Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: j2dac on May 17, 2010, 09:27:48 AM
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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
of AgN03 solution added.
(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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volume1 x moles 1 = volume2 x moles2
M means molarity (concentration)
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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How would you suggest I go about the question?
Like in steps...
What values do I get first?
Thanks for your help.
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Check out
http://www.titrations.info/precipitation-titration-curve-calculation