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 #"HCl"# and titrate the excess with #"NaOH"#, because this titration will give you a sharp endpoint.
EXAMPLE:
You have 25.00 mL of a base #"B"# of unknown concentration.
Titration with 0.1000mol/L #"HCl"# gives an imprecise end point at about 20 mL, so you add 30.00 mL of #"HCl"# and titrate the excess #"HCl"# with 0.1100 mol/L #"NaOH"#.
What is the concentration of the unknown base if the back titration takes 9.64 mL of #"NaOH"#?
SOLUTION
Calculate the volume of excess HCl
#"HCl"color(white)(l) + "NaOH" → "NaCl" + "H"_2"O"#
#"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"#
#"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"#
#"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
#"B"color(white)(l) + "HCl" → "BH"^+"Cl"^"-"#
#"Volume of reacted HCl = 30.00 mL – 10.60 mL = 19.40 mL"#
#"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"#
#"Concentration of B" = "1.940 mmol"/"25.00 mL" = "0.077 60 mol/L"#
