To calculate the pH of a Mg(OH)2 solution, you need to consider the dissociation of the compound into its constituent ions, magnesium ions (Mg2+) and hydroxide ions (OH-). Since Mg(OH)2 is a strong base, it dissociates completely in water.
The balanced chemical equation for the dissociation of Mg(OH)2 is as follows:
Mg(OH)2(s) → Mg2+(aq) + 2OH-(aq)
Given that the concentration of the Mg(OH)2 solution is 0.2 mol∙dm-3, the concentration of magnesium ions (Mg2+) and hydroxide ions (OH-) will also be 0.2 mol∙dm-3.
Since Mg(OH)2 dissociates to give two OH- ions for every one Mg2+ ion, the concentration of hydroxide ions will be twice the concentration of Mg2+ ions. Therefore, [OH-] = 2 × 0.2 = 0.4 mol∙dm-3.
The pH of a solution can be calculated using the equation:
pOH = -log[OH-]
To find the pOH, substitute the concentration of hydroxide ions:
pOH = -log(0.4) ≈ 0.3979
Since pH + pOH = 14 (at 25°C), you can subtract the pOH from 14 to find the pH:
pH = 14 - 0.3979 ≈ 13.6021
Therefore, the pH of the Mg(OH)2 solution with a concentration of 0.2 mol∙dm-3 is approximately 13.6021.