Answer:
5.6x10^–3 M
Explanation:
We'll begin by calculating the pOH of the solution. This can be achieved as shown below:
pH + pOH = 14
pH = 12.05
12.05 + pOH = 14
Collect like terms
pOH = 14 – 12.05
pOH = 1.95
Next, we shall determine the concentration of hydroxide ion OH- in the solution. This is illustrated below:
pOH = – Log [OH-]
pOH = 1.95
1.95 = – Log [OH-]
– 1.95 = Log [OH-]
[OH-] = anti log (– 1.95)
[OH-] = 0.0112 M
Finally, we can obtain the concentration of Mg(OH)2 as follow:
In solution, Mg(OH)2 will dissociates as shown below:
Mg(OH)2 —> Mg2+ + 2OH-
From the balanced equation above,
1 mole of Mg(OH)2 produced 2 moles of OH-.
Therefore, XM of Mg(OH)2 will produce 0.0112 M of OH- i.e
XM of Mg(OH)2 = 0.0112/2
XM of Mg(OH)2 = 5.6x10^–3 M
Therefore, the concentration of Mg(OH)2 is 5.6x10^–3 M
