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Derive the equation of solubility and solubility product of sparingly soluble salt M_x^(p+) X_y^(q-) |
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Answer» <P> Solution :`M_x^(p+) X_y^(q-)` is a sparingly soluble salt and solubility is S mol `L^(-1)`. The amount of this salt is dissolve in solution is totally in the FORM of ions. In solution 1 molecule `XM^(p+)` positive ion and `yX^(q-)` negative ion and form , from this salt and ionic equation is as follows .`M_x^(p+) X_(y(s))^q hArr xM_((aq))^(p+) +yX_((aq))^(q-)` ...(Eq.-i) Where `(x xx p^(+) = yxxq^(-))` The concentration of solid salt mix with `K_(SP)`, So solubility product is as follows . `K_(sp)=[M^(p+)]^x [X^(q-)]^y` But APPLYING stoichiometry and S `K_(sp)=(x S)^x (S)^y` `=x^x xx y^y (S)^((x+y))`...(Eq.-ii) `therefore (S)^((x+y)) = K_(sp)/(x^x y^y)` `therefore S=K_(sp)/((x^x xx y^y)^(1//x+y))`....(Eq.-iii) |
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