in an experiment 5.00ml of 0.1440 mol L^1- aqueos KIO3 was reacted with an excess of aqueous iodide and the molecular iodine I2, Formed was titrated against 0.1000mol L^1- aqueous S2O3^2- the equation for the titration reaction is:
I2(aq)+ 2 S2O3^2-(aq)---->S4O6^2-(aq)+ 2I^-1(aq)
calculate the numbers of moles of sodium thiosulfate (Na2S2O3) and hence the moles of S2O3^2- ions present in the volume of 43mL
that is all the information i have. Can someone explain to me how to do it?
You've picked the problem apart. Your first post (five posts ago, should have had all of the problem and that would have made things simpler for you). As it stands, you're still not at the end. But here goes, for these two parts anyway.
moles KIO3 = M KIO3 x L KIO3 = ?
The last part of the problem you asked is for moles S2O3^2-. That is moles S2O3^2-= M S2O3^2- x L S2O3^2- = y moles S2O3^2-
To solve this problem, we need to follow a step-by-step approach. Here's how you can calculate the number of moles of sodium thiosulfate (Na2S2O3) and the moles of S2O3^2- ions present in the 43 mL volume:
Step 1: Calculate the moles of I2 reacted.
Start with the given volume of aqueous KIO3, which is 5.00 mL. Convert this to liters by dividing by 1000 (1 L = 1000 mL):
5.00 mL ÷ 1000 = 0.005 L
Now, use the given concentration of KIO3, which is 0.1440 mol L^-1, to calculate the moles:
Moles of KIO3 = volume (in L) × concentration
Moles of KIO3 = 0.005 L × 0.1440 mol L^-1 = 0.00072 mol
From the balanced equation, we know that 1 mole of I2 reacts with 2 moles of S2O3^2-. Therefore, the moles of I2 will be equal to twice the moles of S2O3^2-.
Step 2: Calculate the moles of Na2S2O3 and S2O3^2- ions.
Using the mole ratio from the balanced equation, we can determine the moles of Na2S2O3:
Moles of Na2S2O3 = Moles of I2 × (1 mol Na2S2O3/1 mol I2)
Moles of Na2S2O3 = 0.00072 mol × (1 mol Na2S2O3/1 mol I2) = 0.00072 mol
Since one mole of Na2S2O3 contains two moles of S2O3^2-, we can calculate the moles of S2O3^2- ions present:
Moles of S2O3^2- = 2 × moles of Na2S2O3 = 2 × 0.00072 mol = 0.00144 mol
Therefore, the number of moles of sodium thiosulfate (Na2S2O3) present in the 43 mL volume is 0.00072 mol, and the moles of S2O3^2- ions present is 0.00144 mol.