Answer:
concentration of Mg ion = 0.0122 g/L
Explanation:
Given data;
initial concentration of Magnesium in water is 40 mg/l
concentration of [tex](OH^-) = 0.001000[/tex]
we have dissociation reaction Magnesium dioxide
[tex]Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^{-}[/tex]
from above reaction we can conclude
concentration of [tex]Mg(OH)_2 = \frac{OH}{2} = \frac{.0010}{2} = 0.0005 M[/tex]
Mass of magnesium ion is calculated as = Mg mole * molar mass of magnesium
concentration of Mg ion = 0.0005*24.305 g/mol = 0.0122 g/L