Solubility problems

[Harley comic]

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I am having troubles with the solubility problems. I tried to study for it, but it didnt seem to promissing. I am taking the test very soon, do you guys think I should just forget about it and study something else instead? If you guys have a better way of doing it, please help!!!
EX:
How much (CrO4)2- must be addd to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4? (Ksp of AgCl=2.8 x 10^-10, Ksp of Ag2CrO4=1.4x10^-22)
A. 2.5x10^-12
B. 2.5x10^-13
C.5.0x10^-13
D.5.0x10^-12
E.1.0x10^-13
Kaplan practice test.
Thanks a bunch

 

[KiTmAn]

okay..first thinking logically first we need to set up teh equation using the second Ksp..
equation 1 ..Ksp for Ag2CrO4 = [Ag2+]^2 [CrO4 2-]
so now we need to know the concentration of Ag2+ which can be calculated by using the solubility for AgCl -->
equation 2...2.8 x 10^-10 = x^2
no need to calculate this..reason for this will become clear in a minute...
if u think about it x value will be [Ag2+]..after taking the square root which u will have to square it later..when plugged into equation 1...so u can use the same value so in equation 1 ..
1.4x10^-22 = (2.8 x 10^-10) (x)
x = 1.4x10^-22 / 2.8 x 10^-10
x = 0.5 x 10 ^-12
x = 5 x 10^ -13
now i m not sure if answer is C or A b/c adding amt of C will create a saturated solution ...so need to add more so answer can be A..
let me know what the answer is ..

 

[Harley comic]

the answer is C
but you think we have enough time in the real test? This problem take me at least 5 minutes, do you think it is worth trying?

 

[KiTmAn]

now wouldnt C jus create a saturated solution Har..?

 

[KiTmAn]

i guess its C...i doubt we get qts like these on dat...it took me atleast 3 mins....

 

[keibee82]

okay..first thinking logically first we need to set up teh equation using the second Ksp..
equation 1 ..Ksp for Ag2CrO4 = [Ag2+]^2 [CrO4 2-]
so now we need to know the concentration of Ag2+ which can be calculated by using the solubility for AgCl -->
equation 2...2.8 x 10^-10 = x^2
no need to calculate this..reason for this will become clear in a minute...
if u think about it x value will be [Ag2+]..after taking the square root which u will have to square it later..when plugged into equation 1...so u can use the same value so in equation 1 ..
1.4x10^-22 = (2.8 x 10^-10) (x)
x = 1.4x10^-22 / 2.8 x 10^-10
x = 0.5 x 10 ^-12
x = 5 x 10^ -13
now i m not sure if answer is C or A b/c adding amt of C will create a saturated solution ...so need to add more so answer can be A..
let me know what the answer is ..

in the second equation forming Ag2CrO4, there is two moles of Ag+ per
one mole of CrO4. Therefore its 2x so you have to calculate the x value
by square root from the first equation you had and then multiply by 2
and then square. That gives you the value for the Ag+ concentration
and divide Ksp for Ag2CrO4 by this concentration will give you concentration
of CrO4.

 

[kkumalap]

in the second equation forming Ag2CrO4, there is two moles of Ag+ per
one mole of CrO4. Therefore its 2x so you have to calculate the x value
by square root from the first equation you had and then multiply by 2
and then square. That gives you the value for the Ag+ concentration
and divide Ksp for Ag2CrO4 by this concentration will give you concentration
of CrO4.

that's wut I thought too. But Kaplan said otherwise...it's so weird. Can someone please clear this up.

 

[slayerdeus]

If you are trying to find the concentration of Ag+ then you need to multiply it, but since you already know the concentration from the AgCl solution you don't need to multiply it by 2.

 

[KiTmAn]

yea...i knew i had to do that...jus dint kno the reason behind my using it...

 

[JohnDoeDDS]

quick question. I understand this problem but...

In this problem you got KSP= X*X

But howcome sometimes you would say KSP = (2X)^2 *X
Because you have 2 Ag+ so why doesnt it equal (2x)^2 but just X?

 

[fancymylotus]

quick question. I understand this problem but...

In this problem you got KSP= X*X

But howcome sometimes you would say KSP = (2X)^2 *X
Because you have 2 Ag+ so why doesnt it equal (2x)^2 but just X?

cause for AgCl it goes Ksp= [Ag][Cl]...no 2.

 

[DonExodus]

2.8e-10 = [Ag][Cl]. [Ag] = [Cl], so 2.8e-10= [Ag]² = 2.8e-10
Therefore, [Ag] = square root of 2.8e-10

For the second part, AgCl + CrO4 --> Ag2CrO4.
The rate expression is
1.4e-22 = [Ag]²[CrO4]
[Ag]² = 2.8e-10
Plug in, and you get
1.4e-22 = (2.8e-10)[CrO4]

Solving for [CrO4] yields .5e-12, or 5x10^-13

 

[KiTmAn]

k..i still have one last qt on this stupid problem...wouldnt 5 e -13 of CrO4 jus create a saturated solution...so wouldnt u need to add more.. e -12 or something? lemme know pleaseeeeee......