What is the silver concentration in a solution prepared by mixing 500mL of .500M silver nitrate with 400mL of .600M sodium chromate? Given that the ksp of silver chromate 1.2x10^-12
500mL of .500 M silver nitrate
400mL of .600M sodium chromate
The first step that needs to be done is balance the reaction.
2Ag+ (aq) + CrO4^2- ---> Ag2CrO4
The ratio of the reaction is 2:1
mol Ag+ = .500 mol Ag+ 500 mL 1L
_____________ X X ___________ = .25 mol Ag+
L 1000 mL
mol CrO4^2- = .600 mol CrO4^2- 400mL 1L
________________ X X _______ = .24 mol CrO4^2-
L 1000 mL
1/2 (.25 mol Ag+) < .24 mol CrO4^2-
Silver ion is the limiting reactant.
Then use ICE equation
2Ag+(aq) + CrO4^2-(aq) -----> Ag2CrO4(s)
.25 .24
-.25 -(.25/2)
__________________
0 .115
Finding excess CrO4^2-
[CrO4^2-]excess= .115mol CrO4^2-
_______________ = 128 M CrO4^2-
(500+400) mL X 1/1000mL
Ag2CrO4(s) <------> 2Ag=(aq) + CrO4^2-(aq) Ksp= 1.2x10^-12
0M .128
+ 2x +x
__________________________
2xM (.128+x)
Ksp=1.2x10^-12 = [Ag+]^2 [CrO4^2-] = (2x)^2 (.128+x)
1.2x10^-12 = [Ag+]^2 (.128M)
[Ag+] = (1.2x10^-12) ^ 1/2
___________ = 3.06 x 10^-6
(.128M)
*Couldn't find one big bracket for whole equation & all of it is raised to the half, not just the top.
My final answer is 3.06 x 10^-6
