# Solubility problems

Discussion in 'DAT Discussions' started by Harley comic, Aug 9, 2006.

1. ### Harley comic Junior Member 2+ Year Member

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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

2. ### KiTmAn Kit 10+ Year Member

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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 ..

3. OP

### Harley comic Junior Member 2+ Year Member

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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?

4. ### KiTmAn Kit 10+ Year Member

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

5. ### KiTmAn Kit 10+ Year Member

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

6. ### keibee82 Blue_tooth... 2+ Year Member

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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.

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7. ### kkumalap Member 2+ Year Member

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

8. ### slayerdeus Member 2+ Year Member

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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.

9. ### KiTmAn Kit 10+ Year Member

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

10. ### JohnDoeDDS Senior Member 7+ Year Member

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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?

11. ### fancymylotus A Whole New World Dentist 10+ Year Member

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

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12. ### DonExodus Dentist in Virgin Islands 10+ Year Member

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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

13. ### KiTmAn Kit 10+ Year Member

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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......

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