Solution Stoichiometry Worksheet Page 3
ADVERTISEMENT
1
2
3
4b)
Calculate the concentration of NaOH in the solution after the reaction is complete, assume
a negligible volume change.
2 mol NaOH
⎛
⎞
0.0543 mol Na
= 0.0543 mol NaOH
⎜
⎟
2 mol Na
⎝
⎠
0.0543 mol NaOH
= 0.121 M NaOH
0.450 L H
O
2
NOTE: The amount of water that reacts is negligible compared to the amount of
water present, so the volume of solution is the same as the volume of water initially
present. Since the amount of sodium added is very small compared to the amount of
water we will assume the volume of the solution is the same as the volume of water.
We can only do this when the amount of solute is very, very much smaller than the
volume of water.
DCI14.4.
What volume of 0.406 M KOH is required to completely react with 18.50 mL of 0.287 M
H
SO
2
4
Important reaction is
2KOH
+ H
SO
→ K
SO
+ 2H
O
(aq)
4 (aq)
4 (aq)
(l)
2
2
2
We can use an ICE table to solve this problem:
2KOH
+ H
SO
→ K
SO
+ 2H
O
(aq)
4 (aq)
4 (aq)
(l)
2
2
2
Initial
Change
Final
The initial amount of H
SO
can be determined,
2
4
0.287 mol H
SO
⎛
⎞
2
4
-3
⎜
⎟
0.01850 L H
SO
= 5.31 x 10
mol H
SO
2
4
2
4
1 L
⎝
⎠
add this amount of H
SO
to the ICE table
2
4
2KOH
+
H
SO
K
SO
+
2H
O
→
(aq)
4 (aq)
4 (aq)
(l)
2
2
2
-3
Initial
5.31 x 10
mol
0
0
-3
Change
-5.31 x 10
mol
Final
Determine the amount of KOH that will react,
2 mol KOH
⎛
⎞
-3
-2
5.31 x 10
mol H
SO
= 1.06 x 10
mol KOH
⎜
⎟
2
4
1 mol H
SO
⎝
⎠
2
4
and add to the ICE table and complete the ICE table,
2KOH
+
H
SO
K
SO
+
2H
O
→
(aq)
4 (aq)
4 (aq)
(l)
2
2
2
CHEM 1314
3
FALL 2005
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education