Gchem question

[GRAD]

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please explain:

The ksp of AgI in an aqueous soln is 1*10-6 mol/L. If a 1*10-5 soln of AgNO3 is saturated with AgI, what will be the final concentration of the iodide ion??

kinda confused..can someone work it out step ny step? thanks!

 

[GRAD]

anyone?

 

[Manyak222]

since the Ksp of AgI is 1*10^-6, but you already have 1*10^-5 molar solution of AgNO3. This is now a simple common ion effect problem, with ksp=1*10^-6=[Ag], and you know there's already 1*10^-5 of Ag in solution. the answer is thus 1*10^-1 moles per liter

anyone?

 

[dat_student]

please explain:

The ksp of AgI in an aqueous soln is 1*10-6 mol/L. If a 1*10-5 soln of AgNO3 is saturated with AgI, what will be the final concentration of the iodide ion??

kinda confused..can someone work it out step ny step? thanks!

Ksp = 1 * 10^-6 = [Ag+][I-] = [Ag]^2 = [I-]^2

[Ag+] = [I-] = 10^-3 M This is the maximum amount that can dissolve

assuming the final volume doesn't change much:

[I-] = 10^-3 (max. Ag) - 10^-5 (Ag in soln prior to adding AgI) = 9.9 * 10^-4 = AgI or Ag or I that was added

Can someone please check my math & calculations?

 

[dat_student]

since the Ksp of AgI is 1*10^-6, but you already have 1*10^-5 molar solution of AgNO3. This is now a simple common ion effect problem, with ksp=1*10^-6=[Ag], and you know there's already 1*10^-5 of Ag in solution. the answer is thus 1*10^-1 moles per liter

I don't think [I-] = 10^-1. @10^-3 the solution is saturated. [I-] in the solution can't be greater than 10^-3

 

[wireless_doc]

I don't think [I-] = 10^-1. @10^-3 the solution is saturated. [I-] in the solution can't be greater than 10^-3

Yes, it is. Here is how you do it..

--------------AgI --> Ag+ ------+ I-
Original---------------10^-5 (AgNO3 is completely ionized)
From AgI -------------x-----------x
Equilibrium: ----------(x+10^-5)---x

Ksp = (x+10^-5)x = 10^-6

10^-5 >> x. Thus, (x+10^-5) = 10^-5
Thus x = 10^-1

 

[dat_student]

Yes, it is. Here is how you do it..

--------------AgI --> Ag+ ------+ I-
Original---------------10^-5 (AgNO3 is completely ionized)
From AgI -------------x-----------x
Equilibrium: ----------(x+10^-5)---x

Ksp = (x+10^-5)x = 10^-6

10^-5 >> x. Thus, (x+10^-5) = 10^-5
Thus x = 10^-1

Ksp = 10^-6
Ksp = [Ag+][I-] = x * x = x^2 = 10^-6
[Ag+] = [I-] = [x] = 10^-3 This is the maximum amnt of [I-] you can have in the solution

How can you have 10^-1 which is equal to 100 times the max amnt you can have in a saturated solution???!!!

 

[wireless_doc]

Ksp = 10^-6
Ksp = [Ag+][I-] = x * x = x^2 = 10^-6
[Ag+] = [I-] = [x] = 10^-3 This is the maximum amnt of [I-] you can have in the solution

How can you have 10^-1 which is equal to 100 times the max amnt you can have in a saturated solution???!!!

OK, I'm pretty sure what I posted was how it's done for this type of question. But I just realized that my assumption (10^-5 >>x) is not met in this case. So, I guess, we'll have to solve the equation to get the exact answer.
x^2 + 10^-5x - 10^-6 = 0
x comes out to be 10^-3 - (10^-5)/2 which can be rounded up to your 10^-3. However, I don't think you can ignore the 10^-5 (which should be chosen differently). It must be included. Check your g chem book.

