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The solubility product for silver choride is 1.2xx10^(-10) at 298 K. Calculate the solubility of silver chloride at 298 K. |
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Answer» Solution :Silver chloride dissociates according to theequation : `AgCl(s) hArr AgCl(aq) hArr AG^(+) (aq) + CL^(-)(aq)` LET s be the solubility of AgCl in moles per litre. Consequently, the molar concentration of `Ag^(+) and Cl^(-)` will also be s each. Substituting in the expression for solubilityproduct of AgCl, `K_(sp)=[Ag^(+)][Cl^(-)]=s xx s =s^(2)` But `K_(sp)= 1.2xx10^(-10) ` (Given) `:. s^(2)=1.2xx10^(-10) or s = sqrt(1.2xx10^(-19))=1.1xx10^(-5) "mol " L^(-1)` |
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