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

Step-by-step 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=(0.0131mol/L* 21.55)/(1000)=2.8* 10^(-4)mol`

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

`2.8* 10^(-4)` moles of
`KIO_3` produces =
`(3)/(1)* 2.8* 10^(-4)=8.4* 10^(-4)` moles of
`I_2`

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

`8.4* 10^(-4)` moles of
`I_2` uses =
`(2)/(1)* 8.4* 10^(-4)=1.69* 10^(-3)` moles of
`Na_2S_2O_3`

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=(1.69* 10^(-3)* 1000)/(15.65)=0.108M`

The molarity of
`Na_2S_2O_3` is 0.108 M