Question

Question #885cc

Answer 1

`[OH^-]` =`10^-5`

From `Ksp` of `Cu(OH)_2`, it starts precipitating at:
`2.2*10^-20` = `[Cu^(2+)]`*`[OH^-]^2`
`[OH^-]`=`sqrt (0.22*10^-20)`=`0.46*10^(-10)`

and from `Ksp` of `Mg(OH)_2`, it starts precipitating at
`6.3*10^-10` =`[Mg^(2+)]`*`[OH^-]^2`
`[OH^-]`=`sqrt (1.26*10^-10)` =`1.12*10^(-5)`

therefore at `[OH^-]`=`10^(-5)` all `Cu(OH)_2` is precipitated and all `Mg(OH)_2` is in solution