Solving AgCl and Ag2CrO4 Solubility Product Homework

In summary, when using K2CrO4 as an indicator for the titration,it is accurate to a certain degree,though it may not be the most accurate choice.
  • #1
leena19
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Homework Statement



The ksp 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[tex]\rightarrow[/tex] Ag++NO3-

AgCl [tex]\rightarrow[/tex] Ag++ Cl-

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

Ag2CrO4[tex]\rightarrow[/tex]2Ag++ CrO42-
Ksp= [Ag+]2[CrO42-]
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.
 
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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.

leena19 said:
[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 :smile:
 
  • #3
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.
 

FAQ: Solving AgCl and Ag2CrO4 Solubility Product Homework

1) How do you calculate the solubility product (Ksp) for AgCl and Ag2CrO4?

The solubility product is calculated by multiplying the molar concentrations of the dissociated ions in a saturated solution. For AgCl, it would be [Ag+][Cl-], and for Ag2CrO4, it would be [Ag+]^2[CrO4^2-].

2) What factors affect the solubility of AgCl and Ag2CrO4?

The solubility of AgCl and Ag2CrO4 is affected by temperature, pH, and the presence of other ions that can form complexes with silver ions. Additionally, the solubility of Ag2CrO4 is also affected by the common ion effect.

3) How do you determine the molar solubility of AgCl and Ag2CrO4?

The molar solubility can be determined by using the Ksp value and the initial concentration of the salt in the solution. The molar solubility for AgCl is equal to the square root of Ksp, and for Ag2CrO4, it is equal to the cube root of Ksp.

4) What is the significance of the solubility product in relation to the solubility of AgCl and Ag2CrO4?

The solubility product is a measure of the maximum amount of a salt that can dissolve in a particular solvent. It is a constant value that indicates the equilibrium between the undissolved solid and the dissociated ions in a saturated solution. A higher Ksp value indicates a more soluble salt.

5) How can the solubility product be used to predict the precipitation of AgCl and Ag2CrO4?

If the ion product, calculated by multiplying the concentrations of the dissociated ions in a solution, is greater than the solubility product, precipitation will occur. If it is less than the solubility product, no precipitation will occur and the solution is considered undersaturated. If it is equal to the solubility product, the solution is saturated and at equilibrium.

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