Lab: Spectrophotometric determination of K Student: Key Concepts: beer\'s law an

Question

Lab: Spectrophotometric determination of K Student: Key Concepts: beer's law and mixtures, equilibrium constants, pH Pre-Lab Question(s): Elisha wants to determine the Ka of an acid-base indicator. Elisha knows that the solid form of the indicator is protonated, so that Hin H'+ In. Based on this equilibrium in water, what is the expression for K,? Elisha made an acidic solution, a basic solution, and an intermediate solution of the K as directed to in the procedure. The spectra of the intermediate solution was determined as shown below, at pH-4.70. 0.7 g 0.6 0.5 2 0.4 0.3 0.2 0.1 0.0 400 450 500 55 600 650 700 Wavelength (nm) What wavelength(s) should Elisha use to measure all of the samples? Elisha measured the absorbance at the Low and High wavelength for each sample. Given the absorbances recorded below: What is the molar absorptivity of each species of indicator at each wavelength? What is the concentration of each species in the intermediate solution? What is the Ka? AbsorbanceLow High Acid (pH=2.5) 10.119 0.634 Base (pH-10.5) 0.996 0.164 intermediate 0.585 0.385 [Injtotal 0.07000 0.07000

Explanation / Answer

1) For the rxn, HIn <--->H+ +In-

Ka=[H+] [In-]/[HIn] (law of mass action)

2)Maximum absorbance is observed at wavelength=450 nm (lambda max) [from spectra of intermediate]

Elisha should use this value to measure Absorbance (to get sharp peaks)

3)

Using Beer's law ,

Absorbance=e*l*C

C=concentration of absorbing species

where e=absorptivity of species

l=path length of light(1 cm)

So, Absorbance for (Acidic solution with HIn)=0.634 (high)

Abs=C(HIn)* el

or, 0.634=C(HIn)* el=0.070M*el

or, e=0.634/0.070M*(1cm) =9.057 M^-1 cm^-1 [l=1cm)

for low wavelength,

e=0.119/0.070M*(1cm) =1.7 M^-1 cm^-1

for (basic solution),high

Abs=C(In-)*el

or, e=0.996/0.07M(1cm)=14.228 M^-1 cm^-1

for (basic solution),low

Abs=C(In-)*el

or, e=0.164/0.07M(1cm)=2.343 M^-1 cm^-1

4) Intermediate

[In]total=0.07

[HIn]+[In]=[In]total

Abs(total)=C(HIn)*e(HIn)*l +C(In)*e(In)*l

for high wavelength,*e(HIn)=9.057 M^-1 cm^-1

e(In-)=14.228 M^-1 cm^-1

or, 0.585=C(HIn)*(9.057 M^-1 cm^-1)(1cm)+C(In-)*(14.228 M^-1 cm^-1)*(1cm)

or, 0.585=(0.07M-C(In-)*(9.057 M^-1)+C(In-)*(14.228 M^-1)           [C(HIn)=0.07M-C(In-)]

or,0.585=0.634 - C(In-)*(9.057 M^-1) +C(In-)*(14.228 M^-1)  

or, 0.049=5.171 C(In-)

C(In-)=0.049/5.171=0.00947M

C(HIn)=0.07-0.00947=0.0605M

C(HIn)=0.0605M

C(In-)=0.00947M

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