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### Topic: Common Ion Effect Solubility  (Read 895 times)

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#### MilkyCar

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##### Common Ion Effect Solubility
« on: November 22, 2023, 02:50:27 PM »
Stuck on this assignment question:
The Ksp for PbI2 (s) is 1.4 x 10^-8. Calculate solubility of PbI2 (s) in 0.41 M Pb(NO3)2.

I know that the balanced chemical equation for the dissolution of PbI2 is:
PbI2 (s) Pb2+ (aq) + 2I- (aq)

And I know that the Ksp expression is:
Ksp = [Pb2+][I-]^2

I did:
[Pb2+] = 2 x 0.41 M = 0.82 M

[Pb2+] = 0.82 + x M
[I-] = 2 x M

Ksp = (0.82 + x)(2x)^2

I am stuck on this step above and I'm not sure what I'm supposed to do here.

#### mjc123

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##### Re: Common Ion Effect Solubility
« Reply #1 on: November 22, 2023, 03:14:46 PM »
Assume that x << 0.82, so that (0.82 - x) ≈ 0.82. (Check the validity of this assumption at the end of the calculation.)

Hang on, why do you say [Pb2+] = 0.82 M?

#### MilkyCar

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##### Re: Common Ion Effect Solubility
« Reply #2 on: November 22, 2023, 04:32:09 PM »
My mistake, I multiplied 0.41 by 2 for no reason. Would it just be Ksp = (0.41 + x)(2x)^2?

#### MilkyCar

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##### Re: Common Ion Effect Solubility
« Reply #3 on: November 22, 2023, 09:40:07 PM »
And afterwards, I did 1.4x10^-8/0.41 in which I got 3.4 x 10^-8. And then, I square-rooted that to get 1.8 x 10^-4. Is this correct?

#### Borek

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##### Re: Common Ion Effect Solubility
« Reply #4 on: November 23, 2023, 03:10:56 AM »
And afterwards, I did 1.4x10^-8/0.41 in which I got 3.4 x 10^-8. And then, I square-rooted that to get 1.8 x 10^-4. Is this correct?

Close, but as mjc123 wrote: you should check validity of the assumption made.
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#### MilkyCar

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##### Re: Common Ion Effect Solubility
« Reply #5 on: November 23, 2023, 10:23:36 AM »
Alright, I did 1.8 x 10^-4/0.41 x 100 = 0.043 which is less than 5 percent. Correct assumption?

#### mjc123

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##### Re: Common Ion Effect Solubility
« Reply #6 on: November 23, 2023, 05:30:46 PM »
That's 0.043 percent, which is way less than 5%, so assumption is justified.

But wait - look at your calculation - 1.8 x 10-4 = 2x

Be careful. It's a shame to take trouble getting the hard part right, only to lose marks by careless errors in what should be easy.

#### MilkyCar

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##### Re: Common Ion Effect Solubility
« Reply #7 on: November 23, 2023, 08:30:14 PM »
Right, so I divided both sides by 2 after that and I got 9.0 x 10^-5 and then checked once again afterward, in which I got 0.021, which is still way less than 5 percent.

This must be the actual correct answer, right?

#### mjc123

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##### Re: Common Ion Effect Solubility
« Reply #8 on: November 24, 2023, 04:24:53 PM »
It's my answer. But try to get into the habit of always using units - the answer is 9 x 10-5 M. Actually, to 2 s.f. it's 9.2 x 10-5 M - don't round off prematurely.
And 9 x 10-5/0.41 = 0.021 %.