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 ?

Options

A. 9
B. 10
C. 11
D. 8

Answer

B

Explanation

` Mg ( OH ) _ 2 harr Mg ^( ++ ) + 2 OH ^ - `
` K _ (sp) = [Mg^(++) ][OH^ - ]^ 2 `
` 1.0 xx 10^( - 11) = 10 ^( - 3 ) xx [O H ^ - ] ^ 2 `
` [OH^-] = sqrt (( 10^( - 11))/( 10 ^( - 3)) ) = 10 ^( - 4 ) `
` therefore pOH = 4 `
` therefore pH + pOH = 14 `
` therefore pH = 10 `