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`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)` |
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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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