0792-x

we also need Kb, not Ka

Kb = Kw/Ka = (10^-14)/(2.3*10^-11) = 0.0004347

Substitute in Kb

Kb = [HB+][OH-]/[B]

(0.0004347)= x*x/(0.0792-x)

Solve for x, using quadratic equation

x = OH- = 0.00565

pOH = -log(OH-) = -log( 0.00565) = 2.2479

pH = 14-pOH = 14-2.2479

pH = 11.7521

b)

assume total volume = 50 mL.... then

mmol of CH3NH3Cl = M*V = 25*0.1609 = 4.0225 mmol of conjguate

mmol of CH3NH2 = 0.003960 mol = 3.96 mmol of base

pH = pKa+ log(B/BH+)

pKA = -log(Ka) = -log(2.3*10^-11) = 10.638

pH = 10.638 + log(3.96/4.0225) = 10.6311

C)

If we add 0.0512 M of HCl

mmol of HCL = 25*0.0512 = 1.28 mmol of H+

mmol of CH3NH2 = 0.003960 mol = 3.96 mmol of base

after H+`reacts

mmol of CH3NH2 = 3.96-1.28 = 2.68

mmol of CH3NH+ formed = 0 + 1.28 = 1.28

substitute in pH:

pKA = -log(Ka) = -log(2.3*10^-11) = 10.638

pH = 10.638 + log(2.68/1.28) = 10.958