Kaplan Gchem Q plz help

[far123]

---

Members do not see this ad.

what is the amount of Cro2 that must be added to a liter of saturated soln of Agcl in order to precipitate Ag2cro4(KSP of Agcl=2.8x10*-8, and KSP of Ag2Cro4= 1.4x10*-22?😕
does any body know how to solve this problem plz?

 

[DELTA888]

what is the amount of Cro2 that must be added to a liter of saturated soln of Agcl in order to precipitate Ag2cro4(KSP of Agcl=2.8x10*-8, and KSP of Ag2Cro4= 1.4x10*-22?😕
does any body know how to solve this problem plz?

CrO4^2- + 2Ag+ <---> Ag2CrO4 (s)

Ksp= [AgCl]^2[CrO2)
Ksp= x(2x)^2
x=(Ksp/4)^(1/3) = [(2.8X10^-8)/4]^(1/3)= 3.27x10^-8 mole/L


*Hope this calculation is right!🙂

 

[PooyaH]

CrO2^2- + 2Ag+ <---> Ag2CrO4 (s)

Ksp= [AgCl]^2[CrO2)
Ksp= x(2x)^2
x=(Ksp/4)^(1/3) = [(2.8X10^-8)/4]^(1/3)= 3.27x10^-8 mole/L
Amt of CrO2 must added to 1 liter of Agcl:
(3.27E-8 mole/L) x 84g/mole CrO2= 2.75X10^-6 gramCrO2

*Hope this calculation is right!🙂

Your reaction isn't even balanced!!! you have 2 oxygens on the reactant side and 4 oxygens on the product side???

 

[far123]

this is how Kaplan solved it...but I don't think it is right....​

Ksp = [Ag+][Cl&#8211;] = x2​

[Ag+]2 = 2.8 x 10&#8211;10

[CrO42&#8211;] = Ksp/[Ag+]2 = (1.4 x 10&#8211;22)/(2.8 x 10&#8211;10) = 0.5 x 10&#8211;12 = 5.0 x 10&#8211;13​

​

 

[PooyaH]

this is how Kaplan solved it...but I don't think it is right....​

Ksp = [Ag+][Cl–] = x2​

[Ag+]2 = 2.8 x 10–10

[CrO42–] = Ksp/[Ag+]2 = (1.4 x 10–22)/(2.8 x 10–10) = 0.5 x 10–12 = 5.0 x 10–13​

The formula is correct! You take the Ksp of the Ag2CrO4 and divide it by the molar solubility of Ag ( [Ag+] )
But if the Ksp of AgCl = 2.8x10^-8 then [Ag+] [Cl-] = 2.8x10^-8
x^2 = 2.8x10^-8
x = 1.67x10^-4 NOT 2.8 x 10^-10!!!

 

[DELTA888]

this is how Kaplan solved it...but I don't think it is right....​

Ksp = [Ag+][Cl&#8211;] = x2​

[Ag+]2 = 2.8 x 10&#8211;10

[CrO42&#8211;] = Ksp/[Ag+]2 = (1.4 x 10&#8211;22)/(2.8 x 10&#8211;10) = 0.5 x 10&#8211;12 = 5.0 x 10&#8211;13​

I took G. Chemistry 5 years ago and I thought I still remember it! 😀
Anyway, The answer was [CrO42-] but why not [CrO22-]? I thought CrO22- was a typo?

Thanks for correcting me. I forgot about the common ion effect [Ag+]

Ksp = [Ag+][Cl&#8211;] = X^2=2.8 x 10&#8211;10

[Ag+] = 1.67 x10^-5


Ag2CrO4 <---->2 [Ag+] + [ CrO42-]

Initial ----- (1.67 x10^-5)------ 0M

Change +2X ------ +X

Final [(1.67X10^-5) +2x)]------- x


Ksp=[(1.67x10^-5) +2x)^2][x] which is ~ [1.67X10^-5)]^2 *[X]

[CrO42-]=X= (1.4 x 10&#8211;22)/{(2.8 x 10&#8211;10)}=5.0X10^-13M

Far123, assume my calculation is right. My answer is the same as Kaplan!