Back Titrations

A back titration is a titration method where the concentration of an analyte is determined by reacting it with a known amount of excess reagent. The remaining excess reagent is then titrated with another, second reagent. The second titration’s result shows how much of the excess reagent was used in the first titration, thus allowing the original analyte’s concentration to be calculated.

Worked Examples

Example 1

Excess KI solution was added to 10.0 cm³ of saturated bromine solution.

$Br_2 + 2I^- \rightarrow 2Br^- + I_2\\$

The iodine produced required 37.30 cm³ of 0.120M sodium thiosulphate for complete reaction in a titration, using starch indicator. Find the solubility of bromine in water in gm dm^-3.

Step 1

Write the two redox reactions as a table:

${\color{White} ----}Br_2 + 2I^- \rightarrow 2Br^- + I_2\\mass\\molarity\\volume\\moles$

${\color{White} ----}I_2 + 2S_2O_3^2^- \rightarrow 2I^- + 2S_4O_6^2^-\\mass\\molarity\\volume\\moles$

Step 2

Write all the information given and design a route to take you the required answer.

${\color{White} ----}Br_2 +{\color{White} ---}2I^- \rightarrow {\color{White} --}2Br^- + {\color{White} ---}I_2\\mass{\color{White} ---}\underline{{\color{White} -}e{\color{White} -}}\\molarity\\volume{\color{White} --}10cm^-^3\\moles {\color{White} ---}\underline{{\color{White} -}d{\color{White} -}}{\color{White} --------------}\underline{{\color{White} -}c{\color{White} -}}$

${\color{White} ----}I_2 +{\color{White} ----} 2S_2O_3^2^- \rightarrow {\color{White} --}2I^- + {\color{White} ----}2S_4O_6^2^-\\mass\\molarity{\color{White} -------}0.120M \\volume{\color{White} -------}37.3cm^-^3M\\moles{\color{White} ---}\underline{{\color{White} -}b{\color{White} -}}{\color{White} ----}\underline{{\color{White} -}a{\color{White} -}}$

The first step in this question is to find the number of moles of thiosulfate used to neutralise the Iodine formed. From this, the number of moles of Iodine can be found which can then be used to find the moles of Bromine dissolved in water.

In such a question it is important to plan ahead and not just start working the question as it comes. This can both save you a lot of time + help you avoid mistakes.

Step 3 (Finding moles of S₂O₃^2-)

$0.12moles =1000cm^-^3 \\ {\color{white} -\: -\: -- \: -}?=37.3cm^-^3 \\ \frac{37.3*0.12}{1000}=0.004476\, moles$

Step 4 (Finding moles of I₂)

$S_2O_3^2^- : I_2 \\ {\color{white} ----}2:1 \\ 0.004476 : 0.002238 \\ 0.002238\, moles\, of\, I_2$

Step 5 (Finding moles of I₂)

The number of moles of Iodine reacted with thiosulfate are equal to the number of moles of Iodine produced in the first reaction with Bromine.

$\therefore 0.002238\: moles$

Step 6 (Finding moles of Br₂)

${\color{white} -}Br_2 : I_2 \\ {\color{white} ----}2:1 \\ 0.002238 : 0.002238 \\ 0.002238\, moles\, of\, Br_2$

Step 7 (Finding the mass of Br₂)

$1\, mole =160\, g \\ {\color{white} -}0.002238=? \\ \frac{0.002238*160}{1}=0.3581\, g$

0.3581 g was dissolved in 10cm³ which would mean that 35.81g of Bromine dissolve in 1 Litre.

This would mean that the solubility of Br₂ is 35.81g dm^-3.

Example 2

25.0 cm³ of H₂O₂ was diluted to 250 cm³. 25.0 cm³ of this solution required 31.25 cm³ of 0.02M potassium permanganate solution for titration, during which oxygen gas was evolved. Find the molarity of the original hydrogen peroxide solution.

Step 1

${\color{White} ----}5H_2O_2 + 2MnO_4^-+6H^+ \rightarrow 5O_2 + 8H_2O+2Mn^2^+\\mass\\molarity\\volume\\moles$

Step 2

${\color{White} ----}5H_2O_2 +{\color{White} ---}2MnO_4^- \rightarrow {\color{White} --}5O_2 + {\color{White} ---}2Mn^2^+ \\mass\\molarity{\color{White} --}\underline{{\color{White} -}c{\color{White} -}}{\color{White} -----}{\color{White} -}0.02M{\color{White} -}\\volume{\color{White} --}25cm^-^3{\color{White} ----}{\color{White} -}31.25cm^-^3{\color{White} -}\\moles {\color{White} ---}\underline{{\color{White} -}b{\color{White} -}}{\color{White} ------}\underline{{\color{White} -}a{\color{White} -}}$