The solubility product of AgCl is `4.0 xx 10^(-10)` at 298 K . The solubility of AgCl in 0.04 M Ca `Cl_(2)` will be
The solubility product of AgCl is `4.0 xx 10^(-10)` at 298 K . The solubility of AgCl in 0.04 M Ca `Cl_(2)` will be
A. `2.0 xx 10^(-5) M`
B. `1.0 xx 10^(-4)M`
C. `5.0 xx 10^(-9) ` M
D. `2. 0 xx 10^(-4) M`
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Correct Answer - C
In `CaCl_(2),[Cl^(-)]=2 xx 0.04=0.08 mol L^(-1) [Cl^(-)]` from AgCl is too small and is neglected
`K_(sp)=[Ag^(+)][Cl^(-)]`
`4 xx 10^(-10)=[Ag^(+)]xx 0.08`
`4 xx 10^(-10)=[Ag^(+)]xx0.08`
`[Ag^(+)]=(4 xx 10^(-10))/(0.08)(4)/(8)xx10^(-8)`
`=5.0 xx 10^(-9) M`.
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