Question

Ksp problem attached. Number 31. Thanks!

Answer 1

@JFraser @abb0t @aaronq

Answer 2

@chmvijay @wolfe8 Can anyone confirm or correct so far? Thanks so much!

Answer 3

@taramgrant0543664 Any thoughts?

Answer 4

I wanted to work it through myself completely and I got 0.0677M As the concentration for Ag but that doesn't sound right to me

Answer 5

Sure, how did you get that?

Answer 6

I'm getting something that makes less sense, so any thoughts you have would be awesome!

Answer 7

So what I did was I took the concentration for the Ba(NO3)2 assuming since its in a 1:1 ratio the Ba2+ concentration should be the same at .5M I then used the formula Ksp=[Ba2+][CrO42-] using the Ksp for Ba (the 1.2x10^-10) and the concentration of .5M for Ba I solved for the concentration of the chromate. Having the concentration of chromate I then used the formula
[Ag2+]= sqrt(Ksp/[CrO4] where Ksp is 1.1x10^-12 and you square root it because of the stoichometric coefficients

Answer 8

Your method totally makes sense, thanks so much!

Answer 9

Well that's good that it makes sense lol no problem!

Answer 10

Oh, I have a question though. We know we can use the Ksp formula for the Ba because it tells us that it just started to precipitate. We don't know that for the Ag, so can we use that formula?

Answer 11

@taramgrant0543664

Answer 12

I never actually really considered that I just assumed since the Ksp was given

Answer 13

Yeah, I don't think we can use it, since we don't know if it is at the solubility point or not.

Answer 14

What I was thinking from here is to use the concentration of the Chromate as the concentration of the K2CrO4 in the reaction.

Answer 15

But that doesn't really change the concentration of the Ag+ much, so it seems strange.

Answer 16

Well we can still get up to the concentration of chromate and there is the concentration of Ag2CrO4 as .5M could you solve from there for the concentration of Ag?

Answer 17

I think you mean the concentration of AgNO3 is .5M, and yes, that is what I used. But it gives me the concentration of Ag+ to be essentially .5M still.

Answer 18

Haha ya that's what I mean!!! How are you setting up that equation?

Answer 19

Since the concentration of CrO4 is 2.4x10^-10, the concentration of K2CrO4 is the same. The molar ratio requires double the AgNO3, so 4.8x10^-10 M reacts.

Answer 20

Leaving .5-4.8x10^-10 = about .5M

Answer 21

Ya I'm getting the same thing... The math is alright but that answer makes no sense!

Answer 22

Yeah, I hear ya!

Answer 23

the problem does give the KSP for silver chromate, it is \(10^{-12}\) The KSP for silver chromate is less than the KSP for barium chromate, so the silver chromate will have already exceeded its precipitation point once the barium chromate starts to precipitate. this is clever

Answer 24

So I'm thinking since the math all seems right it's either right or it's the way I had said initially. I know it doesn't say that the precipitate formed but I looked on a solubility table and the one I looked at said that AgCrO4 is insoluble in water and makes a precipitate so could you just assume that even if it's not stated?

Answer 25

the barium chromate won't start to precipitate until most of the silver ions have precipitated, because the KSP for silver chromate is less than the KSP for barium chromate

I would find the concentration of silver in saturated silver chromate, using the KSP values and an ICE table approach, starting with the 0.5M \(AgNO_3\) That gives you the starting concentration of the silver ions once the barium ions start to precipitate. As you add more chromate ions, the silver ion concentration will continue to drop (common ion effect) until the concentration reaches the limits of the barium chromate.

Answer 26

So, are you saying that I can use the Ksp of the silver chromate since it will have precipitated already?

Answer 27

Id use the KSP of silver chromate to find the concentration of silver ions at the moment they start precipitating, and then compare that to the barium concentration somehow.

not exactly sure

Answer 28

I'm going to save this problem and give it to my students next year, see what we can do with it