Question

In the titration of 40.00 mL of 0.500 M methyl amine, CH3NH2 (Kb = 4.4 x 10–4) with 0.200 M HCl, calculate the pH of the resulting solution when the titration is 1/4 of the way to the equivalence point.

Answer 1

V = 40 ml = 0.04 L

M = 0.5 methyl amine CH3NH2

Kb = 4.4*10^-4

M = 0.2 Hcl

V = ?

1/4 way of the equilalence point

pH = ?

first,

identify the equivalence point

equivalence point: mol of acid = mol of base

M1V1 = M2V2

0.5*40 = 0.2*V2

V2 = 0.5*40/0.2 = 100 ml of HCl

then

1/4 of way must be 100/4 = 25 ml of acid added

Total volume then = 40 ml base + 25 ml acid = 65 ml

This is a basic buffer since we have

conjugate + base

pOH = pKb + log(CH3NH3+ / CH3NH2 )

CH3NH3+ = 0 + M*V = 25*0.2 = 5 mmol of CH3NH3+

CH3NH2 = (40)(0.5) - MV = 20-5 = 15 mmol of CH3NH2

then

pKb = -log(Kb) = -log(4.4*10^-4) =3.35654

pOH = pKb + log(CH3NH3+ / CH3NH2 )

pOH = 3.35654+ log(5/ 15)

pOH = 2.8794

pH = 14-2.8794 = 11.1206

pH = 11.1206