Calculating Initial Concentrations for Clock Reaction Lab | Trial Data Included

[ashley_s_411]

Reaction Lab (NEED HELP!)

1. This is a clock reaction lab where you have 10mL of KIO3 solution in a beaker, and you are adding 10mL of NaHSO3 solution to it. You then start the stopwatch and wait until the solution turns blue. I need to calculate the initial concentrations of KIO3 in each of the mixtures for part A after mixing. There are trials in total, and I will show my data as well. There is always 10mL of NaHSO3 in the beaker.
Trial 1: Add 10mL KIO3
Trial 2: Add 7mL KIO3 and 3mL Water
Trial 3: Add 5mL KIO3 and 5mL Water
Trial 4: Add 3mL KIO3 and 7 mL Water

My Teacher says to use the formula C1V1=C2V2, but I don't know which values to put where.

*Note: Concentration of KIO3= 0.0200 mol/L
Concentration of NaHSO3= 0.00200 mol/L 3. My Attempts:

I have tried for Trial one and this is what I got:
C1=what I am solving for
V1=10.00 mL
C2=0.00200 mol/L
V2=20.00 mL

c1=c2v2/v1
=0.00200 mol/L x 20.00 mL/10.00mL
=0.004 mol/L

Is this right? I just don't get what to do for the next 3 trials because water is added?

Thanks so much!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[symbolipoint]

You could us the variables subscripted as 1 for initial C and V, and subscripted 2 for final C and V.

 

[Blueshift5]

Hello,

I understand your confusion with the initial concentrations for the clock reaction lab. The formula your teacher provided, C1V1=C2V2, is correct but it may be easier to understand if we break it down for each trial.

For Trial 1, you are correct in your calculation. The initial concentration of KIO3 would be 0.004 mol/L.

For Trial 2, we need to take into account the dilution of the KIO3 solution. Since you are adding 7mL of KIO3 and 3mL of water, the total volume of the solution is now 20mL. We can use the same formula, but now C1=0.0200 mol/L, V1=7.00 mL, C2=0.00200 mol/L, and V2=20.00 mL. Plugging in these values, we get C1=0.0014 mol/L. This is the initial concentration of KIO3 for Trial 2.

For Trial 3, the same concept applies. The total volume of the solution is still 20mL, but now we have 5mL of KIO3 and 5mL of water. Plugging in the values, we get C1=0.001 mol/L.

For Trial 4, the same calculation applies. The total volume is 20mL and we have 3mL of KIO3 and 7mL of water. Plugging in the values, we get C1=0.0006 mol/L.

In summary, the initial concentrations for the four trials would be:
Trial 1: 0.004 mol/L
Trial 2: 0.0014 mol/L
Trial 3: 0.001 mol/L
Trial 4: 0.0006 mol/L

I hope this helps clarify the calculations for you. Just remember to take into account the dilution of the KIO3 solution when adding water. Good luck with your lab!