Henderson Hasselbach to absorbance derivation

[Puchinita5]

Homework Statement

Did a lab where I had to determine the Ka value of an indicator using spectrophotometry.

This is a question I need to answer in my lab report:

Derive the equation

pH = pKa + log([A-Aa]/[Ab-A])

from the equation

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

using Beer's law.

A: measured absorbance at λb at specified pH
Aa: absorbance of acidic form at λb
Ab: absorbance of basic form at λb
λb: wavelength of maximum absorbance for basic form

Homework Equations

The Attempt at a Solution

A1 = ε_1ab[HIn] + ε_1bb[In-]
A2 = ε_2ab[HIn] + ε_2bb[In-]

Aa = ε_ab[HIN]
Ab = ε_bb[IN-]

Ab/Aa = ε_bb[IN-] / ε_ab[HIN]

Rearrange:

[In-] / [HIn] = ε_aAb / ε_bAa

A = Aa+Ab

so

[In-] / [HIn] = ε_a(A-Aa) / ε_b(A-Ab)

I'm stuck. I have been working on this for hours and cannot figure out how to get to the answer. Mainly, I just don't know how to get the molar absorbtivities to help me because they are all different. I'm so confused. Can someone steer me in the right direction? I also tried solving using the first two equations:

A1 = ε_1ab[HIn] + ε_1bb[In-]
A2 = ε_2ab[HIn] + ε_2bb[In-]

which led me to

[In-] / [HIn] = (A1ε_2a-A2ε_1a ) / (A2ε_1b - A1ε_2b)

which doesn't help me either because I can't see how to get rid of the molar absorbtivities.


HELP

Thank you.

 

[mighty2000]

I understand that you are struggling with deriving the equation pH = pKa + log([A-Aa]/[Ab-A]) from the given equation pH = pKa + log([In-]/[HIn]) using Beer's law. Here is a step-by-step guide on how to approach this problem:

1. Start by writing out the Beer's law equation for the acidic form of the indicator at the given wavelength (λb). This will give you: Aa = εab[HIn]

2. Next, write out the Beer's law equation for the basic form of the indicator at the same wavelength (λb). This will give you: Ab = εbb[In-]

3. Since the absorbance (A) at λb is the sum of the absorbance of the acidic and basic forms, you can write the equation A = Aa+Ab.

4. Substitute the equations from steps 1 and 2 into the equation from step 3. This will give you: A = εab[HIn] + εbb[In-]

5. Rearrange this equation to get [In-]/[HIn] on one side and everything else on the other side. This will give you: [In-]/[HIn] = (A-εab[HIn]) / εbb

6. Now, use the given equation pH = pKa + log([In-]/[HIn]) and substitute the expression from step 5 for [In-]/[HIn]. This will give you: pH = pKa + log((A-εab[HIn]) / εbb)

7. Finally, use the molar absorptivities (εab and εbb) for the basic form of the indicator at λb to simplify the equation. This will give you: pH = pKa + log(([A-Aa] / [Ab-A]))

I hope this helps you in deriving the required equation. If you have any further questions, please don't hesitate to ask. Good luck with your lab report!