Question

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?

Answer 1

The molarity of `"I"_3^"-"` is 0.434 mol/L.

Explanation:

1. Write the balanced equation

`"2S"_2"O"_3^"2-" + "I"_3^"-" → "S"_4"O"_6^"2-" + "3I"^"-"`

2. Calculate the moles of `"S"_2"O"_3^"2-"`

`"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 `"I"_3^"-"`

`"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 `"I"_3^"-"`

`"Molarity" = "moles"/"litres" = "0.004 34 mol"/"0.0100 L" = "0.434 mol/L"`