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The solubility of AgCl in 0.2 M NaCl is `[K_(sp) AgCl =1.8 xx 10^(-10)]`A. `1.8 xx 10^(-11)M`B. `9.0 xx 10^(-10)M`C. `6.5 xx 10^(12)M`D. `5.6 xx 10^(-11)M`

Answer» Correct Answer - B
`K_(sp)(AgCl)=[Ag^(+)][Cl^(-)]`
Let the solubility of AgCl in 0.2 M NaCl be s mol `L^(-1)`
`:. K_(sp)=(s) (Cl^(-))`
`1.8 xx 10^(-10)=s xx 0.2`
`:. s=(1.8 xx 10^(-10))/(0.2)=9 xx 10^(-10)M`


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