Question

Determine the pH during the titration of 36.3 mL of 0.281 M trimethylamine ((CH3)3N, Kb = 6.3×10-5) by 0.281 M HBr at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem.

(a) Before the addition of any HBr ?

(b) After the addition of 15.7 mL of HBr ?

(c) At the titration midpoint ?

(d) At the equivalence point ?

(e) After adding 57.7 mL of HBr ?

Answer 1

(a) Before the addition of any HBr ?

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!)

6.3*10^-5 = x*x/(0.281-x)

solve with quadratic formula

x = [OH-] = 0.004176

pOH = -log(Oh-) = -log(0.004176) = 2.38

pH = 14-pOH = 14-2.38= 11.62

(b) After the addition of 15.7 mL of HBr ?

this is a buffer so

mmol of base = MV = 0.281*36.3 = 10.2003

mmol of acid = MV = 0.281*15.7 = 4.4117

mmol of bas eleft = 10.2003 -4.4117 = 5.7886

mmol of conjguate formed = 4.4117

pH = pKa + log(B/BH+)

pH = 14-pKb = (14) - (-log(Kb) = 14 + log(6.3*10^-5) = 9.79

pH = 9.79 + log(5.7886/4.4117)

pH = 9.907

(c) At the titration midpoint ?

in the midpoint, Conjugate acid = weak base

so

pH = pKa + log(B/BH+); B = BH+

then

log(1)

pH = pKa + 0

pH = pKa

pH = 9.79

(d) At the equivalence point ?

mmol of base= MV = 36.3*0.281 = 10.2003

Vbase= mmol/M = 10.2003/0.281= 36.3 mL

this is equivalence point!

mmol of acid = mmol of base

so..

all base/acid reacts

Vtotal = V1+V2 = 36.3 +36.3 = 72.6 mL of total solution

calculate conjugate formed

recalcualte concentration of conjugate

[BH+] = mmol/Vt = 10.2003/72.6 = 0.1405 M of BH+

so..

BH+ + H2O <-> H3O+ + B

so

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

Ka = Kw/Kb = (10^-14)/(6.3*10^-5) = 1.58*10^-10

so

[H3O+] = x = [B]

[BH+] = M-x = 0.1405-x

substitute

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

becomes

1.58*10^-10= x*x/(0.1405-x)

solve for x

x = 3.61*10^-6

[OH-] = x =  3.61*10^-6

pOH = -log(OH-) -log(3.61*10^-6) = 5.44

pH = 14-pOH = 14-5.44= 8.56

ph = 8.56

(e) After adding 57.7 mL of HBr ?

mmol of acid = MV = 57.7*0.281 = 16.2137

mmol of base =  = MV = 36.3*0.281 = 10.2003

mmol of acid left = 16.2137-10.2003 = 6.0134

[H+] 0 mmol/V = (6.0134) / (57.7+36.3) =

[H+] = 0.063

pH = -log(0.063 = 1.20