Question

An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI present. Suppose 15.65 mL of the Na2S2O3 solution is required to titrate the iodine formed from 21.55 mL of 0.0131 M KIO3. What is the molarity of the Na2S2O3 solution

Answer 1

Answer: The molarity of $Na_2S_2O_3$ is 0.108 M

Explanation:

$KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2$

$2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI$

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

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

$\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol$

1 mole of $KIO_3$  produces = 3 moles of   $I_2$

$2.8\times 10^{-4}$ moles of $KIO_3$ produces = $\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}$ moles of $I_2$  

Now 1 mole of $I_2$ uses = 2 moles of $Na_2S_2O_3$

$8.4\times 10^{-4}$ moles of $I_2$ uses =  $\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}$  moles of $Na_2S_2O_3$

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M$

The molarity of $Na_2S_2O_3$ is 0.108 M