1.

The solubility of AgCl(s) with solubility product 1.6xx10^(-10) in 0.1 (M) NaCl solution would be-

Answer»

`1.26xx10^(-5)` (M)
`1.6xx10^(-9)`(M)
`1.6xx10^(-11)` (M)
zero

Solution :Let the solubility of AgCl in 0.1(M) NaCl is x `mol*L^(-1)`
Thus `[AG^+]=x mol*(-L)`
`[Cl^-]=(x+0.1)mol*L^(-1)`
`K_(sp)=[Ag^+][Cl^-]=x(x+0.1)`
or, `x(x+0.1)=1.6xx10^(10)`
or `x^2+0.1x-1.6xx10^(-10)=0`
or `x=1.6xx10^(-9)mol*L^(-1)`


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