Question 4.13: USING A REDOX REACTION TO DETERMINE A SOLUTION’S CONCENTRATI...
USING A REDOX REACTION TO DETERMINE A SOLUTION’S CONCENTRATION
The concentration of an aqueous I_{3}^{-} solution can be determined by titration with aqueous sodium thiosulfate, Na_{2}S_{2}O_{3}, in the presence of a starch indicator, which turns from deep blue to colorless when all the I_{3}^{-} has reacted. What is the molar concentration of I_{3}^{-} in an aqueous solution if 24.55 mL of 0.102 M Na_{2}S_{2}O_{3} is needed for complete reaction with 10.00 mL of the I_{3}^{-} solution? The net ionic equation is
2 S_{2}O_{3}^{2-}(aq) + I_{3}^{-}(aq) → S_{4}O_{6}^{2-}(aq) + 3 I^{-}(aq)The procedure is similar to that outlined in Figure 4.5. We first need to find the number of moles of thiosulfate ion used for the titration:
24.55 \cancel{mL} × \frac{1 \cancel{L}}{1000 \cancel{mL}} × \frac{0.102 mol S_{2}O_{3}^{2-}}{1 \cancel{L}} = 2.50 × 10^{-3} mol S_{2}O_{3}^{2-}According to the balanced equation, 2 mol of S_{2}O_{3}^{2-} ion react with 1 mol of I_{3}^{-} ion.
Thus, we can find the number of moles of I_{3}^{-} ion:
Knowing both the number of moles of mol) and the volume of I_{3}^{-} (2.25 × 10^{-3} mol) the I_{3}^{-} solution (10.00 mL) then lets us calculate molarity:
\frac{1.25 × 10^{-3} mol I_{3}^{-}}{10.00 \cancel{mL}} × \frac{10^{3} mL}{1 L} = 0.125 M
The molarity of the I_{3}^{-} solution is 0.125 M.
BALLPARK CHECK
According to the balanced equation, the amount of S_{2}O_{3}^{2-} needed for the reaction (2 mol) is twice the amount of I_{3}^{-} (1 mol). The titration results indicate that the volume of the S_{2}O_{3}^{2-} solution (24.55 mL) is a little over twice the volume of the I_{3}^{-} solution (10.00 mL). Thus, the concentrations of the two solutions must be about the sameapproximately 0.1 M.
The reddish I_{3}^{-} solution turns a deep blue color when it is added to a solution containing a small amount of starch.
