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What is the ratio of moles of `Mg(OH)_(2)` and `Al(OH)_3` present in 1 lit saturated aqueous solution of `Mg(OH)_2` & `Al(OH)_3` (`K_(SP)` of `Mg(OH)_2=4xx10^(-12)` and (`K_(SP)` of `Al(OH)_3=1xx10^(-33)` .Give answer by multiply by `10^(-16)`. |
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Answer» Correct Answer - 80 `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` |
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