# How to find the mass of dissolved AgCl when placed in a solution of NaCl?

I think the following problem in my textbook is malformed:

Solid $$\ce{AgCl}$$ is placed in $$1\,\mathrm{L}$$ of $$0.55\,\mathrm{M}$$ $$\ce{NaCl}$$. Find the mass of $$\ce{AgCl}$$ which dissolves.

I know the method to solve the problem. I think the problem is malformed: When we get an answer, say $$x$$ grams $$\ce{AgCl}$$ dissolves, but if we placed less than $$x$$ grams solid $$\ce{AgCl}$$ in the solution initially, then it would be impossible for $$x$$ grams to dissolve.
So we obviously have to make certain assumptions, but I am looking at the solution in my textbook and I don't see in which part of the solution is the assumption used (implicitly). And by the way, how do we solve the problem if we have place less than $$x$$ grams solid $$\ce{AgCl}$$?

• Is that all the data that was provided, or were any equilibrium constants given? Mar 24 '15 at 7:47
• When you have solved the problem, you will notice that there is less than 1 g of AgCl dissolved (probably even less than 1 mg). I think the question should be reworded to ask What is the maximum amount of AgCl that can dissolve in 1 L of 0.55 M NaCl (at 25°C)?
– LDC3
Jun 20 '15 at 16:07

The question is most likely badly worded and should have been

What is the maximum mass of $\ce{AgCl}$ that can be dissolved?

As we all know, $\ce{AgCl}$ is damn insouble in aquaeous solutions — and due to the equilibrium constant it doesn’t get better at all if you add chloride ions (reducing the silver ion concentration). So without having done the maths, I’m going to assume that a small spec of nothingness is all that will dissolve. That spec is so tiny that you will have a hard time weighing less so you can actually perform the experiment you suggested.

Because it’s just a flick of dust, and because the authors likely

1. knew that, too
2. assumend you knew that, too

they may have gone with a less obvious wording. I mean it’s clear: You can’t dissolve what you didn’t put in, and to-be-dissolved silver ions aren’t going to magically appear with Alakazam.

Since the already present sodium chloride highly ionic, it's dissociation can be considered 100% and thus the contribution of chloride form silver chloride would be thus negligible (in comparison to rock salt's) or suppressed. Thus we can treat the final chloride concentration as $0.55\rm M$. Now the solubility product is given as: $$K_{\rm sp,AgCl}=\ce{[Ag+][Cl- ]}$$ Now we can find out $\ce {[Ag+]=\frac{K_{\rm sp,AgCl}}{[Cl- ]}}$. This would actually be all from silver chloride and hence the solubility of it.

• What about the second part of my question (the last sentence)? May 30 '15 at 16:37