Question

Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 of unknown concentration. The endpoint was observed to occur at 16.50 mL. 1. How many moles of KIO3 were titrated? ___________________ Show work! 2. How many moles of Na2S2O3 did this require? _____

Answer 1

Answer:

For 1: The amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2: The amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

Explanation:

• For 1:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of $KIO_3$ solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

$0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol$

Hence, the amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

• For 2:

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = $\frac{1}{2}\times 0.00025=0.000125mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles