Question

Consider the titration of 50.0 mL of 0.0500 M C₂H₅NH₂ (a weak base; K_b = 0.000640) with 0.100 M HIO₄. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 6.3 mL pH =	(c) 12.5 mL pH =
(d) 18.8 mL pH =	(e) 25.0 mL pH =	(f) 30.0 mL pH =

Answer 1

pH of a Monoprotic Base

This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibirum Kb:

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

let "x" be OH- in solution

in equilibrium due to sotichiometry:

[HB+]= x= [OH-]

Acocunt for the dissolved base in solution vs. not in solution:

[B] = 0.05-x

Substitute in Kb

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

Kb = 0.000640 = 6.4*10^-4

6.4*10^-4= x*x/(0.05-x)

Solve for x, using quadratic equation

x = OH- = 0.00534

pOH = -log(OH-) = -log(0.00534) = 2.2724

pH = 14-pOH = 14-2.2724

pH = 11.7276

b)

this will form a buffer

pKb = -log(Kb) = -log(0.00064) = 3.193820

pOH = pKb + log(HB+/B)

pOH = 3.193820 + log(HB+/B)

mmol of acid = MV = 0.1*6.3 = 0.63 mmol of H+

mmol of HB+ formed = 0 + 0.63 = 0.63

mmol of B left = 50*0.05 - 0.63 = 1.87

pOH = 3.193820 + log(0.63/1.87) = 2.7213

pH = 14-2.7213 = 11.2787

c)

V = 12.5

this will form a buffer

pOH = pKb + log(HB+/B)

pOH = 3.193820 + log(HB+/B)

mmol of acid = MV = 0.1*12.5= 1.25 mmol of H+

mmol of HB+ formed = 0 + 1.25 = 1.25

mmol of B left = 50*0.05 - 1.25 = 1.25

pOH = 3.193820 + log(1.25 /1.25 ) = 3.193820

pH = 14-3.193820 = 10.80618

d)

V = 18.8

this will form a buffer

pOH = pKb + log(HB+/B)

pOH = 3.193820 + log(HB+/B)

mmol of acid = MV = 0.1*18.8= 1.88 mmol of H+

mmol of HB+ formed = 0 + 1.88 = 1.25

mmol of B left = 50*0.05 - 1.88 = 0.62

pOH = 3.193820 + log(1.25 /0.62) = 3.4983

pH = 14-3.4983= 10.5017

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