Chem 3411 Solution Set 5 - Mindanao State University - Iligan Institute Of Technology - 2010 Page 9
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14CHEM 3411, Fall 2010
Solution Set 5
We can construct an expression for b in terms of the initial concentration b
and the moles n of KNO
added to the
0
3
solution volume M
.
solv
n
b = b
+
0
M
solv
Additionally, as we’re interested in mass of KNO
we can replace moles n with mass m using molecular weight
3
M
.
w,KNO
3
m
b = b
+
0
M
M
solv
w,KNO
3
This can be substituted back into our expression for ionic strength I and the resulting equation can be solved for m
(mass of KNO
to add).
3
)
⊖
m =
Ib
b
M
M
0
solv
w,KNO
(
)
3
$ $
mol ¨ ¨ ¨
mol ¨ ¨ ¨
1.00 $ $
0.110 $ $
kg × 101.11 g $ $
× 0.500 & &
1
1
1
=
kg
kg
mol
= 44.994 g
+2
For part (b) we’ll once again begin with the definition of ionic strength and this time we’ll include the ion Ba
with
z = +2.
[
]
b
b
1
b
b
Ba(NO
)
Ba(NO
)
2
KNO
2
KNO
2
2
3
2
3
2
3
3
I
=
(+1)
+ ( 1)
+ (+2)
+ ( 1)
+
⊖
⊖
⊖
⊖
2
b
b
b
b
3b
b
Ba(NO
)
KNO
3
2
3
=
+
⊖
⊖
b
b
where in this equation b
is the initial (and constant) concentration of potassium nitrate and b
is the
KNO
Ba(NO
)
3
3
2
concentration of barium dinitrate created by adding Ba(NO
)
to the solution.
3
2
Similarly to part (a) we’ll construct an expression for b
in terms of mass of Ba(NO
)
added m.
3
2
Ba(NO
)
3
2
m
b
=
Ba(NO
)
3
2
M
M
solv
w,Ba(NO
)
3
2
Substituting this into our expression for I and solving for m gives
⊖
(Ib
b
) M
M
KNO
solv
w,KNO
3
3
m =
3
(
)
$ $
mol ¨ ¨ ¨
mol ¨ ¨ ¨
1.00 $ $
0.110 $ $
kg × 261.32 g $ $
1
1
× 0.500 & &
1
kg
kg
mol
=
3
= 38.762 g
Solution
• (a) m = 45.0 g
• (b) m = 38.8 g
9
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