- #1
firebirds0707
- 4
- 0
Solution A: is a dilute soultion of potassium iodate, KIO3 which is the source of iodate, IO2(1-)
Solution B: contains some starch and the other reacting species, the hydrogen sulfite ion, HSO3(1-)
1. In a small beaker obtain about 50mL of solution A (0.02 M IO3(-)) and in another small beaker obtain about 70mL of soulition B (HSO3(-))
Trial - Contents of Test tube 1 - Contents of Test Tube 2 - Time (s) - Concentraion IO3(-)
1 10.0mL A + 0mL water 10.0 mL B 6.9 (my guess 0.02)
2 9.0 mL A + 1mL water 10.0 mL B 8.5 (my guess 0.018)
3 8.0 mL A + 2mL water 10.0mL B 9.1 (my guess 0.016)
- My question is, am I getting the concentration right? I don't think I am and I am not sure how to figure it out.
My friend with the same lab results said he got:
1. 0.01
2. 0.09
3. 0.08
so basiclaly what I got divide by 2
Please help me out
thxx
Solution B: contains some starch and the other reacting species, the hydrogen sulfite ion, HSO3(1-)
1. In a small beaker obtain about 50mL of solution A (0.02 M IO3(-)) and in another small beaker obtain about 70mL of soulition B (HSO3(-))
Trial - Contents of Test tube 1 - Contents of Test Tube 2 - Time (s) - Concentraion IO3(-)
1 10.0mL A + 0mL water 10.0 mL B 6.9 (my guess 0.02)
2 9.0 mL A + 1mL water 10.0 mL B 8.5 (my guess 0.018)
3 8.0 mL A + 2mL water 10.0mL B 9.1 (my guess 0.016)
- My question is, am I getting the concentration right? I don't think I am and I am not sure how to figure it out.
My friend with the same lab results said he got:
1. 0.01
2. 0.09
3. 0.08
so basiclaly what I got divide by 2
Please help me out
thxx