1.

`K_(sp)` for `Cr(OH)_(3)` is `2.7 xx 10^(-31)`. What is its solubility in moles/litreA. `1 xx 10^(-8)`B. `8 xx 10^(-8)`C. `1.1 xx 10^(-8)`D. `0.18 xx 10^(-8)`

Answer» Correct Answer - A
`{:(Cr(OH)_(3),rarr,Cr^(+3)+,3OH^(-)),(,,x,3x):}`
`K_(sp) = x.(3x)^(3) = 27 x^(4)`
`x = 4sqrt((K_(sp))/(27)) , x = 4sqrt((2.7 xx 10^(-31))/(27))`
`x = 1 xx 10^(-8)` mole/litre.


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