[yg7s7]

A solution was prepared by mixing 0.025 L of 0.0800 M aniline, 0.025 L of 0.0600 M sulfanilic acid, and 0.001 L of 1.23 * 10^-4 M HIn (protonated indicator) and then diluting to 0.1 L. The absorbance measured at 550 nm in a 5.00 cm was 0.110. Find the conc. of HIn and In- and pKa for HIn.

For HIn:
ε325 = 2.45 * 10^4
ε550 = 2.26 * 10^4
For In-:
ε325 = 4.39 * 10^3
ε550 = 1.53 * 10^4

I have no idea how to do this. Any ideas?
I set up equation 0.110 = [HIn]*5.00*(2.26*10^4) + [In-]*5.00*(1.5*10^4), but I'm stuck. How do you find the absorbance at wavelength = 325? The answer key says pKa = 4.00

[ARGOS++]

Dear yg7s7;

You’re right: One equation with two unknowns is not solvable; but you may have forgotten that
       

[HIn] + [In^-] = c_total of Ind.

Don’t forget the dilution steps!

I hope I have been of some help to you.
Good Luck!
                    ARGOS⁺⁺

[yg7s7]

so you mean I have to do (1.23*10^-4) * 0.001 L / 0.1 L and use it as [HIn] + [In-] = 1.23*10^-6?

[ARGOS++]

Dear yg7s7;

That’s exactly correct!
And now have you any suggestion, how you find the ratio:  [HIn] / [In-], and from that the individual concentration's of both?

Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear yg7s7;

Could this additional hint guide you into the right direction?:
What would you read on your spectrometer as A_550nm:
     - if  [HIn]  =  c_total of Ind?;  and
     - if  [In^-]   =  c_total of Ind?
And finally:   What ratio [HIn] / [In^-]  must result if  A_550nm is equal 0.110?

May this be of any help to you?
Good Luck!
                    ARGOS⁺⁺