2.4.6Inorganic Chemistry: Part 1
How to Prepare a Soluble Salt: Method 2
5 min read 6 parts Lesson 18 of 20 Unit quiz
Section 1 of 6When to Use Method 2
Soluble alkali
- This method is used when the base is soluble in water, such as sodium hydroxide or potassium hydroxide.
No excess to filter
- A soluble alkali cannot be added in excess and filtered off, because the leftover alkali would stay in the solution.
Exact neutralisation
- The acid and alkali must therefore be added in exactly the right amounts to neutralise each other.
Indicator
- A titration with an indicator finds the exact volume of acid that reacts with a fixed volume of alkali.
Example
- Sodium hydroxide and hydrochloric acid give sodium chloride and water: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).
Section 2 of 6Method
1
Use a pipette to transfer a fixed volume of alkali, such as 25.0 cm³, into a conical flask and add a few drops of indicator.
2
Fill a burette with the acid and record the starting volume.
3
Add the acid slowly, swirling the flask, until the indicator just changes colour, then record the final volume.
4
Repeat with the same volume of alkali and the same volume of acid, but with no indicator.
5
Warm the solution to evaporate some of the water, leave it to cool and crystallise, then pour off the liquid and dry the crystals.
Section 3 of 6Choosing Indicators and Avoiding Errors
Choosing Indicators and Avoiding Errors
Indicator choice
- Phenolphthalein turns from pink to colourless and methyl orange from yellow to orange or red in a strong acid-alkali titration.
Repeat without indicator
- The indicator would contaminate the salt, so it is not used in the final mixture.
Partial evaporation
- Some water is left so that crystals form on cooling and the salt does not overheat or decompose.
Drying crystals
- The crystals are dried by gently patting with filter paper or leaving them in a warm place.
Section 4 of 6Calculations
Moles of alkali
- Moles = concentration (mol/dm³) × volume (dm³), using the pipette volume converted to dm³.
Mole ratio
- The balanced equation shows that one mole of NaOH reacts with one mole of HCl.
Theoretical yield
- The mass of salt expected is moles of salt × its relative formula mass.
Percentage yield
- Compare the mass of dry crystals obtained with the theoretical mass to find the percentage yield.
| Salt | Acid | Alkali |
|---|---|---|
| Sodium chloride | Hydrochloric acid | Sodium hydroxide |
| Potassium nitrate | Nitric acid | Potassium hydroxide |
| Sodium sulfate | Sulfuric acid | Sodium hydroxide |
Section 5 of 6Worked Example
Question
- 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide is exactly neutralised by 20.0 cm³ of hydrochloric acid.
- Calculate the concentration of the acid and the maximum mass of sodium chloride that can be obtained.
- (Ar: Na = 23, Cl = 35.5)
- Moles of NaOH = 0.200 × 0.0250 = 0.00500 mol.
- NaOH and HCl react 1 : 1, so moles of HCl = 0.00500 mol.
- Concentration of HCl = 0.00500 ÷ 0.0200 = 0.250 mol/dm³.
- Moles of NaCl = 0.00500 mol, and Mr of NaCl = 58.5, so mass = 0.00500 × 58.5 = 0.29 g (to 2 significant figures).
Answer The acid has a concentration of 0.250 mol/dm³, and at most 0.29 g of sodium chloride can be obtained.
Section 6 of 6Exam Checklist & Tips
2.4.6 – Key Points for the Exam
- Soluble salt A salt that dissolves in water, so it can be prepared in solution and crystallised.
- Alkali A soluble base that forms hydroxide ions in water.
- Titration A method of finding the exact volumes of two solutions that react together.
- Burette A graduated glass tube used to add a measured, variable volume of liquid.
- Crystallisation The formation of solid crystals as a hot, concentrated solution cools.
Common Mistakes
- Adding excess alkali and filtering, which does not work because the alkali is soluble.
- Keeping the indicator in the mixture that is crystallised, so the salt is not pure.
- Evaporating all the water instead of stopping at the saturation point.
- Using a different volume of acid in the second step from the volume found in the titration.