How do you calculate molar solubility and ksp?

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    Molar solubility represents the number of moles of that can be dissolved per liter of solution before the solution becomes saturated.
    The solubility product constant or expresses the product of the of ions raised to the power of their respective stoichiometric coefficients from the equilibrium reaction.
    Usually you’ll get 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’s ##K_(sp)## is
    ##K_(sp) = [A]^(a) * [B]^(b)##
    Let’s say you know ##K_(sp)##. You can determine the molar solubility of both ions by using the fact that both concentrations increase proportional to their respective stoichiometric coefficients.
    For A the increase in the concentration of the ion will be ##a*x## (there are ##a## moles of ##A## dissociated in solution); for ##B## this increase will be ##b * x##. Therefore
    ##K_(sp) = (a*x)^(a) * (b * x)^(b) = a^(a) b^(b) * x^((a+b))##
    Since ##a## and ##b## are derived from the equilibrium equation solving for ##x## will produce the molar solubilities of both ions.
    LIkewise if you know the concentrations of the dissociated ions ##K_(sp)## can then be determined 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-2-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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