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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: msted14 on April 28, 2008, 01:43:05 PM

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 1.00cm 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!

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

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

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

pH = log(1/[H+]) ?
i really don't understand this question am i supposed to use the A=Ebc?

Ka = [H+][In]/[HIn]
Divide both sides of the equation by [H+].

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 1.00cm 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]

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.

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

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^{3}M solution. It later dissociated because of pH changes, but the formal (or analytical) concentration has not changed.

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]

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?

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?

Looks much better now :)

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)

pH is given by the aniline buffer.