Question #ffbc6

1 Answer
Jan 1, 2016

Answer:

See below.

Explanation:

I presume that you are titrating a solution of #"Na"_2"S"_2"O"_3# against a primary standard like #"KIO"_3#.

You react a known mass of #"KIO"_3# with excess #"KI"#, and then titrate the liberated #"I"_2# with your #"Na"_2"S"_2"O"_3# solution.

The equation for the iodate half-reaction is

#"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

#"Equivalent weight" = "molar mass"/"number of electrons per mole"#

So, for #"KIO"_3#,

#"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 #"KIO"_3# with excess #"KI"# and that it took 25.40 mL of #"Na"_2"S"_2"O"_3# solution to titrate the liberated iodine.

#"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#

#"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.