How is a back titration lab carried out?

1 Answer
Mar 16, 2016

Answer:

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 #"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"#