Calculate Ksp From Molar Solubility

How do you summate molar solubility and ksp?

Molar solubility represents the number of moles of that tin be dissolved per liter of solution earlier the solution becomes saturated.

The solubility product abiding, or , expresses the product of the of ions raised to the power of their corresponding stoichiometric coefficients from the equilibrium reaction.

Usually, you'll go either the value of ##K_(sp)## and have to determine the molar solubility of the dissociated ions, or vice versa. I'll demonstrate both cases. A general equation of a salt dissolved in aqueous solution is

##A_aB_(b(s)) rightleftharpoons aA_((aq)) + bB_((aq))##

The expression for this reaction'due south ##K_(sp)## is

##K_(sp) = [A]^(a) * [B]^(b)##

Let's say you lot know ##K_(sp)##. Yous can decide the molar solubility of both ions past using the fact that both concentrations increment proportional to their respective stoichiometric coefficients.

For A, the increase in the concentration of the ion volition be ##a*x## (there are ##"a"## moles of ##"A"## dissociated in solution); for ##"B"##, this increase volition be ##b * x##. Therefore,

##K_(sp) = (a*x)^(a) * (b * x)^(b) = a^(a) b^(b) * 10^((a+b))##

Since ##"a"## and ##"b"## are derived from the equilibrium equation, solving for ##10## will produce the tooth solubilities of both ions.

Besides, if you know the concentrations of the dissociated ions, ##K_(sp)## can then be adamant by plugging these values in the equation

##K_(sp) = [A]^(a) * [B]^(b)##

Here are some links to actual examples:

http://socratic.org/questions/find-the-max-concentration-of-mg-ii-ions-permissible-in-1l-0-01m-naoh-solution-if

http://socratic.org/questions/how-can-you-use-the-solubility-product-constant-to-calculate-the-solubility-of-a

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