Solving AgCl and Ag2CrO4 Solubility Product Homework

[leena19]

Homework Statement

The k_sp values of AgCl and Ag₂CrO₄ are 2*10^-10 mol²dm^-6 and 3*10^-12mol³dm^-9 respectively.100cm³ of an aqueous solution contains NaCl and K₂CrO₄ only,whose cocentrations are 0.1M and 0.1M respectively.To this solution,is added AgNO₃ dropwise while mixing.

(1) Calculate and show whether AgCl or Ag₂CrO₄ will be precipitated first?

(2) When the second precipitate begins to appear,what is the concentration of the anion of the Ag salt that precipitated first ,which still remains in the solution?

(3) Using the above answer,explain briefly the accuracy of K₂CrO₄ as an indicator when used in the above titration.

Homework Equations

The Attempt at a Solution

I managed to do part (1) of the problem but m having trouble with the rest

For(1),
AgNO₃\rightarrow Ag⁺+NO₃^-

AgCl \rightarrow Ag⁺+ Cl^-

Ksp = [ Ag⁺] [Cl^-]
2*10^-10/0.1 = [ Ag⁺]
[ Ag⁺]=2*10^-9M

Ag₂CrO₄\rightarrow2Ag⁺+ CrO_42-
Ksp= [Ag⁺]²[CrO_42-]
3*10^-12/0.1 = [Ag⁺]²
[Ag⁺] = 5.477*10^-6M

since the minimum concentration of Ag+ needed to precipitate AgCl is less than that needed to precipitate Ag chromate,AgCl would precipitate 1st.

(2) I'm not sure I understand the question,

When the second precipitate begins to appear

which means the [Ag+] present in the solution is now, 5.477*10^-6M,right? but what do I do now?Do I just substitute this value to the ksp equation of AgCl and find the[chloride] in the precipitate and then substract this value from 0.1M?

THANK YOU.

 

[Borek]

Compare

precipitation titration end point detection

After all Cl^- was precipitated in form of AgCl you still add more solution of AgNO₃ till the concentration of Ag⁺ is high enough for the Ag₂CrO₄ to precipitate.

[Ag+] present in the solution is now, 5.477*10^-6M,right? but what do I do now?Do I just substitute this value to the ksp equation of AgCl and find the[chloride]

At this moment you are done

 

[leena19]

Thanx for the very helpful link.
So the [chloride] of AgCl would be=2*10^-10/5.477*10^-6

​

=0.365*10^-4M

Thanks again.