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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: UG on April 11, 2010, 05:54:33 AM

Title: Solubility of silver acetate
Post by: UG on April 11, 2010, 05:54:33 AM
Hey everyone, I'm stuck on another one of these questions  ???
It goes: Silver acetate is a slightly soluble salt of a weak acid (Ka=1.75 x 10–5). At 20 oC, 100 g of water dissolves 1.04 g of crystalline silver acetate. Compare the solubility of silver acetate in pure water with that in 0.10 M nitric acid.

So, the first part is fairly straightforward, n(AgAc) = 1.04 g/166.87 g mol-1
n = 6.23 x 10-3 mol in 100 grams
So in one litre the concentration is 0.0623 mol L-1
Ksp = 0.06232 = 3.88 x 10-3
The next part I am a bit unsure, I know that:
Ksp = [Ag+][CH3CO2-] and
Ka = [H3O+][CH3CO2-]/[CH3COOH]
Solubility = [Ag+]
[CH3COO-]dissolved = [CH3CO2-] + [CH3COOH]

Is my approach to this question correct so far? Are there more equations I can write? I am not sure how to proceed from here on. 
Title: Solubility of silver acetate
Post by: Borek on April 11, 2010, 06:46:07 AM
So far so good. What does sum of HAcetate and H+ equal?

You really should start a new thread on that, I will split it later.
Title: Solubility of silver acetate
Post by: UG on April 11, 2010, 06:50:53 AM
So far so good. What does sum of HAcetate and H+ equal?
[H+] + [CH3CO2H] = 0.1 ?
Title: Solubility of silver acetate
Post by: Borek on April 11, 2010, 08:40:18 AM
Yes.

You have three equations and three unknowns now, I think it should be solvable.
Title: Solubility of silver acetate
Post by: UG on April 11, 2010, 06:14:23 PM
Yes.

You have three equations and three unknowns now, I think it should be solvable.
I do? These three? But what about [CH3COO-]dissolved? Is that not a fourth variable?

[H+] + [CH3CO2H] = 0.1

Ka = [H3O+][CH3CO2-]/[CH3COOH]

[CH3COO-]dissolved = [CH3CO2-] + [CH3COOH]

Title: Solubility of silver acetate
Post by: Borek on April 11, 2010, 06:54:53 PM
No, there is no separate value for dissolved acetate.

Ah, I think I see where you have confused yourself.

Amount of acetic acid and acetate equals solubility of the salt.
Title: Solubility of silver acetate
Post by: UG on April 11, 2010, 07:03:46 PM
Amount of acetic acid and acetate equals solubility of the salt.
Is this the bit I got wrong? Or is this the right answer?  :-\
Title: Solubility of silver acetate
Post by: Borek on April 12, 2010, 03:17:13 AM
Solubility = [Ag+]
[CH3COO-]dissolved = [CH3CO2-] + [CH3COOH]

[CH3COO-]dissolved is just a solubility, isnt it? If so, it must equal concentration of [Ag+]...
Title: Solubility of silver acetate
Post by: UG on April 12, 2010, 03:23:09 AM
[CH3COO-]dissolved is just a solubility, isnt it? If so, it must equal concentration of [Ag+]...

Okay, I understand that part, on second thought, I still can't see the three variable  :-\

Title: Solubility of silver acetate
Post by: Borek on April 12, 2010, 04:25:16 AM
What variables do you see?
Title: Solubility of silver acetate
Post by: UG on April 12, 2010, 04:36:24 AM
Right now I see [Ag+] [CH3COOH] [CH3CO2-] and [H+]
Title: Solubility of silver acetate
Post by: UG on April 12, 2010, 05:00:18 AM
This is what I am doing, if I let x = [CH3COOH], then
Ka = (0.1 - x)(CH3CO2-) / x
I don't know what to do with the acetate from here.
Title: Solubility of silver acetate
Post by: Borek on April 12, 2010, 05:08:23 AM
Argh, my mistake. Not three in three but four in four. But you have correctly listed all equations so far, and now you have correctly listed all unknowns.

Unfortunately, that leads to a full blown 3rd degree polynomial. Solvable, but not easily solvable.

There is probably a way of simplifying the system, although it will require checking afterwards.

We start with a very low pH, around 1. At so low pH acetic acid is almost completely protonated (see 2nd equation here: http://www.titrations.info/acid-base-titration-indicators - discussion is about indicators, but it is correct for any weak acid). That means [HA] >> [A-] - do you see how it can be used to modify equations?

Trick is, after finishing calculations we have to check if pH is really low enough. Acetate protonation consumes H+ and raises pH, concentrations of strong acid present and amount of acetate are comparable, so we need to be cautious.
Title: Solubility of silver acetate
Post by: UG on April 12, 2010, 07:12:26 AM
Okay, see if I have the right idea, we have

[H+] + [CH3CO2H] = 0.1

Ka = [H3O+][CH3CO2-]/[CH3COOH]

[CH3COO-]dissolved = [CH3CO2-] + [CH3COOH]

[CH3COO-]dissolved = [Ag+]  :-\

Ksp = [CH3COO-][Ag+]

If the acetate is very low, can I assume that [Ag+] ~ [CH3COOH] ?

And can I put it into the Ka like before ??

Ka = (0.1 - x)(Ksp/x) / x

(3.88 x 10-4 - 3.88 x 10-3 x) / x = 1.75 x 10-5 x
Calculating gives x = 0.0999549 M which would give [H+] = 4.51 x 10-5 M and pH = 4.3
Does this mean the approximation is not valid?
Title: Solubility of silver acetate
Post by: Borek on April 12, 2010, 08:00:18 AM
[CH3COO-]dissolved = [Ag+]  :-\

Correct. This is just a mass balance for silver acetate.

Quote
Ksp = [CH3COO-][Ag+]

If the acetate is very low, can I assume that [Ag+] ~ [CH3COOH] ?

And can I put it into the Ka like before ??

Ka = (0.1 - x)(Ksp/x) / x

What is x? I can guess, but I prefer to learn from you to be sure.
Title: Solubility of silver acetate
Post by: UG on April 12, 2010, 08:03:34 AM
x is ethanoic acid
Title: Solubility of silver acetate
Post by: Borek on April 12, 2010, 08:46:02 AM
(3.88 x 10-4 - 3.88 x 10-3 x) / x = 1.75 x 10-5 x
Calculating gives x = 0.0999549 M which would give [H+] = 4.51 x 10-5 M and pH = 4.3
Does this mean the approximation is not valid?

I am afraid so. Unfortunately, that also means you should solve full 3rd degree equation. Correct result doesn't have to be far from what you have calculated, but it is hard to predict how far it is. From my calculations pH is around 3.3.
Title: Solubility of silver acetate
Post by: UG on April 12, 2010, 05:41:41 PM
But the answer says that the solubility is 0.130 mol L-1 and pH is equal to 4.23  ???
Title: Solubility of silver acetate
Post by: Borek on April 13, 2010, 02:56:56 AM
Unfortunatelty that most likely means they used wrong approximation and have not checked its validity later.

Try to use equations you already have and the answer given as real to calculate concentrations of all ions and check if all mass balances and reaction quotients have correct values. My bet is that you will find they are inconsistent with question data.
Title: Re: Solubility of silver acetate
Post by: UG on April 13, 2010, 03:45:39 AM
Hmm... you're right, the answers come out all funny. So the way to solve this is to just try and find a way to express an equation using only one of the 4 variables?
Title: Re: Solubility of silver acetate
Post by: Borek on April 13, 2010, 04:40:09 AM
Yes. Unfortunately that leads to a 3rd degree polynomial.