The solubility of a sparingly soluble salt `A_xB_y` in water at `25^@C=1.5xx10^(-4)M`.The solubility product is `1.1xx10^(-11)` The possibilities are
The solubility of a sparingly soluble salt `A_xB_y` in water at `25^@C=1.5xx10^(-4)M`.The solubility product is `1.1xx10^(-11)` The possibilities are
A. x=1 ,y=2
B. x=2,y=1
C. x=1,y=3
D. x=3,y=1
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Correct Answer - A,B
`K_(sp)=1.1xx10^(-11)=(1.4 xx 10^(-4))^(x+y)x^x. y^y`
so we have x+y=3 (by comparing values )
so, `x^x .y^y=(1.1xx10^(-11))/(1.4xx1.4xx1.4xx10^(-12))=110/(1.6xx1.4)=4`
Hence x=1, y=2
y=1, x=2
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