1.

The solubility of AgCl(s) with solubility product 1.6 times 10^-10 in 0.1 M NaCl solution would be …………

Answer»

`1.26 times 10^-5 M`
`1.6 times 10^-9M`
`1.6 times 10^-11 M`
Zero

Solution :`AGCL(s)=Ag^+(aq)+CL^(-) (aq)`
`NaCl to Na^++Cl^-`
`K_(sp)=1.6 times 10^-10`
`K_(sp)=[Ag^+][Cl^-]`
`K_(sp)=(s) (s+0.1)1`
`0.1 gt gt gt s`
`THEREFORE s+0.1 approx 0.1`
`therefore s=(1.6 times 10^-10)/0.1=1.6 times 10^-9`


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