# Question #ffbc6

Jan 1, 2016

See below.

#### Explanation:

I presume that you are titrating a solution of ${\text{Na"_2"S"_2"O}}_{3}$ against a primary standard like ${\text{KIO}}_{3}$.

You react a known mass of ${\text{KIO}}_{3}$ with excess $\text{KI}$, and then titrate the liberated ${\text{I}}_{2}$ with your ${\text{Na"_2"S"_2"O}}_{3}$ solution.

The equation for the iodate half-reaction is

$\text{2IO"_3^(-) + "12H"^+ + "10e"^(-) → "I"_2 + "6H"_2"O}$

We see that five moles of electrons are transferred per mole of iodate.

.

The formula for the "equivalent weight" of a compound in a redox reaction is

$\text{Equivalent weight" = "molar mass"/"number of electrons per mole}$

So, for ${\text{KIO}}_{3}$,

$\text{Equivalent weight" = "214.0 g"/5 = "42.80 g}$

The usefulness of equivalents is that one equivalent of anything reacts with one equivalent of anything else.

Assume that you have reacted 0.1100 g of ${\text{KIO}}_{3}$ with excess $\text{KI}$ and that it took 25.40 mL of ${\text{Na"_2"S"_2"O}}_{3}$ solution to titrate the liberated iodine.

${\text{Milliequivalents of KIO"_3 = 110.0 color(red)(cancel(color(black)("mg KIO"_3))) × ("1 meq KIO"_3)/(42.80 color(red)(cancel(color(black)("mg KIO"_3)))) = "2.570 meq KIO"_3 = "2.570 meq Na"_2"S"_2"O}}_{3}$

$\text{Normality" = "equivalents"/"litres" = "milliequivalents"/"millilitres" = "2.570 meq"/"25.40 mL" = "0.1012 N}$

Repeat the titration and the calculations several times and average the results.