Core Practical: Performing an Accurate Acid-Alkili Titration
Specification Point 5.9C:
Carry out an accurate acid-alkali titration, using burette, pipette and a suitable indicator.
- To perform an acid-alkali titration using suitable equipment and technique
- Acid, alkali, suitable indicator
- Clamp and stand, burette and volumetric pipette, conical flask, white tile
Experimental set-up for an acid-alkali titration
- Place the conical flask on a white tile and few drops of a suitable indicator to the solution in the conical flask.
- Perform a rough titration by taking a burette reading and running in the solution in 1 – 3 cm3 portions, while swirling the flask vigorously.
- Quickly close the tap when the end-point is reached (sharp colour change) and record the volume (being sure that you place your eye level with the meniscus).
- Now repeat the titration with a fresh batch of alkaline.
- As the rough end-point volume is approached, add solution from the burette one drop at a time until the indicator changes colour.
- Repeat until you achieve two concordant results (two results that are within 0.1cm3 of each other).
Analysis of results:
- Readings should be recorded to two decimal places, ending in 0 or 5 (where the liquid level is between two graduations on the burette) and in a suitable table.
- The titre is the volume added (the difference between the end and start readings).
- Calculate the mean titre and use this for subsequent calculations.
Specification Point 5.10C:
Carry out simple calculations using the results of titrations to calculate an unknown concentration of a solution or an unknown volume of solution required.
- Once a titration is completed and the average titre has been calculated, you can now proceed to calculate either the unknown variable using the formula triangle:
- The formula triangle can be used to derive equations for concentration and volume:
Formula triangle showing the relationship between concentration, number of moles and volume of liquid
- From the triangle:
- Concentration = moles ÷ volume
- Volume = moles ÷ concentration
Equation: concentration = moles ÷ volume
25.0 cm3 of 0.050 mol / dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration, in mol / dm3 of the hydrochloric acid.
Step 1 Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing by 1000 to convert cm3 to dm3
Amount of Na2CO3 = (25.0 x 0.050) ÷ 1000 = 0.00125 mol
Step 2 Calculate the amount, in moles, of hydrochloric acid reacted
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
1 mol of Na2CO3 reacts with 2 mol of HCl, so the Molar Ratio is 1 : 2
Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl
Step 3 Calculate the concentration, in mol / dm3 of the Hydrochloric Acid
1 dm3 = 1000 cm3
Volume of HCl = 20 ÷ 1000 = 0.0200 dm3
Concentration HCl (mol / dm3) = 0.00250 ÷ 0.0200 = 0.125
Concentration of Hydrochloric Acid = 0.125 mol / dm3
Equation: volume = moles ÷ concentration
Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm3 that is required to react completely with 2.5g of calcium carbonate.
Step 1 Calculate the amount, in moles, of calcium carbonate that reacts
Mr of CaCO3 is 100
Amount of CaCO3 = (2.5 ÷ 100) = 0.025 mol
Step 2 Calculate the moles of hydrochloric acid required
CaCO3 + 2HCl → CaCl2 + H2O + CO2
1 mol of CaCO3 requires 2 mol of HCl
So 0.025 mol of CaCO3 Requires 0.05 mol of HCl
Step 3 Calculate the volume of HCl Required
Volume = (Amount of Substance(mol) ÷ Concentration (mol / dm3)
= 0.05 ÷ 1.0
= 0.05 dm3 (the moles cancel out above and below the line)
Volume of Hydrochloric Acid = 0.05 dm3
Edexcel GCSE Chemistry Notes
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