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!!
mols KI = mols I^- = M x L = ?
Then (I^-) = mols/L of solution. Total L = 25 + 48 + 1 + 1 + 25
jjkjn
To calculate the concentration of I^(-) and S208^(2-), you need to take into account the stoichiometry of the reaction and initial volumes of the reactants.
Given:
- 25 mL of 0.2 M KI
- 48.00 mL of 0.2 M KNO3
- 1 mL of 0.4 M Na2S2O3
- 25 mL of 0.2 M (NH4)2S2O8
1. Start by converting the volume of each solution to liters, as concentrations are typically given in moles per liter (M).
- 25 mL of 0.2 M KI = 0.025 L * 0.2 mol/L = 0.005 mol KI
- 48.00 mL of 0.2 M KNO3 = 0.048 L * 0.2 mol/L = 0.0096 mol KNO3
- 1 mL of 0.4 M Na2S2O3 = 0.001 L * 0.4 mol/L = 0.0004 mol Na2S2O3
- 25 mL of 0.2 M (NH4)2S2O8 = 0.025 L * 0.2 mol/L = 0.005 mol (NH4)2S2O8
2. Now, consider the balanced chemical reaction for the given equation to determine the stoichiometry between the reactants and products. Without the full reaction equation, I can't provide an exact answer, but I can explain the concept.
For example, if the reaction involves 2 moles of KI to produce 1 mole of I^(-), then you would multiply the number of moles of KI by the stoichiometric coefficient of I^(-) to find the initial concentration of I^(-) produced.
3. Calculate the initial concentration of I^(-) and S208^(2-) using the stoichiometry and initial moles of reactants found in step 1.
Remember that the initial concentrations will depend on the stoichiometry of the reaction, not the molality of the compounds.
If you have a specific balanced chemical reaction for this experiment, I can help you calculate the initial concentrations more accurately.