The solubility of Mg(OH)2 at 298k is 1.5x10^-4
Calulate the solubility product
Answers
Explanation:
Magnesium hydroxide,
Mg
(
OH
)
2
, is considered Insoluble in aqueous solution, which implies that the ions it forms will be in equilibrium with the solid.
Magnesium hydroxide dissociates only partially to form magnesium cations,
Mg
2
+
, and hydroxide anions,
OH
−
Mg
(
OH
)
2(s]
⇌
Mg
2
+
(aq]
+
2
OH
−
(aq]
For an insoluble compound, its molar solubility tells you how many moles of the compound can be dissolved per liter of aqueous solution before reaching saturation.
In your case, a molar solubility of
s
=
1.6
⋅
10
−
4
means that you can only dissolve
1.6
⋅
10
−
4
moles of magnesium in a liter of water at that temperature.
Take a look at the dissociation equilibrium. Notice that every mole of magnesium hydroxide that dissociates produces
1
mole of magnesium cations and
2
moles of hydroxide anions.
This tells you that if you successfully dissolve
1.6
⋅
10
−
4
moles of magnesium hydroxide in one liter of solution, number of moles of each ion will be
n
M
g
2
+
=
1
×
1.6
⋅
10
−
4
=
1.6
⋅
10
−
4
moles Mg
2
+
and
n
O
H
−
=
2
×
1.6
⋅
10
−
4
=
3.2
⋅
10
−
4
moles
Since we're working with one liter of solution, you can cay that
[
Mg
2
+
]
=
1.6
⋅
10
−
4
M
[
OH
−
]
=
3.2
⋅
10
−
4
M
By definition, the solubility product constant,
K
s
p
, will be equal to
K
s
p
=
[
Mg
2
+
]
⋅
[
OH
−
]
2
Plug in these values to get
K
s
p
=
1.6
⋅
10
−
4
⋅
(
3.2
⋅
10
−
4
)
2
K
s
p
=
1.6
⋅
10
−
11
→
rounded to two sig figs
The listed value for magnesium hydroxide's solubility product is
1.6
⋅
10
−
11
, so this is an excellent result