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?

1 Answer
Ernest Z.
Jul 7, 2016

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"

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