Question

A water initially contains 40 mg · L−1 of Mg2+. The pH of the water is increased until the concentration of hydroxide ion (OH−) is 0.001000 M. What is the concentration of magnesium ion in this water at this pH? Give your answer in milligrams per liter. Assume that the temperature of the solution is 25◦

Answer 1

Answer:

concentration of Mg ion  = 0.0122 g/L

Explanation:

Given data;

initial concentration of Magnesium in water is 40 mg/l

concentration of $(OH^-) = 0.001000$

we have dissociation reaction  Magnesium dioxide

$Mg(OH)_2 \rightarrow Mg^{2+} + 2OH^{-}$

from above reaction we can conclude

concentration of $Mg(OH)_2 = \frac{OH}{2} = \frac{.0010}{2} = 0.0005 M$

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