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)]`

Answer

80

Explanation

`Mg(OH)_2hArrMg^(+2)+2OH^(-)`
`x " " 2x+3y " " K_(SP)` of `Mg(OH)_2 gt K_(SP)"of" Al(OH)_3`
`Al(OH)_3hArrAl^(3+)+3OH^(-)`
`y " " 3y+2x" " so " " xgtgty`
`2x+3y=2x`
`because 4xx10^(-12)=[Mg^(2+)][OH^(-)]^2`
`=x xx (2x)^2`
`x=10^(-4)`
`1xx10^(-33)=[Al^(3+)][OH^(-)]^3`
`1xx10^(-33)=(y)(2x)^3`
`1xx0^(-33)=yxx(10^(-4)xx2)^3`
`y=10^(-21)/8 " " so " " x/y=8xx10^17 " " 8xx10^17xx10^(-16)=80`