# Solubility Problem

#### caragheen

In class, we've been working on predicting equations given particular chemicals, and later, actually testing some of them.
I've been doing just fine, but I'm stuck on one of them--

1. The problem statement, all variables and given/known data + attempted solutions

I was given: AgNO3 + Na2SO4, and was told to predict what the products would be.
Here is what my equation looks like--I know it's a double replacement reaction, and I matched up the ions, balanced them, and determined their solubility(using my rules):

2AgNO3(aq) + Na2SO4(aq) yields 2NaNO3 (aq) + Ag2SO4 (precipitate)

However, when I did the lab, the 2 solutions, when mixed together, did not form a precipitate/did not form a solid. Thus, this means that the products of this reaction should all be aqueous, so that Ag2SO4 (precipate) should actually be aqueous.
This does not seem to make sense to me though, as in my rules, it states that "sulfate salts are soluble/aqueous, but there are exceptions with Ag, Hg, Pb, Ca, Ba, and Sr"

I repeated the lab 3 more times, getting the same results.
Is there some sort of extra exception in this solubility rule? Why didn't I get a precipitate when my equation states that I should?

Homework Helper

#### caragheen

Hmm...
The concentration is something I'm not really sure of--we were told to mix 2 drops of each solution with each other.

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#### Borek

Mentor
Part of what you wrote is right, part is either wrong or my English fails me.

In general classification of substances as soluble/insoluble is only an approximation. All substances are soluble in all solvents. However, they differ in solubility. Saturated solution of AgI is about 10-9M, saturated solution of calcium sulfate is about 10-2M - both are considered insoluble, but solubility of one is 107 times larger. There is no such thing as "low saturated solution" - solution either is saturated, or is not. If solution is not saturated, there is no precipitate (or, in other words, there is no solid in equilibrium with the solution).

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#### chemisttree

Homework Helper
Gold Member
Silver sulfate has a solubility of 1.2 g/100 mL: http://en.wikipedia.org/wiki/Silver_sulfate
It is actually 0.84 g/100mL water at 25oC. Wiki is wrong a bit. Your point is still valid.

Caraqheen, are you sure you used sodium sulfate and not ammonium sulfate?

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#### caragheen

@ chemisttree--
Yes, I am positive that I used sodium sulfate.
Also, just to be on the safe side, I checked with one of my classmates today and we both had the same results--a predicted product side where one was a precipitate while in the actual lab, there was no precipitate evident.

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#### Borek

Mentor
Thank you for clearing up some of the parts that were confusing to me--I apologize for any inconveniences I gave you with my conclusion.
No need to apologize, as long as you are learning you have every right to be occasionally wrong

Thus, the silver sulfate turns out to be aqueous because the solution was not saturated.
Not sure if that's a correct way of stating things, or at least I have never seen it stated this way. I would not say "sulafte is aqueous", rather "sulfate was dissolved".

If the AgSO4 solution had been saturated, or more concentrated, it's number of ions would have enabled a precipitate to be formed.
I don't like the wording, but the idea behind is not incorrect.

And the reason why it was unsaturated in the first place was because the two original solutions (the silver nitrate and the sodium sulfate) weren't saturated?
No, it doesn't matter. You can mix two unsaturated solutions and observe precipitate, that's quite common. Solution is saturated for each salt (pair of ions) separately.

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#### caragheen

Okay--thanks for answering my side question.

Just as a note, if a solution has more ions dissolved than it can hold, wouldn't that be called a supersaturated solution?

If not 100% correct, at least I know I'm heading in the right path. :)

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#### Borek

Mentor
Looks much better now

And yes, that would be a supersaturated solution.

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#### caragheen

Thank you so much for helping me, Borek!
My question has been answered--thanks to all who helped me here as well.

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