Question

part a:

Consider the titration of a 23.4 −mL sample of 0.125 MRbOH with 26.6mL 0.110 M HCl.

1a. If the pH oafter 5.8 mL of added acid is 12,89 what is the pH after adding 5.8 mL of acid beyond the equivalence point?

part b

Consider the titration of a 28.0-mL sample of 0.170 MCH3NH2 with 0.155 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

1b. what is the pH at 6.0 mL of added acid?

2b. what is the pH after adding 6.0 mL of acid beyond the equivalence point?

Answer 1

A)

V = 23.4 mL of M = 0.125 M of RbOH

V = 26.6 mL of M = 0.11 M of HCl

1a)

First, find the equivalence point of the base:

mmol of base = MV = (0.125)(23.4*10^-3) = 0.002925 mmol of base

then, equivalence point for acid = 0.002925 mmol of acid

Volume = mmol/M = (0.002925)/(0.11) = 0.026590 Liters = 26.59 mL

after addition of 5.8 mL of acid:

Vtotal acid = 26.59 +5.8 = 32.39

Vtotal solution = Vbase+ Vacid = (23.4 +32.39) = 55.79

mmol of HCl added = (MV) = (5.8*0.11) = 0.638 mmol

[H+] = mmol/V = (0.638)/55.79 = 0.011435 M

pH = -log(H) = -log(0.011435) = 1.94

b)

titration of weak base + storng acid

1a

pH when V = 6 mL of acid is added

This is in a Buffer Region, a Buffer is formed (weak base and its conjugate)

Use Henderson-Hassebach equations!

mmol of acid added = MV = 0.155*6 = 0.93 mmol of acid

mmol of base present = MV = (0.170)(28) = 4.76 mmol of base

now, after reaction

mmol of bas eleft after reaction = 4.76 - 0.93 = 3.83 mmol

mmol of conjugate formed = 0 + 0.93 = 0.93 mmol

this is a BASIC buffer,

pOH = pKb + log(HB+/B)

substitute data

pKb = -log(Kb=) -log(4.4*10^-4) = = 3.36

pOH = 3.36 + log(0.93 /3.83)

pOH = 2.745

pH = 14-pOH = 14-2.745

pH = 11.255

2b

pH after adding 6 mL beyond the equivalnece point

base left = 0

acid left = MV = (0.155)(6) = 0.93 mmol of acid

Total volume added: volume of base + volume of acid

volume of base = 28 mL

volume of acid = Vequivalence + Vextra = (Mbase*Vbase/Macid) + 6 = (28*0.17/0.155) + 6 = 36.709 mL

so

Vtotal = 28 + 36.709 = 64.709 mL

[acid] = mmol of acid /( vtotal )= (0.93 )/(64.709) = 0.014372 M

so

[H+] = [HBr] = 0.014372

pH = -log(H) = -log(0.014372 = 1.84248