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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.

Compose the equation that is balanced.

#"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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Answer 2

Caleb Adams

The molarity of I₃⁻(aq) is 0.868 M.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.