Question

At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which `pH`, will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of `0.001 M Mg^(2+)` ions ?
A. `9`
B. `10`
C. `11`
D. `8`

Answer 1

Correct Answer - B
`Mg(OH)_(2)hArrMg^(2+)+2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^(-)]^(2)`
`[OH^(-)] = sqrt((K_(sp))/([Mg^(2+)]))=sqrt((1xx10^(-11))/(0.001))`
`pOH= -log[OH^(-)]= -log[10^(-4)]`
`pOH=4` and `pH=10`
`:. pH=14-pOH=14-4=10`