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#### j2dac

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##### Precipitation Titration question
« on: May 17, 2010, 03:27:48 AM »

Hey guys,

Been trying for the past week to get this done. Ive come to a conclusion but theres no way of knowing its correct as I didn't get an answer key. Anyway, heres the question, followed by what Ive done. Thanks.

Consider the precipitation titration of 12.5 ml of a solution that is 0.2 M
in KI and a solution that is 0.05 M in AgN03. Calculate the concentration
of I- and thus Ag+ in solution after the addition of 15.0 mL, 25.0 mL, 50.0
mL and 65.0 mL of the AgN03 solution.

Use these data to construct a titration curve, plotting pAg against volume

(Ksp for AgI = 8.3 x 1 0-17)

So what I've done:

The equation is:

KI + AgNO3 -> AgI + KNO3

I first get the starting volume of AgNO3 using volume1 x moles 1 = volume2 x moles2 and I get 50 ml of AgNO3 to start with.

I then get the amount of moles for each solution of AgNO3:

at 50ml - 1.5x10^21

at 65ml - 3.045x10^22

at 75ml - 2.25x10^21

and the other 2 solutions....

Now that I have this information, how do I calculate the concentration of I- and Ag+ ions in AgI?

Its been confusion for me for ages! Thanks to anyone that helps.

Cheers.
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#### AWK

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##### Re: Precipitation Titration question
« Reply #1 on: May 17, 2010, 04:30:49 AM »

Quote
volume1 x moles 1 = volume2 x moles2

M means molarity (concentration)

Quote
at 50ml - 1.5x10^21

at 65ml - 3.045x10^22

at 75ml - 2.25x10^21

Horrible numbers

Where you accounted reaction of silver cation with iodide anion?

You have an excess of iodide after addition of 15 and 25 ml of Ag+ solution and the excess of Ag+ after addition of 65 ml of solution
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AWK

#### j2dac

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##### Re: Precipitation Titration question
« Reply #2 on: May 17, 2010, 04:45:58 AM »

How would you suggest I go about the question?

Like in steps...

What values do I get first?

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#### Borek

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##### Re: Precipitation Titration question
« Reply #3 on: May 17, 2010, 05:21:29 AM »
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