Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kaylove095 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?

[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 K_{sp}, what is the concentration of Cl^{}? Verify that this is a very small fraction of the original Cl^{}, so that the simplification is justified.

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?

Looks OK to me  but put the units in!

thank you so much!