# A student mixed the following reactants, using the general procedure for the experiment:

10.0 mL of 0.10 M KI, 10.0 mL of 0.0010 M Na2S2O3, 20.0 mL of 0.040 M K2S2O8, and 10.0 mL of 0.10M HCl. It took about 75 seconds for the mixture to turn blue.

a. By using the dilution equation (M1V1=M2V2), calculate the concentrations of each
reactant in the mixture after all of the solutions have been combined:
[I-] = ________ [S2O32-] = ________ [H+] = ________

## To calculate the concentrations of each reactant in the mixture, we can use the dilution equation (M1V1 = M2V2). This equation states that the initial molarity multiplied by the initial volume is equal to the final molarity multiplied by the final volume.

Let's calculate the concentration of [I-]:

Initial volume and concentration of KI: V1 = 10.0 mL, M1 = 0.10 M
Final volume of the mixture: V2 = 50.0 mL (10.0 mL KI + 10.0 mL Na2S2O3 + 20.0 mL K2S2O8 + 10.0 mL HCl)

Using the dilution equation: M1V1 = M2V2
0.10 M * 10.0 mL = M2 * 50.0 mL

Solving for M2, the concentration of [I-]:
M2 = (0.10 M * 10.0 mL) / 50.0 mL

Calculating M2:
M2 = 0.02 M

Therefore, the concentration of [I-] in the mixture is 0.02 M.

Now let's calculate the concentration of [S2O32-]:

Initial volume and concentration of Na2S2O3: V1 = 10.0 mL, M1 = 0.0010 M
Final volume of the mixture: V2 = 50.0 mL

Using the dilution equation: M1V1 = M2V2
0.0010 M * 10.0 mL = M2 * 50.0 mL

Solving for M2, the concentration of [S2O32-]:
M2 = (0.0010 M * 10.0 mL) / 50.0 mL

Calculating M2:
M2 = 0.00020 M

Therefore, the concentration of [S2O32-] in the mixture is 0.00020 M.

Finally, let's calculate the concentration of [H+]:

Initial volume and concentration of HCl: V1 = 10.0 mL, M1 = 0.10 M
Final volume of the mixture: V2 = 50.0 mL

Using the dilution equation: M1V1 = M2V2
0.10 M * 10.0 mL = M2 * 50.0 mL

Solving for M2, the concentration of [H+]:
M2 = (0.10 M * 10.0 mL) / 50.0 mL

Calculating M2:
M2 = 0.02 M

Therefore, the concentration of [H+] in the mixture is 0.02 M.

Summarizing the results:

[I-] = 0.02 M
[S2O32-] = 0.00020 M
[H+] = 0.02 M