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What is the ratio of moles of Mg(OH)(2) ...

What is the ratio of moles of `Mg(OH)_(2)` and `Al(OH)_(3)`, present in 1L saturated solution of `Mg(OH)_(2)` and `Al(OH)_(3) K_(sp)` of `Mg(OH)_(2)=4xx10^(-12)` and `K_(sp)` of `Al(OH)_(3)=1xx10^(-33)`.[Report answer by multiplying `10^(-18)]`

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The correct Answer is:
8
What is the ratio of …………….
`{:(Mg(OH)_(2),,hArr,Mg^(2+)+2OH^(-)),(,x, 2x+3y),(AI(OH)_(3),hArr,AI^(3+)+3OH^(-)),(,y,3y+2x):}`
Since `K_(ap)` of `Mg(OH)_(2)gt K_(ap)` of `AI(OH)_(3)`
`:. X gtgt y` , `:. 2x+3y~=2x`
`4xx10^(-12)= [Mg^(2+)][OH^(-)]^(2)`
`=x+(2x)^(2)`
`:. x= 10^(-4)`
Similary `1xx10^(-33)=[AI^(3+)][OH^(-)]^(3)`
`1xx10^(-33)=yxx(2x)^(3)`
`:. y = (10^(-21))/(8)`
Thus `(x)/(y) = 8xx10^(17)`
`:.` Ans. `= 8xx10^(+17) = 8`
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