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^(2+)+2OH^(-)hArr Mg(OH)_(2)`
`K_(SP)=[Mg^(2+)][OH^(-)]^(2)`
`[OH^(-)]= sqrt((K_(SP))/([Mg^(2+)]))=sqrt((1.0xx10^(-11))/(10^(-3)))=10^(-4)`
`pOH= -log[OH^(-)]= -log 10^(-4)=4`
`pH= 14-pOH=10`