 

[Manyak222]

You already have your Ag in solution. it is not x as you are stating. the ksp of AgI is 1*10^-6, but you already have 1*10^-5 moles of Ag. All you can add is 10^-1 mol of Ag, which means that 10^-1 mole of I must accompany it.

I don't think [I-] = 10^-1. @10^-3 the solution is saturated. [I-] in the solution can't be greater than 10^-3

 

[dat_student]

OK, I'm pretty sure what I posted was how it's done for this type of question. But I just realized that my assumption (10^-5 >>x) is not met in this case. So, I guess, we'll have to solve the equation to get the exact answer.
x^2 + 10^-5x - 10^-6 = 0
x comes out to be 10^-3 - (10^-5)/2 which can be rounded up to your 10^-3. However, I don't think you can ignore the 10^-5 (which should be chosen differently). It must be included. Check your g chem book.

I didn't ignore 10^-5. This was my original solution:

Ksp = 1 * 10^-6 = [Ag+][I-] = [Ag]^2 = [I-]^2

[Ag+] = [I-] = 10^-3 M This is the maximum amount that can dissolve

assuming the final volume doesn't change much:

[I-] = 10^-3 (max. Ag) - 10^-5 (Ag in soln prior to adding AgI) = 9.9 * 10^-4 = AgI or Ag or I that was added

Can someone please check my math & calculations?

*********************************************************
*********************************************************

You already have your Ag in solution. it is not x as you are stating. the ksp of AgI is 1*10^-6, but you already have 1*10^-5 moles of Ag. All you can add is 10^-1 mol of Ag, which means that 10^-1 mole of I must accompany it.

10^-1 = 10000 * 10^-5 = 100 * 10^-3

P.S. I'll be back home later tonight. I'll be sure to check this thread.

 

[cryptozoologist]

people seem to be confused whether there should be first a limit to the saturated iodine ion concentration as a ceiling concentration for I-.

just set it up like your supposed to:
AgI --> Ag+ + I-
na 1e-5 0
na x x

equation: 1e-6 = x(1e-5+x)

For the added AgI to participate in the reaction, you have to see that for every I- that dissociates, there will be another Ag+ and you're already near the ceiling for Ag+. Because x and the original concentration are close, the only way to solve is via quadratic equation. which probably won't ever come up on the DAT.

good practice though. Answer is 1e-3 for I- at point of saturation.

 

[Manyak222]

Im pretty sure the answer is 1*10^-1. I just did this problem the long, protracted way, and assuming x is small, you get 10^-1. however, in this case x is not small, so you'd have to go into foiling to figure this out. I checked Kaplan and they have a similar question, but where the Ksp of AgI is 1*10^-16, which when worked out makes x small.

 

[wireless_doc]

people seem to be confused whether there should be first a limit to the saturated iodine ion concentration as a ceiling concentration for I-.

just set it up like your supposed to:
AgI --> Ag+ + I-
na 1e-5 0
na x x

equation: 1e-6 = x(1e-5+x)

For the added AgI to participate in the reaction, you have to see that for every I- that dissociates, there will be another Ag+ and you're already near the ceiling for Ag+. Because x and the original concentration are close, the only way to solve is via quadratic equation. which probably won't ever come up on the DAT.

good practice though. Answer is 1e-3 for I- at point of saturation.

Yep, that's how I approached too.

 

[lifeisgood]

Where did this question come from? This exact problem is on page 286 in the Kaplan Blue book but the Ksp is given to be 1 x 10^-16 mol/L.

I think the way this particular from the OP is presented is not representative of how one would see it on the dat. If the solution already contained 1 x 10^-5 mol Ag+, it is already supersaturated and I don't believe any appreciable amount of I- will dissociate.

But 1 x 10^-1 is WAY too much I- to dissociate b/c your Ksp would then be

Ksp = ((1 x 10^-5) + (1 x 10^-1)*(1 x 10^-1) = 1 x 10^-2


BTW, EVEN 1 x 10^-3 is TOO HIGH...

