# Given that it requires 31.0 mL of 0.280 M Na_2S_2O_3(aq) to titrate a 10.0-mL sample of I_3^- (aq), how would you calculate the molarity of I_3^- (aq) in the solution?

Jul 7, 2016

The molarity of $\text{I"_3^"-}$ is 0.434 mol/L.

#### Explanation:

1. Write the balanced equation

$\text{2S"_2"O"_3^"2-" + "I"_3^"-" → "S"_4"O"_6^"2-" + "3I"^"-}$

2. Calculate the moles of $\text{S"_2"O"_3^"2-}$

$\text{Moles of S"_2"O"_3^"2-" = 0.0310 color(red)(cancel(color(black)("L S"_2"O"_3^"2-"))) × ("0.280 mol S"_2"O"_3^"2-")/(1 color(red)(cancel(color(black)("L S"_2"O"_3^"2-")))) = "0.008 68 mol S"_2"O"_3^"2-}$

3. Calculate the moles of $\text{I"_3^"-}$

$\text{Moles of I"_3^"-" = "0.008 68" color(red)(cancel(color(black)("mol S"_2"O"_3^"2-"))) × ("1 mol I"_3^"-")/(2 color(red)(cancel(color(black)("mol S"_2"O"_3^"2-")))) = "0.004 34 mol I"_3^"-}$

4. Calculate the molarity of $\text{I"_3^"-}$

$\text{Molarity" = "moles"/"litres" = "0.004 34 mol"/"0.0100 L" = "0.434 mol/L}$