Calculations Involving Solutions

Specification Point 5.9C:

• Carry out an accurate acid-alkali titration, using burette, pipette and a suitable indicator.

Titration

Aim

• To perform an acid-alkali titration using suitable equipment and technique

Materials:

• Acid, alkali, suitable indicator
• Clamp and stand, burette and volumetric pipette, conical flask, white tile

Experimental set-up for an acid-alkali titration

Method: 

• 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

Calculating 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

Calculating Volume

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

M_r 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