My prelab is telling me to find the initial concentrations of I(^-) and S208(^2-) given the following reaction:
25 mL of 0.2 M KI + 48.00 mL 0.2 M KNO3 + 1 mL 0.4 M Na2S2O3 + 1 mL starch + 25 mL 0.2 M (NH4)2S2O8 + 1 drop EDTA
How do I calculate the concentration of I^-1 when I am given the molality of KI? Is the molality of I^-1 a percentage of this number of the total? Same goes for S2O8(^-2) in (NH4)2S2O8.
Thanks!!
To calculate the initial concentration of I^(-) and S2O8^(2-), we need to consider the molarities provided for the different solutions and the volumes used for each component of the reaction.
Let's break down the given information and use it to determine the initial concentrations of I^(-) and S2O8^(2-):
1. We have a solution of KI with a molarity of 0.2 M and a volume of 25 mL.
- This means that we have (0.2 mol/L) * (0.025 L) = 0.005 mol of I^(-) in the KI solution.
2. We have a solution of (NH4)2S2O8 with a molarity of 0.2 M and a volume of 25 mL.
- This implies that we have (0.2 mol/L) * (0.025 L) = 0.005 mol of S2O8^(2-) in the (NH4)2S2O8 solution.
Now, let's calculate the total volume of the solution, considering all the components used:
- 25 mL KI + 48.00 mL KNO3 + 1 mL Na2S2O3 + 1 mL starch + 25 mL (NH4)2S2O8 + 1 drop EDTA
Using the given volumes, the total volume is:
25 mL + 48.00 mL + 1 mL + 1 mL + 25 mL + 1 drop
Note: Typically, the volume of a drop is about 0.05 mL, but this can vary. You may need to consult your lab manual or instructor to determine the exact volume of a drop in your experiment.
Finally, when you have the total volume, you can calculate the initial concentrations of I^(-) and S2O8^(2-):
- Concentration of I^(-) = (0.005 mol) / (total volume in liters)
- Concentration of S2O8^(2-) = (0.005 mol) / (total volume in liters)
By following these steps, you should be able to determine the initial concentrations of I^(-) and S2O8^(2-) based on the given information in your prelab.