A dilution is made during the preparation of a reaction mixture. The compounds are: 2.00 ml of 0.325 KMnO4, 7.50 ml of 0.155 M Oxalic acid, and 10.00 ml of H2O. What are the final concentrations of the compounds?
compute moles of each, eg
molesKMnO4=.325*.002=...
now, with those moles computed, all all the volumes .002+.0075+.010 l, and divide each of the moles you found by the new total volume.
To find the final concentrations of the compounds after the dilution, you need to use the dilution formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Let's calculate the final concentrations of each compound:
1. KMnO4:
Initial concentration (C1) = 0.325 M
Initial volume (V1) = 2.00 ml
Final volume (V2) = 2.00 ml + 7.50 ml + 10.00 ml = 19.50 ml
Plug these values into the dilution formula:
(0.325 M)(2.00 ml) = C2(19.50 ml)
C2 = (0.325 M)(2.00 ml) / (19.50 ml)
C2 ≈ 0.033 M
Therefore, the final concentration of KMnO4 is approximately 0.033 M.
2. Oxalic acid:
Initial concentration (C1) = 0.155 M
Initial volume (V1) = 7.50 ml
Final volume (V2) = 2.00 ml + 7.50 ml + 10.00 ml = 19.50 ml
Using the dilution formula:
(0.155 M)(7.50 ml) = C2(19.50 ml)
C2 = (0.155 M)(7.50 ml) / (19.50 ml)
C2 ≈ 0.059 M
Therefore, the final concentration of oxalic acid is approximately 0.059 M.
3. Water (H2O):
The concentration of water is assumed to be 0 M since it does not contribute to the molar concentration of the solution.
So, the final concentration of water remains 0 M.
In summary, the final concentrations of the compounds after the dilution are approximately:
- KMnO4: 0.033 M
- Oxalic acid: 0.059 M
- Water: 0 M (unchanged)