Question

1.Calculate the rate of reaction for the solution to turn from colorless to blue-black after each addition.

Volume Added to Solution 1	Reaction time (s)	Total Volume (L)	Rate (mol/L·s)
25 mL (NH4)2S2O8	69	.100	.032
1 mL Na2S2O3	105	.101	9.96X10^-4
1 mL Na2S2O3	80	.102	.0013
1 mL Na2S2O3	75	.103	.0014

2.What is the average rate of the reaction for Solution 1 (include units)?                   .009 mol/L*s              

3.What is the concentration of KI and (NH₄)₂S₂O₈ in the reaction flask?

                                                                                                          M KI                                      M (NH₄)₂S₂O<sub>8

How do I figure out #3?</sub>

_{There is}
25.0 mL of .2 M KI
25.0 mL of .2 M (NH4)2S2O8

Answer 1

3. Given the volume of KI = 25.0 mL = 0.025 L

Concentration of KI = 0.2M

Hence moles of KI in the solution = MxV = 0.2 M x 0.025L = 0.005 mol

Given the total volume of the solution, Vt = 0.100 L

Concentration of KI = moles of KI / Vt = 0.005mol / 0.100 = 0.05 M(answer)

Given the volume of (NH4)2S28 = 25.0 mL = 0.025 L

Concentration of (NH4)2S28 = 0.2M

Hence moles of (NH4)2S28 in the solution = MxV = 0.2 M x 0.025L = 0.005 mol

Given the total volume of the solution, Vt = 0.100 L

Concentration of (NH4)2S28 = moles of (NH4)2S28 / Vt = 0.005mol / 0.100 = 0.05 M(answer)