Calculate the concentrations of S2O8 2-, I-, S2O32 2- in moles per litre of reaction mixture given the following:
20mL of 0.1 M (NH4)2S2O8
added to:
20mL of 0.2M KI
10mL of 0.01M Na2S2O3
5mL 0.2% starch
Also, calculate -(delta)S2O8 2-
that is, how much S2O8 2- is used up (consumed) from the time the above solutions are mixed until a blue colour appears, in moles per litre of reaction mixture.
To calculate the concentrations of S2O8 2-, I-, and S2O3 2- in moles per liter of the reaction mixture, we first need to find the moles of each substance.
For (NH4)2S2O8:
Volume = 20 mL = 0.020 L
Concentration = 0.1 M
So, Moles of (NH4)2S2O8 = Concentration x Volume
= 0.1 M x 0.020 L
= 0.002 moles
For KI:
Volume = 20 mL = 0.020 L
Concentration = 0.2 M
So, Moles of KI = Concentration x Volume
= 0.2 M x 0.020 L
= 0.004 moles
For Na2S2O3:
Volume = 10 mL = 0.010 L
Concentration = 0.01 M
So, Moles of Na2S2O3 = Concentration x Volume
= 0.01 M x 0.010 L
= 0.0001 moles
The concentrations of each substance can be calculated by dividing the number of moles by the total volume of the reaction mixture, which is the sum of the volumes of all the solutions.
Total Volume = 20 mL + 20 mL + 10 mL + 5 mL
= 0.020 L + 0.020 L + 0.010 L + 0.005 L
= 0.055 L
Concentration of (NH4)2S2O8 = Moles of (NH4)2S2O8 / Total Volume
= 0.002 moles / 0.055 L
= 0.036 M
Concentration of KI = Moles of KI / Total Volume
= 0.004 moles / 0.055 L
= 0.073 M
Concentration of Na2S2O3 = Moles of Na2S2O3 / Total Volume
= 0.0001 moles / 0.055 L
= 0.0018 M
Since starch is not involved in the reaction, we don't need to calculate its concentration.
To calculate the -(delta)S2O8 2-, we need to determine the stoichiometry of the reaction.
(NH4)2S2O8 + 2KI + Na2S2O3 → K2S2O8 + (NH4)2SO4 + NaI
From the balanced equation, we see that 1 mole of (NH4)2S2O8 consumes 1 mole of KI. Therefore, the moles of S2O8 2- consumed will be the same as the moles of KI used.
-(delta)S2O8 2- = Moles of KI = 0.004 moles
So, the concentration of -(delta)S2O8 2- is the same as the concentration of KI:
Concentration of -(delta)S2O8 2- = 0.073 M
To calculate the concentrations of the different species, we need to determine the moles of each species in the reaction mixture and then divide by the total volume of the mixture.
Let's start by calculating the moles of (NH4)2S2O8 added:
Moles of (NH4)2S2O8 = 20 mL * 0.1 M = 2 mmol
Next, let's calculate the moles of KI added:
Moles of KI = 20 mL * 0.2 M = 4 mmol
Now, let's calculate the moles of Na2S2O3 added:
Moles of Na2S2O3 = 10 mL * 0.01 M = 0.1 mmol
Finally, let's calculate the moles of S2O8^2-, I-, and S2O3^2- in the reaction mixture:
Moles of S2O8^2- = Moles of (NH4)2S2O8 = 2 mmol
Moles of I- = Moles of KI = 4 mmol
Moles of S2O3^2- = Moles of Na2S2O3 = 0.1 mmol
Now, we can calculate the concentrations of each species:
Concentration of S2O8^2- = (2 mmol / (20 mL + 20 mL + 10 mL + 5 mL)) * 1000 mL = 20/55 M = 0.36 M
Concentration of I- = (4 mmol / (20 mL + 20 mL + 10 mL + 5 mL)) * 1000 mL = 80/55 M = 1.45 M
Concentration of S2O3^2- = (0.1 mmol / (20 mL + 20 mL + 10 mL + 5 mL)) * 1000 mL = 2/55 M = 0.036 M
Now, let's calculate the change in the concentration of S2O8^2- until the blue color appears. Since we don't have the rate of the reaction, we'll assume that from the time the above solutions are mixed until the blue color appears, all of the S2O8^2- is consumed.
So, the change in the concentration of S2O8^2- = Concentration of S2O8^2- = 0.36 M
Therefore, the -(delta)S2O8^2- is 0.36 M.