Suppose 10.7 g of lead(II) nitrate is dissolved in 150 ml. of a 0.60 M aqueous solution of ammonium sulfate.
Calculate the final molarity of lead (II) cation in the solution.
You can assume the volume of the solution doesn't change when the lead(II) nitrate is dissolved it.
Answers
Explanation:
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Explanation: We can solve this problem using some molarity calculations:
molarity
=
mol solute
L soln
We should convert the given mass of
(NH
4
)
2
SO
4
to moles using its molar mass (calculated to be
132.14
g/mol
):
10.8
g (NH
4
)
2
SO
4
⎛
⎝
1
l
mol (NH
4
)
2
SO
4
132.14
g (NH
4
)
2
SO
4
⎞
⎠
=
0.0817
mol (NH
4
)
2
SO
4
This is the quantity present in
100
mL soln
, so let's calculate the molarity of the solution (converting volume to liters):
molarity
=
0.0817
l
mol (NH
4
)
2
SO
4
0.100
l
L soln
=
0.817
M
10
mL
of this solution is added to
50
mL H
2
O
, which makes a
60
-
mL
total solution.
We can now use the dilution equation
M
1
V
1
=
M
2
V
2
to find the molality of the new,
60
-
mL
solution:
(
0.0817
M
)
(
10
l
mL
)
=
(
M
2
)
(
60
l
mL
)
M
2
=
(
0.817
M
)
(
10
mL
)
60
mL
=
0.136
M
This means that there are
0.136
moles of
(NH
4
)
2
SO
4
per liter of solution.
Let's recognize that
1
mol (NH
4
)
2
SO
4
contains
2
mol NH
+
4
1
mol SO
2
−
4
The concentrations of each ion is thus
(
2
)
(
0.136
M
)
=
0.272
M
NH
+
4
(
1
)
(
0.136
M
)
=
0.136
M
SO
2
−
4