An acid-base indicator, Hln, dissociates according to the following reaction in

Question

An acid-base indicator, Hln, dissociates according to the following reaction in an aqueous solution. HIn(aq) In^- (aq) + H^+ (aq) The protonated form of the Indicator, Hln, has a molar absorptivity of 2509 M^-1.cm^-1 and the deprotonated form, In^-, has a molar absorptivity of 17190 M^-1-.cm^-1 at 440 nm. The pH of a solution containing a mixture of Hln and In^- is adjusted to 5.84. The total concentration of Hln and In^- is 0.000158 M. The absorbance of this solution was measured at 440 nm in a 1.00 cm cuvette and was determined to be 0.856. Calculate pKa for Hln.

Explanation / Answer

we know that

according to beers law

Absorbance = molar absorptivity x path length x concentration

A = e x l x c

here path length is 1 cm

so

A = e x c

for a solution with many species

A = e1 x c1 + ( e2 x c2)

given

e1 = 2509

c1 = [HIn]

e2 = 17190

c2 = [In-]

A = 0.856

so

0.856 = 2509 [HIn] + 17190 [In-]

now

given

[Hin] + [In-] = 0.000158

so

[In-] = 0.000158 - [HIn]

so

0.856 = 2509 [HIn] + 17190 [In-]

0.856 = 2509 [HIn] + 17190 ( 0.000158 - [HIn]

0.856 = 2509 [HIn] + 2.71602 - 17190 [HIn]

[HIn] = 1.267 x 10-4

so

[In-] = 0.000158 - 1.267 x 10-4

[IN-] = 3.13 x 10-5

now

according to hasselbach hendersen equation

pH = pKa + log [In-] / [HIn]

given

pH = 5.84

so

5.84 = pKa + log 3.13 x 10-5 / 1.267 x 10-4

pKa = 6.447

so

the pKa for HIn is 6.447

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