Precipitation Titration Worksheet Name (1)______
Updated 1/28/2005
Show units on all answers, use the proper number of significant figures. This is a homework assignment. It has been set up in the order one would use to solve this type of problem. Fill in right on this sheet.
This exercise will be the titration of 30.0 mL of 0.0100M sodium chloride with 0.0200 M Silver nitrate.
How many moles of chloride ion is initially present? (2) 3.00 x 10-4 moles
The equivalence point will be at (2) 15.0 mL
Calculating the titration curve.
The titration curve is plotted with pAg on the y-axis and the volume of AgNO3 added on the x-axis. To plot, there are regions in which there are different methods to solve for pAg.
There are four regions for this titration curve, Start, before endpoint, at the endpoint and after the endpoint.
Start
Since we are adding Ag ion to a Chloride ion solution there will be no silver and the pAg is undefined. (It would be the log of zero) (A reading can be made but it is an artifact)
Before equivalence point
At the point where 1.00 mL of the AgNO3 is added, how many moles of Ag+ has been added? (2) 2.00 x 10-5 moles
All the silver ion added forms precipitate and the only silver ion present is from dissolution of AgCl that is suppressed by the common ion effect.
How many moles of Cl- ion will be left in solution? (2)
30.0 x 10-5 – 2.0 x 10-5 = 28.0 x 10-5
What is the total volume of this solution at this point? (2)
31.0 mL
What is the chloride ion concentration? (2) 28 x 10-5 / .031 = 9.03 x 10-3 M
What is the silver ion concentration? (2) Ksp = [Ag][Cl]
1.8 x 10-10/ 9.03 x 10-3 = 2.0 x 10-8
Assume that Ksp for AgCl is 1.8 x 10-10
What is the pAg for this point? (2) pAg = -log(Ag) = 7.70
At equivalence point
At the equivalence point all the chloride has been used up and all the silver ion has been precipitated. The only source of these ions is the dissolution of the solid. This is now a straight forward Ksp calculation.
What is the concentration of Ag+ ion? (2) [Ag] = [Cl] Ksp = [Ag]2
1.8 x 10-10 = [Ag]2 [Ag] = 1.3 x 10-5
What is the pAg? (2) 4.87
Why don't we need to worry about volume for this point??? (2) Saturated solution
After the equivalence point
All silver added now stays in solution since all the chloride is gone.
When a total of 20.0 mL has been added how many moles of Ag has been totally added to the beaker? (2) 20 x 10-3 x 0.0200 = 4.00 x 10-4
How many moles of silver were lost in the precipitate? (2) 3.00 x 10-4 moles Ag
How many moles of silver are left in solution? (2) (4.00 – 3.00) x 10-4 Ag = 1.00 x 10-4 moles
What is the total volume of the solution? (2) 50.0 mL
What is the Ag+ concentration? (2) 1.00 x 10-4 / 0.050 L = 2.00 x 10-3
What is the pAg? (2) 2.70
Ksp Titration Worksheet Page 1