At what pH, the precipitate La(OH), will occur when 0.010M La acidic solution is treated with NaOH? The Ksp of La(OH), is 2×10²¹
Answers
Answer:
The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions,
OH
−
, would cause the solid to precipitate out of solution.
As you know, the dissociation equilibrium for magnesium hydroxide looks like this
Mg
(
OH
)
2(s]
⇌
Mg
2
+
(aq]
+
2
OH
−
(aq]
The solubility product constant,
K
s
p
, will be equal to
K
s
p
=
[
Mg
2
+
]
⋅
[
OH
−
]
2
Rearrange to find the concentration of the hydroxide anions
[
OH
−
]
=
⎷
K
s
p
[
Mg
2
+
]
Plug in your values to get
[
OH
−
]
=
√
1.0
⋅
10
−
11
0.1
=
√
1.0
⋅
10
−
10
=
10
−
5
M
As you know, you can use the concentration of hydroxide anions to find the solution's pOH
pOH
=
−
log
(
OH
]
−
)
pOH
=
−
log
(
10
−
5
)
=
5
Finally, use the relationship that exists between pOH and pH at room temperature
pOH
+
pH
=
14
to find the pH of the solution
pH
=
14
−
5
=
9
So, for pH values that are below
9
, the solution will be unsaturated. Once the pH of the solution becomes equal to
9
, the solution becomes saturated and the magnesium hydroxide starts to precipitate.
Answer:
if it doesn't have any activity to be able or able and not only do it in a short amount and not just
Explanation:
he has protons to be in a position that is a bit of the same as a saddle
the density is a saddle and g is a bit high for class and it has been very popular since most
the density is a saddle and g
the density is a saddle