**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.

**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 cm^{3}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 0.1cm^{3}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.

**Titration 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 cm^{3} of 0.050 mol / dm^{3} sodium carbonate was completely neutralised by 20.00 cm^{3} of dilute hydrochloric acid. Calculate the concentration, in mol / dm^{3} 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 cm^{3} to dm^{3}

Amount of Na_{2}CO_{3} = (25.0 x 0.050) ÷ 1000 = 0.00125 mol

**Step 2 **Calculate the amount, in moles, of hydrochloric acid reacted

#### Na_{2}CO_{3} + 2HC*l* → 2NaC*l* + H_{2}O + CO_{2}

1 mol of Na_{2}CO_{3} reacts with 2 mol of HC*l*, so the Molar Ratio is 1 : 2

Therefore 0.00125 moles of Na_{2}CO_{3} react with 0.00250 moles of HC*l*

**Step 3 **Calculate the concentration, in mol / dm^{3} of the Hydrochloric Acid

1 dm^{3} = 1000 cm^{3}

Volume of HC*l* = 20 ÷ 1000 = 0.0200 dm^{3}

Concentration HC*l* (mol / dm^{3}) = 0.00250 ÷ 0.0200 = 0.125

** Concentration of Hydrochloric Acid =** 0.125 mol / dm^{3}

**Calculating Volume**

**Equation:** volume = moles ÷ concentration

Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm^{3} 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 CaCO_{3} is 100

Amount of CaCO_{3} = (2.5 ÷ 100) = 0.025 mol

**Step 2 **Calculate the moles of hydrochloric acid required

#### CaCO_{3} + 2HC*l* → CaC*l*_{2} + H_{2}O + CO_{2}

1 mol of CaCO_{3} requires 2 mol of HC*l*

So 0.025 mol of CaCO_{3} Requires 0.05 mol of HC*l*

**Step 3 **Calculate the volume of HC*l* Required

Volume = (Amount of Substance(mol) ÷ Concentration (mol / dm^{3})

= 0.05 ÷ 1.0

= 0.05 dm^{3 }(the moles cancel out above and below the line)

** Volume of Hydrochloric Acid =** 0.05 dm^{3}

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### Author: Morgan

Morgan’s passion for the Periodic Table begun on his 10th birthday when he received his first Chemistry set. After studying the subject at university he went on to become a fully fledged Chemistry teacher, and now works in an international school in Madrid! In his spare time he helps create our fantastic resources to help you ace your exams.

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