# Solubility equilibrium [closed]

This is a worked example about solubility equilbrium I am new to this subject and I am really confused first of all I know that Q is the product of concentration of Ag and Cl in the solution and ksp is the product of molar solubility of Ag and Cl then why the difference of these two is not amount of precipitation second: if we are not given ksp value of AgCl and we are given only the last part ( part b) how can we find the ksp value?
P.s: after change the amount of cl become zero and amount of Ag become 99*10^-6 so it means that after mixing two solution precipitate is formed and amount of dissolved Ag ions become 99*10^-6 and all of Cl ion is precipitated am I right?
(Sorry if my questions doesn't make sense for you it is because I am new to subject and I don't have any teacher I really tried hard on this example and I saw lots of videos about this subject but I didn't understand it)

## closed as off-topic by Mithoron, Jon Custer, airhuff, Tyberius, Todd MinehardtMar 2 '18 at 22:47

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Q is known as the ionic product which is defined as product of concentration of ions of the electrolyte, each raised to the power of their coefficients in the balanced chemical equation in solution at any concentration. $\pu{K_{sp}}$ is the same but it is applicable only for a saturated solution.

To find the solubility for the salt with the given $\pu{K_{sp}}$ value, For $\ce{AgCl}$, we have the equilbrium, $$\ce{ AgCl(s) <=> Ag+(aq) + Cl-(aq)}$$ If initally we have $‘a ‘$ moles of AgCl, at equilbrium we will have $’a-s ‘$ and we have $\ce{[Ag+]=[Cl-]=s}$. Hence we get the relation as, $$\pu{K_{sp}} = s^2$$ This is the maximum concentration of the ions present in solution at equilbrium.(because $\pu{K_{sp}}$ = Q). When Q just crosses $\pu{K_{sp}}$, salt is precipitated out.

Since they are products of ionic concentration, we can’t directly subract Q and $\pu{K_{sp}}$ to find the amount of salt precipitated.

I’m just working backwards to get $\ce{K_{sp}}$.

Let mass of $\ce{AgCl}$ precipitated be $’m ‘$ g.
No. of moles precipitated $n = \frac{m}{M}$. Let the volume be $’V’$ litres,
$\ce{[AgCl]}$ precipitated $x = \frac{n}{V}$

$$\ce{Ag+ + Cl- -> AgCl}$$

We know initial concentrations of the ions. Let $\ce{[Ag+]}$ and $\ce{[Cl-]}$ be $’p‘$ and $’q‘$ respectively.
Then solubility product is,
$$\ce{K_{sp}} = (p-x)(q-x)$$

Hope that makes sense.

• And all the $\ce{Cl-}$ is precipitated, what you said is correct. – MollyCooL Mar 2 '18 at 9:31
• you mean when Q=Ksp s=amount of precipitated ions? And also can you please do the calculation to find ksp value from part b? Because when I did the calculation for ksp It didn't come up to be 1.6*10^-10 which is the actual ksp value for AgCl – Elize Mar 2 '18 at 11:09
• When Q=$\pu{K_{sp}}$, at this point addition of more salt results in precipitation. Solubility is the maximum moles of salt that can be dissolved per Litre of solution at a given temperature. I ll make my answer more clear. Please list out what we are given exactly and I shall try to find $\pu{K_{sp}}$. Thanks – MollyCooL Mar 2 '18 at 11:37
• Suppose that we are given the mass of precipitated Agcl and the volume of solution and we know the initial mole number of Ag and Cl – Elize Mar 2 '18 at 11:46
• so only in case that Q>ksp preciption occur then if Q=ksp we don't have any precipitation and hence No equilbrium right? – Elize Mar 2 '18 at 12:15