Calculate the initial concentration of the ions below in the reaction mixture fo
ID: 1060717 • Letter: C
Question
Calculate the initial concentration of the ions below in the reaction mixture for each reaction (Runs 1 – 5) using the volumes and concentrations given in Table 1. Show a sample calculation for Run 1.
Solution A
Solution B
Run No
0.2% Starch, drops
0.001 MNa2S2O3
ML
0.01 M
KL, mL
H2O, mL
0.04 M
KBrO3, ml
0.1 M HCL, mL
1
4 drops
2
2
2
2
2
2
4 drops
2
4
0
2
2
3
4 drops
2
2
0
4
2
4
4 drops
2
2
0
2
4
5
4 drops
2
1
0
3
4
Run No
Initial [so2O32-]
Initial [I-] (M)
Initial [Bro3-] (m)
Initial [H+}] (M)
1
2
3
4
5
2. During the experiment, after each trial, where will the reaction mixture and runoff of rinsing your glassware be collected
3. If different than your answer to question 2, after the experiment where will the reaction mixture and runoff of rinsing your glassware be collected?
Solution A
Solution B
Run No
0.2% Starch, drops
0.001 MNa2S2O3
ML
0.01 M
KL, mL
H2O, mL
0.04 M
KBrO3, ml
0.1 M HCL, mL
1
4 drops
2
2
2
2
2
2
4 drops
2
4
0
2
2
3
4 drops
2
2
0
4
2
4
4 drops
2
2
0
2
4
5
4 drops
2
1
0
3
4
Explanation / Answer
Ans
Just use the equation: M(1) x V(1) = M(2) x V(2) where the M's stand for Concentrations, V's stand for volume.
the 1's stand for initial, 2's stand for final. remember that V(2) will be the total volume, 60ml. the M(1) will be the
concentrations before the mixture is made, and M(2) is what your looking for, the final concentration of that ion.
your going to have to do this 3 times, for water you cant find the M(2) because its DI (im guessing that stands for
deionized) water, its only going to take part in diluting the ions further (making the M(2) lower).
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