[UG]

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:
K_sp = [Ag⁺][CH₃CO₂^-] and
K_a = [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. 

[Borek]

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.

[UG]

So far so good. What does sum of HAcetate and H⁺ equal?

[H⁺] + [CH₃CO₂H] = 0.1 ?

[Borek]

Yes.

You have three equations and three unknowns now, I think it should be solvable.

[UG]

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

K_a = [H₃O⁺][CH₃CO₂-]/[CH₃COOH]

[CH₃COO^-]_dissolved = [CH₃CO₂-] + [CH₃COOH]

[Borek]

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.

[UG]

Amount of acetic acid and acetate equals solubility of the salt.

Is this the bit I got wrong? Or is this the right answer? 

[Borek]

Solubility = [Ag⁺]
[CH₃COO^-]_dissolved = [CH₃CO₂-] + [CH₃COOH]

[CH₃COO^-]_dissolved is just a solubility, isnt it? If so, it must equal concentration of [Ag⁺]...

[UG]

[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  

[Borek]

What variables do you see?

[UG]

Right now I see [Ag⁺] [CH₃COOH] [CH₃CO₂^-] and [H⁺]

[UG]

This is what I am doing, if I let x = [CH₃COOH], then
K_a = (0.1 - x)(CH₃CO₂^-) / x
I don't know what to do with the acetate from here.

[Borek]

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 3^rd 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 2^nd 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.

[UG]

Okay, see if I have the right idea, we have

[H⁺] + [CH₃CO₂H] = 0.1

K_a = [H₃O⁺][CH₃CO₂-]/[CH₃COOH]

[CH₃COO^-]_dissolved = [CH₃CO₂-] + [CH₃COOH]

[CH₃COO^-]_dissolved = [Ag⁺] 

K_sp = [CH₃COO^-][Ag⁺]

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

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

K_a = (0.1 - x)(K_sp/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?

[Borek]

[CH₃COO^-]_dissolved = [Ag⁺] 

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

K_sp = [CH₃COO^-][Ag⁺]

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

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

K_a = (0.1 - x)(K_sp/x) / x

What is x? I can guess, but I prefer to learn from you to be sure.