# Solubility problems

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

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

Joined:
Jul 25, 2006
Messages:
25
0
Status:
Pre-Dental
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

Joined:
Apr 7, 2005
Messages:
157
0
Status:
Dental Student
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

Joined:
Jul 25, 2006
Messages:
25
0
Status:
Pre-Dental
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

Joined:
Apr 7, 2005
Messages:
157
0
Status:
Dental Student
now wouldnt C jus create a saturated solution Har..?

5. ### KiTmAn Kit 10+ Year Member

Joined:
Apr 7, 2005
Messages:
157
0
Status:
Dental Student
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

Joined:
May 8, 2006
Messages:
476
2
Status:
Pre-Dental
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.

Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
7. ### kkumalap Member 2+ Year Member

Joined:
Apr 15, 2006
Messages:
55
0
Status:
Pre-Dental
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

Joined:
Jul 19, 2006
Messages:
61
0
Status:
Pre-Dental
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

Joined:
Apr 7, 2005
Messages:
157
0
Status:
Dental Student
yea...i knew i had to do that...jus dint kno the reason behind my using it...

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

Joined:
Dec 26, 2004
Messages:
663
0
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

Joined:
Jul 21, 2004
Messages:
36,046
17,663

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

Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
12. ### DonExodus Dentist in Virgin Islands 10+ Year Member

Joined:
Jul 6, 2006
Messages:
121
1
Status:
Dentist
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

Joined:
Apr 7, 2005
Messages:
157
0
Status:
Dental Student
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......

We’ve been on the Internet for over 20 years doing just one thing: providing career information for free or at cost. We do this because we believe that the health education process is too expensive and too competitive.

We believe that all students deserve the same access to high quality information. We believe that providing high quality career advice and information ensures that everyone, regardless of income or privilege, has a chance to achieve their dream of being a doctor.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}