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### Topic: Absorbance with HIn and In  (Read 14749 times)

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#### msted14

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##### Absorbance with HIn and In
« 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 1.00-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

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##### Re: Absorbance with HIn and In
« Reply #1 on: April 28, 2008, 02:02:54 PM »

What is pH of the solution? How does ratio of In-/HIn depend on the pH?
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #2 on: April 28, 2008, 02:07:04 PM »
That's the problem, I can't get there. I have HIn <=> H+ + In- and Ka = [H+][In-]/[HIn].

#### Borek

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##### Re: Absorbance with HIn and In
« Reply #3 on: April 28, 2008, 02:39:28 PM »
Close. What will happen if you will divide both sides of the equation by [H+]?
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #4 on: April 28, 2008, 02:54:55 PM »
pH = log(1/[H+]) ?

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

#### Borek

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##### Re: Absorbance with HIn and In
« Reply #5 on: April 28, 2008, 03:19:03 PM »
Ka = [H+][In-]/[HIn]

Divide both sides of the equation by [H+].
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #6 on: April 28, 2008, 03:29:29 PM »
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

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##### Re: Absorbance with HIn and In
« Reply #7 on: April 28, 2008, 03:44:57 PM »
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.
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #8 on: April 28, 2008, 03:54:29 PM »
Sorry, but how do i know [HIn]+[In-]? is that [HIn] = 1.00 x 10^-3?

#### Borek

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##### Re: Absorbance with HIn and In
« Reply #9 on: April 28, 2008, 04:12:29 PM »
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.
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #10 on: April 28, 2008, 04:22:39 PM »
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

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##### Re: Absorbance with HIn and In
« Reply #11 on: April 28, 2008, 04:36:07 PM »
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?
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #12 on: April 28, 2008, 04:40:37 PM »
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

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##### Re: Absorbance with HIn and In
« Reply #13 on: April 28, 2008, 04:55:48 PM »
Looks much better now
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#### msted14

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##### Re: Absorbance with HIn and In
« Reply #14 on: April 28, 2008, 05:11:11 PM »
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)