Question

2.

For the titration of 20 mL of 0.1026 M NH₃ with 0.09747 M HCl, calculate the pH before the addition of titrant, at 10 mL prior to the equivalence point, at the equivalence point, and at 10 mL after the equivalence point. K_a (NH₄⁺) = 5.7 x 10^-10

Answer 1

a)

before addition

NH3 + H2O <-> NH4+ + OH-

Kb = [NH4][OH-]]/[NH3]

Kb = Kw/Ka = (10^-14)/(5.7*10^-10) = 1.78*10^-5

1.78*10^-5 = x*x/(0.1026-x)

x = OH = 0.00134

pH = 14+pOH = 14+ log(0.00134) = 11.127

b)

10 mL prior equiv.

Veq = M1*V1/M2 = (0.1026*20)/(0.09747 ) = 21.05 mL

10 mL before --> 21.05-10 = 11.05 mL

so

this is a basic buffer

pOH = pKb + log(NH4+/NH3)

pKB = -log(Kb= -log(1.8*10^-5) = 4.75

mol of acid = Macid*Vacid = 0.09747 *11.05

mmol of NH3 = 20*0.1026 - 1.0770435 = 0.9749 mmol of NH3

mmol of NH4+ = 0+1.0770435 = 1.0770435

pOH = 4.75 + log(1.0770435 / 0.9749 ) = 4.7932

pH = 14´-4.7932 = 9.2068

c)

equiv. point:

[NH4+] = mmol of NH4 / total V = (2.052) / (20+21.05) = 0.04998

Ka = [NH3][H+]/[NH4+]

5.7*10^-10 = x*x/(0.04998-x)

x = H+ = 5.337*10^-6

pH= -log( 5.337*10^-6

pH = 5.27

d)

after extra acid is added

mmol of acid = MV = 10*0.09747 = 0.9747 mmol of H+ extra

Total V = V1+V2 = 20 + (10+21.05) = 51.05

[H+] = mmol/V = 0.9747/51.05 = 0.01909 M

pH = -log(0.01909) = 1.71919