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Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as
\[
\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q .)+\mathrm{Cl}^{-}(a q .) ; \quad K_{s p}=\left[\mathrm{Ag}^{+}(a q .)\right]\left[\mathrm{Cl}^{-}(a q .)\right]
\]
\( P \)
Where \( K_{s p} \) is called the solubility product constant or simply the solubility
W product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation.
For concentrations of ions that do not necessarily correspond to equilibrium conditions we use the reaction quotient \( (Q) \) which is called the ion or ionic product \( (Q) \), to predict whether a precipitate will form. Note that \( Q \) has the same form as \( K_{s p} \).
The possible relationships between \( Q \) and \( K_{s p} \) are
\( QK_{s p} \quad \) Unsaturated solution
\( Q=K_{s p} \quad \) Saturated solution
\( QK_{s p} \) Supersaturated solution; precipitate will form
Will a precipitate form if 1 volume of \( 0.1 \mathrm{MPb}^{2+} \) ion solution is mixed with 3 volume of \( 0.3 \mathrm{M} \)
\( \mathrm{Cl}^{-} \)ion solution? [Given : \( K_{s p}\left(\mathrm{PbCl}_{2}\right)=1.7 \times 10^{-5} \mathrm{M}^{3} \) ]
(a) Yes
(b) No
(c) Ionic product is less than solubility product, hence precipitate will form
(d) Data insufficient
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