# How is a back titration lab carried out?

Mar 16, 2016

You use an excess of titrant and then titrate the excess titrant.

#### Explanation:

In back titration you find the concentration of a species by reacting it with an excess of another reactant of known concentration. Then you titrate the excess reactant.

For example, you may want to determine the concentration of a base, but the endpoint is not sharp enough for a precise titration.

You could then add excess $\text{HCl}$ and titrate the excess with $\text{NaOH}$, because this titration will give you a sharp endpoint.

EXAMPLE:

You have 25.00 mL of a base $\text{B}$ of unknown concentration.

Titration with 0.1000mol/L $\text{HCl}$ gives an imprecise end point at about 20 mL, so you add 30.00 mL of $\text{HCl}$ and titrate the excess $\text{HCl}$ with 0.1100 mol/L $\text{NaOH}$.

What is the concentration of the unknown base if the back titration takes 9.64 mL of $\text{NaOH}$?

SOLUTION

Calculate the volume of excess HCl

$\text{HCl"color(white)(l) + "NaOH" → "NaCl" + "H"_2"O}$

$\text{moles of NaOH" = 9.64 color(red)(cancel(color(black)("mL"))) × "0.1100 mmol NaOH"/(1 color(red)(cancel(color(black)("mL")))) = "1.060 mmol NaOH}$

$\text{moles of HCl" = 1.060 color(red)(cancel(color(black)("mmol NaOH"))) ×"1 mmol HCl"/(1 color(red)(cancel(color(black)("mmol NaOH")))) = "1.060 mmol HCl}$

$\text{Volume of excess HCl" = 1.060 color(red)(cancel(color(black)("mmol HCl"))) × "1 mL HCl"/(0.1000 color(red)(cancel(color(black)("mmol HCl")))) = "10.60 mL HCl}$

Calculate the concentration of unknown base

$\text{B"color(white)(l) + "HCl" → "BH"^+"Cl"^"-}$

$\text{Volume of reacted HCl = 30.00 mL – 10.60 mL = 19.40 mL}$

$\text{Moles of HCl" = 19.40 color(red)(cancel(color(black)("mL"))) × "0.1000 mmol HCl"/(1 color(red)(cancel(color(black)("mL")))) = "1.940 mmol HCl}$

"Moles of B" = 1.940 color(red)(cancel(color(black)("mmol HCl"))) × "1 mmol B"/(1 color(red)(cancel(color(black)("mmol HCl")))) = "1.940 mmol B"

$\text{Concentration of B" = "1.940 mmol"/"25.00 mL" = "0.077 60 mol/L}$