[msted14]

I'm having real trouble with this one problem!

A solutionwas prepared by mixing 50.00 mL of 0.0800 M aniline (a monoprotic weak base), 25.00 mL of 0.100 M HCl, and 1.00 mL of 1.00 X 10^-3 M HIn indicator and diluting to 100 mL (HIn stands for protonated indicator). The absorbance measured at 550 nm in a 1.00-cm cell was 0.202. The molar extinction coefficients of HIn and In- are 2.26 X 10^4 1/(M)(cm) and 1.52 X 10^4 1/(M)(cm), respectively. The pKb of aniline is 9.399.

Find the concentrations of HIn and In- in species.

Find the pKa of HIn.

Thanks!

[Borek]

Please read forum rules.

What is pH of the solution? How does ratio of In-/HIn depend on the pH?

[msted14]

That's the problem, I can't get there. I have HIn <=> H+ + In- and Ka = [H+][In-]/[HIn].

[Borek]

Close. What will happen if you will divide both sides of the equation by [H+]?

[msted14]

pH = log(1/[H+]) ?

i really don't understand this question- am i supposed to use the A=Ebc?

[Borek]

Ka = [H+][In-]/[HIn]

Divide both sides of the equation by [H+].

[msted14]

But i don't have Ka?
Ka/[H+] = [HIn][In-]

A solutionwas prepared by mixing 50.00 mL of 0.0800 M aniline (a monoprotic weak base), 25.00 mL of 0.100 M HCl, and 1.00 mL of 1.00 X 10^-3 M HIn indicator and diluting to 100 mL (HIn stands for protonated indicator). The absorbance measured at 550 nm in a 1.00-cm cell was 0.202. The molar extinction coefficients of HIn and In- are 2.26 X 10^4 1/(M)(cm) and 1.52 X 10^4 1/(M)(cm), respectively. The pKb of aniline is 9.399.


i have [H+] = 0.100 M, [HIn] = 1.00 x 10^-3
and pKb of aniline = 9.399

and A=EbC
0.202 = 2.26 X 10^4 1/(M)(cm) x [HIn] + 1.52 X 10^4 1/(M)(cm) x [In-]

[Borek]

You know [HIn]+[In-] from initial concentration of indicator. You know 0.202 = 2.26 X 10^4 1/(M)(cm) x [HIn] + 1.52 X 10^4 1/(M)(cm) x [In-]. This gives you two equations in two unknwon. This means you KNOW both [HIn] and [In-].

Forget about the division, brain fart.

[msted14]

Sorry, but how do i know [HIn]+[In-]? is that [HIn] = 1.00 x 10^-3?

[Borek]

Sorry, but how do i know [HIn]+[In-]? is that [HIn] = 1.00 x 10^-3?

You have used known amount of indicator - 1 mL of 10^-3M solution. It later dissociated because of pH changes, but the formal (or analytical) concentration has not changed.

[msted14]

I don't think I'm doing this right- i got [In-] to be 30.68 M

is that [HIn]+[In-] = 1.00 X 10^-3 M?
 and 0.202 = 2.26 X 10^4 1/(M)(cm) x [HIn] + 1.52 X 10^4 1/(M)(cm) x [In-]

[Borek]

is that [HIn]+[In-] = 1.00 X 10^-3 M?

No.

How many moles of indictaor have been put into the solution?

What is solution volume?

[msted14]

moles of indictaor = 1 x 10^-6
in 100 mL of solution - does that make it 1 x 10^-5 M and is that [HIn]+[In-] = 1.00 X 10^-5 M?

[Borek]

Looks much better now

[msted14]

ok what about the pKa part? shouldn't i use Ka= [H+][In-]/[HIn] and to get [H+] =0.100M HCl x 25mL / (1000mL x 100mL) ...and then pKa = -log(Ka)