IB DP Chemistry: SL复习笔记1.2.9 Titrations

• 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 end 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
  • Multiple trials are carried out until concordant results are obtained

Recording and processing titration results

• Both the initial and final burette readings should be recorded and shown to a precision of  ±0.05 cm³, the same as the uncertainty

A typical layout and set of titration results

• The volume delivered (titre) is calculated and recorded to an uncertainty of ±0.10 cm³
  • The uncertainty is doubled, because two burette readings are made to obtain the titre (V final – V initial), following the rules for propagation of uncertainties (you can find more about this in Topic 11)
• Concordant results are then averaged, and non-concordant results are discarded
  • Concordance is usually considered to be a consistency of ±0.05 between results, depending on the quality of the burette
• The calculation then follows the steps given in 1.2.8 Concentration calculations

Answer:

Step 1: Write the equation for the reaction:

HA (aq) + NaOH (aq) → NaA (aq) + H₂O (l)

Step 2: Calculate the number of moles of the NaOH

Step 3: Deduce the number of moles of the acid

Since the acid is monoprotic the number of moles of HA is also 1.21 x 10^-3 mol

This is present in 25.0 cm³ of the solution

Step 4: Scale up to find the amount in the original solution

Step 5: Calculate the molar mass

Back titration

• A back titration is a common technique used to find the concentration or amount of an unknown substance indirectly
• The principle is to carry out a reaction with the unknown substance and an excess of a further reactant such as an acid or an alkali
• The excess reactant, after reaction, is then analysed by titration and the mole ratios are used to deduce the moles or concentration of the original substance being analysed

Answer:

Step 1: Write the equation for the titration reaction:

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

Step 2: Calculate the number of moles of the NaOH

n(NaOH) = 0.02380 dm³ x 0.100 mol dm^-3 = 2.380 x 10^-3 mol

Step 3: Deduce the number of moles of the excess acid

Since the reacting ratio is 1:1 the number of moles of HCl is also 2.380 x 10^-3 mol

Step 4: Find the amount of HCl in the original solution and then the amount reacted

n(HCl)_original =  0.02720 dm³ x 0.200 mol dm^-3 = 5.440 x 10^-3 mol

n(HCl)_reacted =  5.440 x 10^-3 mol – 2.380 x 10^-3 mol = 3.060 x 10^-3 mol

Step 5: Write the equation for the reaction with the calcium carbonate

2HCl (aq) + CaCO₃ (s) →  CaCl₂ (aq) + CO₂ (g) + H₂O (l)

Step 6: Deduce the number of moles of the calcium carbonate that reacted

Since the reacting ratio is 2:1 the number of moles of CaCO₃ is (3.060 x 10^-3 mol) ÷ 2

n(CaCO₃) = 1.530 x 10^-3 mol

Step 7: Calculate the mass of calcium carbonate in the sample of marble

mass = moles x molar mass = 1.530 x 10^-3 mol x 100.09 g mol^-1 = 0.1531g

Step 8: Calculate the percentage of calcium carbonate in the marble