Question

A solution containing 50 mL of 2 x 10 -3 M of an indicator H 2 In is measured to have an absorbance of 0.240 at 540 nm with a 1 cm thick cuvette. After 50 mL of 2.8 x 10 -3 M NaOH was added to the solution, the absorbance became 0.048. A second solution containing 50 mL of 2 x 10 -3 M of the same indicator in its basic form, In = , is measured to have an absorbance of 0.060 at 540 nm. After 50 mL of 6 x 10 -4 M HC1 was added, the absorbance became 0.039. The indicator is known to have dissociation constants of K 1 = 4.0 x 10 -7 , K 2 = 6.67 x 10 -8 .

a) What is the pH of the original solution of H 2 In?

b) What is the pH after the NaOH was added to it?

c) If I mix the first and second solutions together (i.e. the H 2 In solution with the NaOH added plus the In = solution with the HC1 added), what pH will I get?

d) What are the molar absorptivities of H 2 In, HIn - , and In = ?

e) What is the absorbance of the mixed solution in part c)?

Answer 1

a) the dissociation equation will be

                      H2In --> H+ + HIn-

Initial conc       0.002    0        0

change           -x           +x     +x

equilibrium     0.002-x     x      x

Ka = 4.0 x 10 -7 = x^2 / 0.002-x

we may ignore x in denominator

4.0 x 10 -7 = x^2 / 0.002

x = 2.82 X 10^-5 = [H+]

pH = -log[H+] = 4.55

b) Moles of indicator present = Molarity X volume = 0.002 X 50 X 10^-3 = 0.0001 moles

Moles of base added = Molarity X volume = 0.0028 X 50 X 10^-3 = 0.00014 moles

one mole of indicator will react with two moles of base

so the moles of base reacted = 0.00014,

so 0.0001 moles of base will neutralize H2In to HIn-

Then 0.00004 moles of base will react with HIn- to give 0.00004 moles of In-2

The moles of HIn- left = 0.0001 - 0.00004 = 0.00006

this will make a buffer

pH = pKa2 + log [salt] / [base]

pKa2 = -logKa2 = -log6.67 x 10 -8 = 7.18

pH = 7.18 + log [0.00004]/ [ 0.00006] = 7.003

c) moles of In-2 present = molarity X volume = 0.002 X 50 X 10^-3 = 0.0001

Moles of acid added = 6 X 10^-4 X 50 X 10^-3= 0.00003 moles

The moles of In-2 reacted = 0.00003 moles

Moles of In-2 left = 0.00007 moles

pH = pKa + log [salt] / [acid]

pH = 7.18 + log [0.00007 / 0.00003] = 7.55

d) We know from Beer Lambert's law

Absorbance = Concentration X molar absroptitivity X length

Molar absorptivity = Absrobance X path length / concentration

(i) H2In :

Molar absorptivity = 0.240 X 1 cm / 0.002 = 120 cm mole^-1 L

(ii) In-2

Moalr absroptivity = 0.06 X 1 / 0.002 = 30120 cm mole^-1 L