Titrations (AQA A Level Chemistry)

• Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (a standard solution) to determine the concentration of another unknown solution
• The technique most commonly used is a titration
• The volumes are measured using two precise pieces of equipment, a volumetric or graduated pipette and a burette

Equipment used to measure volumes precisely in titrations

• Burettes are usually marked to a precision of 0.10 cm³
  • Since they are analogue instruments, the uncertainty is recorded to half the smallest marking, in other words to ±0.05 cm³

• The stoichiometric point or equivalence point occurs when the two solutions have reacted completely and is shown with the use of an indicator

The steps in a titration

• The steps in a titration are:
  • Measuring a known volume (usually 20 or 25 cm³) of one of the solutions with a volumetric or graduated pipette and placing it into a conical flask
  • The other solution is placed in the burette
  • A few drops of the indicator are added
  • The tap on the burette is carefully opened and the solution added, portion by portion, to the conical flask until the indicator just changes colour(this is the end point)
  • Multiple trials are carried out until concordant results are obtained

• Titration calculations are used to find the concentration of unknown solutions
• They can also be used to calculate the pH after a given point during a titration

Answer

Step 1: Find the number of moles of acid

moles of acid = concentration x volume in dm³

moles of acid = 0.05 x 25/1000 = 1.25 x 10^-3 mol

Step 2: Deduce the number of moles of alkali

The equation for the reaction shows the mole ratio is 1:1

HCl (aq) + NaOH (aq)  → NaCl (aq) + H₂O (l)

∴ moles of alkali  = 1.25 x 10^-3 mol

Step 3: Work out the concentration of the alkali

concentration = moles/volume in dm³

concentration = 1.25 x 10^-3/0.0085 = 0.15 mol dm^-3

Answer

Step 1: Find the number of moles of acid

moles of acid = concentration x volume in dm³

moles of acid = 0.15 x 25/1000 = 3.75 x 10^-3 mol

Step 2: Deduce the number of moles of alkali added

The equation for the reaction shows the mole ratio is 1:1

HCl (aq) + NaOH (aq)  → NaCl (aq) + H₂O (l)

moles of alkali added = 0.10 x 45/1000 =  4.50 x 10^-3 mol

∴ moles of alkali in excess = (4.50 x 10^-3- 3.75 x 10^-3) = 7.5 x 10^-4 mol

Step 3: Work out the concentration of the alkali

concentration = moles/volume in dm³

concentration = 7.5 x 10^-4/0.070 = 0.0107 mol dm^-3

Step 4: Use K_w to find the concentration of H⁺

K_w = [H⁺][OH^-]

[H⁺] = K_w /[OH^-] = 1.00 x 10^-14/0.0107 = 9.35 x 10^-13

Step 5: Find the pH

-log[H⁺] = -log(9.35 x 10^-13)

pH = 12.03