The solubility of `Mg( OH)_2` in pure water is ` 9.57 xx 10 ^(-3) g L^(-1) ` .Calculate its solubility I ` (g L^(-1))` in 0.02 M Mg `(NO_3)_2` solutions.

Options

A. `1.48 xx 10^(-5)`
B. `9.57 xx 10^(-3)`
C. `0.01`
D. `8.63 xx 10^(-4)`

Answer

4

Explanation

Solubility of Mg `(OH)_(2)` in pure water in moles/litre .
`= (9.57 xx 10^(-3))/(58.31) = 1.64 xx 10^(-4)` [Molecular Mass of Mg `(OH)_(2) = 58.31`]
`[Mg^(2+)] = 6.64 xx 10^(-4) " and " [OH^(-)] = 2 xx 1.64 xx 10^(-4)`
`K_("sp")[Mg(OH)_(2)] = [Mg^(2+)][OH^(-)]^(2)= (1.64 xx 10^(-4) xx 10^(-4))^(2) = 1.764 xx 10^(-11)`
Let the solubility of `Mg(OH)_(2) ` in `0.02 M Mg(NO_(3))_(2) ` solution be 's' moles per litre . It this case.
`[Mg^(+2)] = s + 0.02`
`[OH^(-)] = 2s`
` K_("sp") = (s + 0.02) (2s)^(2) = 1.764 xx 10^(-11)`
` s = 1.48 xx 10^(-5)` mole per litre
` = 1.48 xx 10^(-5) xx 58.31 " g per litre " = 8.63 xx 10^(-4) ` g per letre