1/2 litre of 2xx10^(-3) M AlCl_3 and 1/2 litre of 4 xx 10^(-2) M solution of NaOH are mixed and the solution is diluted to 10^2 litres with water at room temperature. Will a precipitate form ? Given K_(sp)Al(OH)_3=5xx10^(-33)

1/2 litre of 2xx10^(-3) M AlCl_3 and 1/2 litre of 4 xx 10^(-2) M solut

1.	1/2 litre of 2xx10^(-3) M AlCl_3 and 1/2 litre of 4 xx 10^(-2) M solution of NaOH are mixed and the solution is diluted to 10^2 litres with water at room temperature. Will a precipitate form ? Given K_(sp)Al(OH)_3=5xx10^(-33)
Answer» Solution :The solubility equilibrium of `Al(OH)_3` is `Al(OH)_3 hArr Al^(3+) + 3OH^(-)` On mixing `1/2` litre of `AlCl_3`and `1/2` litre of NaOH, the total solution is made `10^2` litres by adding WATER. Concentration of `AlCl_3=(2xx10^(-3))/10^2 xx1/2=1xx10^(-5)` M Concentration of NaOH =`(4xx10^(-2))/10^2 xx1/2=2xx10^(-4)` M Since `AlCl_3` IONISES as : `AlCl_3 hArr Al^(3+) +3Cl^-` `[Al^(3+)]=[AlCl_3]=1xx10^(-5)` M Since NaOH ionises as : `NaOH hArr NA^(+) + OH^(-)` `[OH^-]=[NaOH]=2xx10^(-4)` Now, ionic product =`[Al^(3+)][OH^-]^3` `=1xx10^(-5)xx(2xx10^(-4))^3 =8xx10^(-17)` As the product the concentration of IONS of the salt is more than `K_(sp)`, the salt will be precipitated .