1.

K_(sp)of AgCl is 1.8 times 10^-10. Calculate molar solubility in 1M AgNO_3

Answer»

Solution :`AGCL(s) leftrightarrow Ag^+(aq)+Cl^-(aq)`
X= SOLUBILITY of AgCl in 1 M `AgNO_3`
`AgNO_3(s) leftrightarrow Ag^+(aq)+NO_3^- (aq)`
`[Ag^+]=x+1 approx 1M (because x lt lt 1)`
`[Cl^-]=x`
`K_(SP)=[Ag^+][Cl^-]`
`1.8 times 10^-10 =(1) (x)`
`x=1.8 times 10^-10 M`


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