Question

Consider the titration of 30.0 mL of 0.0700 M (CH₃)₃N (a weak base; K_b = 6.40e-05) with 0.100 M HCl. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH = _________ (b) 5.3 mL pH = __________ (c) 10.5 mL pH = ____________
(d) 15.8 mL pH = _____________ (e) 21.0 mL pH =___________ (f) 39.9 mL pH = __________

Answer 1

For:

(CH3)3N; let us assume (CH3)3N= B for simplicity

so

(CH3)3N+ H2O <--> (CH3)3NH+ + OH- becomes:

B + H2O <-> HB+ + OH-

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

so....

a)

no volume of acid

B + H2O <-> HB+ + OH-

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

in equilibrium

for 1 mol of OH- we hava always 1 mol of HB+

[HB+] = [OH-] = x (the dissociation fraction of acid)

[B] = M-x = 0.07-x (Account for dissociation!)

6.4*10^-5 = x*x/(0.07-x)

solve with quadratic formula

x = [OH-] = 0.002084

pOH = -log(Oh-) = -log(0.002084) = 2.68110

pH = 14-pOH = 14-2.68110 = 11.3189

b)

5.3 mL of acid:

mmol of acid added = MV = 0.1*5.3 = 0.53 mmol of acid

this reacts with base:

mmol of base = MV = 30*0.07 = 2.1 mmol of base

mmol of base after reaction = 2.1 -0.53 = 1.57 mmol of base

and conjugate forms:

mmol of conjugate formed = 0 + 0.53 = 0.53

this becomes a buffer, sicne there is weak base+ conjugate acid

pOH = pKb + log(BH+/B)

pKB = -log(Kb) = -log(6.4*10^-5) = 4.1938

so

pOH = 4.1938 + log(0.53/1.57 )

pOH = 3.722

pH = 14-pOH = 14-3.722 = 10.278

c)

V = 10.5 mL of acid

mmol of acid = MV 0 0.1*10.5 = 1.05 mmol of acid

mmol of base after reaction = 2.1-1.05 = 1.05

mmol of conjugate formed = 0 + 1.05 = 1.05

this is also a buffer... special one , since base = conjugate acid

so

pOH = pKb + log(BH+/B)

pOH = 4.1938 + log(1.05/1.05 )

pOH = pKb = 4.1938

pH = 14-4.1938 = 9.8062

d)

V = 15.8 mL of acid

mmol oacid = MV = 15.8*0.1 = 1.58 mmol of acid

mmol of base after reaction = 2.1-1.58 = 0.52 mmol of base

mmol of conjguate formed = 0 + 1.58 = 1.58

then

pOH = pKb + log(BH+/B)

pOH = 4.1938 + log(1.58/0.52 )

pOH = 4.676

pH = 14-pOH = 14-4.676

pH = 9.324

e)

V = 21 mL of acid

mmol of acid = MV = 21*0.1 = 2.1

this is equivalence point!

mmol of acid = mmol of base

so..

all base/acid reacts

Vtotal = V1+V2 = 21+30 = 51 mL of total solution

calculate conjugate formed

mmol of conjguate formed = MV = 21*0.1 = 2.1 mmol of conjugate

recalcualte concentration of conjugate

[BH+] = mmol/Vt = 2.1/51 = 0.041176 M of BH+

so..

BH+ + H2O <-> H3O+ + B

so

Ka = [H3O+][B]/[BH+]

Ka = Kw/Kb = (10^-14)/(6.4*10^-5) = 1.562*10^-10

so

[H3O+] = x = [B]

[BH+] = M-x = 0.041176 -x

substitute

Ka = [H3O+][B]/[BH+]

becomes

1.562*10^-10 = x*x/(0.041176 -x)

solve for x

x = 2.535*10^-6

[OH-] = x =  2.535*10^-6

pOH = -log(OH-) -log( 2.535*10^-6 ) = 5.59602203

pH = 14-pOH = 14-5.59602203 = 8.4039

f)

finally...

mmol of acid added = MV = 39.9*0.1 = 3.99 mmol of acid

mmol of base = 2.1

mmol of acid left = 3.99-2.1 = 1.89 mmol of acid left

Vtotal = 39.9+30 = 69.9 mL

[acid] = 1.89/69.9 = 0.02703M of acid

pH = -log(H+) = -log(0.02703)

pH = 1.568153