 

[lifeisgood]

people seem to be confused whether there should be first a limit to the saturated iodine ion concentration as a ceiling concentration for I-.

just set it up like your supposed to:
AgI --> Ag+ + I-
na 1e-5 0
na x x

equation: 1e-6 = x(1e-5+x)

For the added AgI to participate in the reaction, you have to see that for every I- that dissociates, there will be another Ag+ and you're already near the ceiling for Ag+. Because x and the original concentration are close, the only way to solve is via quadratic equation. which probably won't ever come up on the DAT.

good practice though. Answer is 1e-3 for I- at point of saturation.

Can't solve via quadratic b/c the value of b2 - 4ac is a negative number.

 

[fancymylotus]

Can't solve via quadratic b/c the value of b2 - 4ac is a negative number.

negative b plus or minus the squaaaaaaare root of b squared....all over 2A

 

[dat_student]

people seem to be confused whether there should be first a limit to the saturated iodine ion concentration as a ceiling concentration for I-.

just set it up like your supposed to:
AgI --> Ag+ + I-
na 1e-5 0
na x x

equation: 1e-6 = x(1e-5+x)

For the added AgI to participate in the reaction, you have to see that for every I- that dissociates, there will be another Ag+ and you're already near the ceiling for Ag+. Because x and the original concentration are close, the only way to solve is via quadratic equation. which probably won't ever come up on the DAT.

good practice though. Answer is 1e-3 for I- at point of saturation.

very close to 1e-3 because Ag+ reaches the saturation point (i.e. 10^-3 or 1e-3) before I- does. See my calculations above.

..

If the solution already contained 1 x 10^-5 mol Ag+, it is already supersaturated

No because 10^-5 < 10^-3 (the saturation point)

BTW, EVEN 1 x 10^-3 is TOO HIGH

it's pretty close to 10^-3 (the saturation point). [See my calculations above]

 

[wireless_doc]

dat_student,
Read page 922 in the white kaplan. There's one example just like this question where they implicitly made the same assumption as I did. Fortunately, it came out to match their assumption. It's OK to be wrong. 😉

 

[dat_student]

dat_student,
Read page 922 in the white kaplan. There's one example just like this question where they implicitly made the same assumption as I did. Fortunately, it came out to match their assumption. It's OK to be wrong. 😉

My room is a mess and I can't find the white book (moving to LA). I agree with you 100%. It's ok to be wrong. I just don't see how you can have 10^-1 when saturation is reached at a much lower concentration (i.e. 10^-3). If you think that makes sense fine I am wrong. I am not going to post anything else regarding the question. I give up.

 

[lifeisgood]

My room is a mess and I can't find the white book (moving to LA). I agree with you 100%. It's ok to be wrong. I just don't see how you can have 10^-1 when saturation is reached at a much lower concentration (i.e. 10^-3). If you think that makes sense fine I am wrong. I am not going to post anything else regarding the question. I give up.

Think about it....if the AgNO3 sol'n already contains 1x10-5 M Ag+,his is 10 times the value of the Ksp... it is supersaturated wrt Ag+ and therefore AgI will dissociate to a negligible degree.

B/c if I- dissociates to 10-3 then so does Ag+. There is NO WAY Ag+ would dissociate to that degree in a solution where the concentration is already above the Ksp.

 

[GRAD]

Personally i think its 1*10-1...isnt it just 6-5?

 

[dat_student]

Personally i think its 1*10-1...

we start with [Ag+] = 10^-5, [I-] = 0

solution is saturated with AgI (i.e. w/ Ag+ & I-, 1 mole of AgI = 1 mole of (Ag+) + 1 mole of (I-))

[I-] = 10^-1 means we added AgI = 10^-1 OR Ag+=10^-1 & I-= 10^-1 because we started with [I-] = 0

ksp = 10^-6 = [Ag+][I-]

but 10^-6 < (10^-1 + 10^-5)(10^-1)
10^-6 < approx. 10^-2

10^-2 is much larger than 10^-6. This means if we add 10^-1 we go beyond the saturation point