6.9 - Preparation of Salts

Solubility of Salts in Water — Experiment 6.5

• Aim: investigate whether different salts dissolve in water.
• Hypothesis: some salts dissolve in water and some do not.
• Variables: salt type (manipulated), solubility (responding), water volume and temperature and salt mass (fixed).
• Compare nitrate, sulphate, chloride, carbonate and ammonium salts using distilled water, a beaker, glass rod and electronic balance, as shown in the apparatus arrangement.
• Conclusion: salts are classified as soluble or insoluble at room temperature, following the solubility rules.
• Precaution: do not taste salts; some are poisonous.

Solubility Rules

• A soluble salt dissolves in water at room temperature; an insoluble salt does not.

Salt type	Solubility in water
Nitrates	All soluble
Sulphates	Soluble except $\mathrm{CaS}{\mathrm{O}}_4$, $\mathrm{BaS}{\mathrm{O}}_4$ and $\mathrm{PbS}{\mathrm{O}}_4$
Chlorides	Soluble except $\mathrm{AgCl}$, $\mathrm{PbC}{\mathrm{l}}_2$ and $\mathrm{H}{\mathrm{g}}_2\mathrm{C}{\mathrm{l}}_2$
Carbonates	Insoluble except $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$, ${\mathrm{K}}_2\mathrm{C}{\mathrm{O}}_3$ and $(\mathrm{N}{\mathrm{H}}_4{)}_2\mathrm{C}{\mathrm{O}}_3$
Ammonium, sodium and potassium salts	All soluble

• $\mathrm{PbC}{\mathrm{l}}_2$ and $\mathrm{Pb}{\mathrm{I}}_2$ are insoluble in cold water but dissolve in hot water and recrystallise on cooling.

Choosing a Preparation Method

Target salt	Method
Soluble ammonium, sodium or potassium salt	Acid–alkali titration, then crystallisation
Other soluble salt	Acid + excess reactive metal, metal oxide or metal carbonate; filter and crystallise
Insoluble salt	Double decomposition (precipitation) between two soluble salts

Soluble Ammonium, Sodium and Potassium Salts

Potassium chloride:

1. Titrate $25\mathrm{c}{\mathrm{m}}^3$ of $2.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ $\mathrm{KOH}$ with $2.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ $\mathrm{HCl}$ using phenolphthalein; record the neutralising acid volume, $V$.
2. Repeat with fresh $\mathrm{KOH}$ and exactly $V\mathrm{c}{\mathrm{m}}^3$ acid without indicator, so the crystals are not contaminated.
3. Evaporate gently to saturation, cool to crystallise, filter, rinse with a little distilled water and dry between filter papers.

• Saturation test: crystals forming on a glass rod after dipping and removing it indicate a saturated solution. Rinse crystals with only a little distilled water to limit dissolution.

Other Soluble Salts

Copper(II) nitrate from acid and metal oxide:

1. Warm $20\mathrm{c}{\mathrm{m}}^3$ of $2.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ nitric acid; add $\mathrm{CuO}$ gradually until some remains undissolved, ensuring all acid has reacted.
2. Filter off excess $\mathrm{CuO}$; heat the filtrate to saturation.
3. Cool, filter, rinse with a little distilled water and dry the crystals.

$2HN{\mathrm{O}}_3(\mathrm{aq})+\mathrm{CuO}(\mathrm{s})\to \mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

• The acid–metal route uses the same excess-solid method with sulphuric acid and zinc to form $\mathrm{ZnS}{\mathrm{O}}_4$; copper is unsuitable because it is below hydrogen in the reactivity series and does not react with dilute sulphuric acid.
• The acid–carbonate route uses an acid and an insoluble metal carbonate; $\mathrm{C}{\mathrm{O}}_2$ is released and confirmed by cloudy limewater.

Purification by Recrystallisation

1. Dissolve the impure soluble salt in the minimum amount of hot distilled water.
2. Filter the hot solution to remove insoluble impurities.
3. Evaporate the filtrate to saturation and cool to crystallise.
4. Filter, rinse crystals with a little distilled water and dry between filter papers.

Insoluble Salts by Double Decomposition

• Mix two soluble salt solutions containing the required ions; their ions exchange partners and an insoluble precipitate forms.
• Filter the precipitate, wash it with distilled water to remove soluble impurities and dry it.

$\mathrm{BaC}{\mathrm{l}}_2(\mathrm{aq})+\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})+2NaCl(\mathrm{aq})$

$\mathrm{B}{\mathrm{a}}^{2+}(\mathrm{aq})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})$