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### Topic: [Cl-] from Ksp  (Read 2884 times)

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#### kaylove095

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##### [Cl-] from Ksp
« on: January 11, 2015, 03:32:17 PM »
you are given a 1.00 mL solution of .100M Cl- and told to verify the chloride concentration by gravimetric analysis. 1.00 mL of 0.500M Ag+ is added to this solution and the resultant AgCl quantitatively precipitated. based on the ksp expression calculate the concentration of the cl- that remains in solution. Ksp of AgCl is 1.82x10^-10 at 25 degrees celsius.

attempt at solution:
Ksp=[Ag+][Cl-]
1.82x10^-10=[Ag+][Cl-]

[Ag+]= (.001L)(o.500mol)/(0.002L)=0.250M
[Cl-]=(1.82x10^-10)/0.250=7.28x10^-10 M?

#### mjc123

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##### Re: [Cl-] from Ksp
« Reply #1 on: January 11, 2015, 04:40:56 PM »
Quote
[Ag+]= (.001L)(o.500mol)/(0.002L)=0.250M
This is the concentration of Ag+ in the mixed solution before precipitation. You need the concentration left in solution after precipitation has occurred. You could set up a quadratic equation, but because the solubility of AgCl is very low you can simplify the problem by assuming that virtually all the Cl- is precipitated as AgCl.
How many moles of Cl- is this?
So how many moles of Ag+ are precipitated, and how many moles are left in solution?
Using Ksp, what is the concentration of Cl-? Verify that this is a very small fraction of the original Cl-, so that the simplification is justified.

#### kaylove095

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##### Re: [Cl-] from Ksp
« Reply #2 on: January 11, 2015, 11:40:39 PM »
so then there are (0.00100L)(0.100mol Cl-)=1x10^-4 mol Cl- and 5.00x10^-4 mol Ag+ at the beginning of the rxn.
making Cl- the limiting reagent so (5.00x10^-4)-(1.00x10^-4)=4x10^-4 mol Ag+ in solution/0.002L= 0.2M Ag+
[Cl-]=(1.82x10^-10)/0.2= 9.10x10^-10?

#### mjc123

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##### Re: [Cl-] from Ksp
« Reply #3 on: January 12, 2015, 04:56:08 AM »
Looks OK to me - but put the units in!

#### kaylove095

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##### Re: [Cl-] from Ksp
« Reply #4 on: January 12, 2015, 11:50:08 AM »
thank you so much!