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7.3.2 Insoluble salts and water of crystallisation

Preparing insoluble salts

Definition

Precipitation

A reaction in which two solutions are mixed and an insoluble solid forms.

Definition

Precipitate

An insoluble solid that forms when two solutions react together.

  1. An insoluble salt cannot be made by crystallising it from a solution, because it does not dissolve to form one.
  2. Instead, it is made by precipitation, which means mixing two solutions so that the salt forms as a solid.
  3. Two soluble compounds are chosen, one supplying the positive ion and the other supplying the negative ion of the salt you want.
  4. When the solutions are mixed, these two ions meet and join straight away to form a solid precipitate.
  5. The other two ions stay dissolved in the solution, so they can be washed away.
  6. Nitrates and sodium or potassium compounds make good starting materials, because they are always soluble.
  7. The precipitate is separated by filtration, so it collects on the filter paper as the residue.
  8. The residue is washed with distilled water to remove the dissolved ions of the other product.
  9. Finally the solid is dried in a warm oven or between filter papers.
  10. The ionic equation for a precipitation shows only the two ions that form the solid, with the state symbol (s) for the precipitate.
Example

Choosing reactants for barium sulfate

  • Barium sulfate is insoluble, so it needs Ba2+\text{Ba}^{2+}Ba2+ ions and SO42−\text{SO}_4^{2-}SO42−​ ions brought together from two solutions.
  • Barium nitrate is soluble because all nitrates are, so it supplies the barium ions.
  • Sodium sulfate is soluble because all sodium compounds are, so it supplies the sulfate ions.
Ba(NO3)2(aq)+Na2SO4(aq)→BaSO4(s)+2NaNO3(aq) \text{Ba(NO}_3)_2\text{(aq)} + \text{Na}_2\text{SO}_4\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} + 2\text{NaNO}_3\text{(aq)} Ba(NO3​)2​(aq)+Na2​SO4​(aq)→BaSO4​(s)+2NaNO3​(aq)
  • The sodium nitrate stays dissolved and is washed away from the white precipitate.
Ba2+(aq)+SO42−(aq)→BaSO4(s) \text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)} Ba2+(aq)+SO42−​(aq)→BaSO4​(s)
Example

Choosing reactants for silver chloride

  • Silver chloride is one of the insoluble chlorides, so it needs Ag+\text{Ag}^{+}Ag+ ions and Cl−\text{Cl}^{-}Cl− ions.
  • Silver nitrate solution supplies the silver ions and sodium chloride solution supplies the chloride ions.
AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq) \text{AgNO}_3\text{(aq)} + \text{NaCl(aq)} \rightarrow \text{AgCl(s)} + \text{NaNO}_3\text{(aq)} AgNO3​(aq)+NaCl(aq)→AgCl(s)+NaNO3​(aq)
  • The ionic equation shows only the ions that form the white precipitate.
Ag+(aq)+Cl−(aq)→AgCl(s) \text{Ag}^{+}\text{(aq)} + \text{Cl}^{-}\text{(aq)} \rightarrow \text{AgCl(s)} Ag+(aq)+Cl−(aq)→AgCl(s)
Practical

Preparing barium sulfate by precipitation

  • Aim: to prepare a pure, dry sample of insoluble barium sulfate by precipitation.
  • Apparatus: two 25 cm325\ \text{cm}^325 cm3 measuring cylinders, 100 cm3100\ \text{cm}^3100 cm3 beaker, glass rod, barium nitrate solution, sodium sulfate solution, filter funnel, filter paper, conical flask, wash bottle of distilled water, watch glass, warm oven and eye protection.
  • Method:
    • Measure 20 cm320\ \text{cm}^320 cm3 of barium nitrate solution into the beaker.
    • Add 20 cm320\ \text{cm}^320 cm3 of sodium sulfate solution and stir with the glass rod, and a white precipitate appears at once.
    • Fold a filter paper into a cone, fit it in the funnel over the conical flask, and pour the mixture through.
    • Rinse the beaker with a little distilled water and pour the rinsings into the funnel, so that no precipitate is left behind.
    • Wash the residue on the filter paper two or three times with distilled water from the wash bottle to remove the dissolved sodium nitrate.
    • Open the filter paper carefully, spread it on a watch glass, and dry the solid in a warm oven.
  • Results: a fine white powder of barium sulfate is left on the filter paper, while the filtrate passing through is colourless.
  • Safety: wear eye protection, and wash your hands after the practical because soluble barium compounds such as barium nitrate are toxic if swallowed.

Water of crystallisation

Definition

Water of crystallisation

The water molecules that are chemically bound within the structure of hydrated crystals.

  1. When many salts crystallise from water, a fixed number of water molecules is built into the crystal structure for each formula unit of the salt.

  2. This chemically bound water is called water of crystallisation.

  3. The formula of a hydrated salt shows the water after a dot, for example CuSO4⋅5H2O\text{CuSO}_4{\cdot}5\text{H}_2\text{O}CuSO4​⋅5H2​O.

  4. The dot means that the water is combined with the salt, so it does not mean multiplication.

  5. Hydrated copper(II) sulfate, CuSO4⋅5H2O\text{CuSO}_4{\cdot}5\text{H}_2\text{O}CuSO4​⋅5H2​O, has five water molecules for every CuSO4\text{CuSO}_4CuSO4​ and forms blue crystals.

  6. Hydrated cobalt(II) chloride, CoCl2⋅6H2O\text{CoCl}_2{\cdot}6\text{H}_2\text{O}CoCl2​⋅6H2​O, has six water molecules for every CoCl2\text{CoCl}_2CoCl2​ and forms pink crystals.

  7. Heating removes the water of crystallisation and leaves the anhydrous salt.

    CuSO4⋅5H2O(s)→CuSO4(s)+5H2O(g) \text{CuSO}_4{\cdot}5\text{H}_2\text{O(s)} \rightarrow \text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(g)} CuSO4​⋅5H2​O(s)→CuSO4​(s)+5H2​O(g) CoCl2⋅6H2O(s)→CoCl2(s)+6H2O(g) \text{CoCl}_2{\cdot}6\text{H}_2\text{O(s)} \rightarrow \text{CoCl}_2\text{(s)} + 6\text{H}_2\text{O(g)} CoCl2​⋅6H2​O(s)→CoCl2​(s)+6H2​O(g)

  8. The water of crystallisation must be included when working out the relative formula mass, MrM_rMr​, of a hydrated salt.

Example

Relative formula mass of hydrated copper(II) sulfate

  • Add the relative atomic masses for the anhydrous part, using ArA_rAr​ values of Cu=64\text{Cu} = 64Cu=64, S=32\text{S} = 32S=32, O=16\text{O} = 16O=16 and H=1\text{H} = 1H=1.
Mr(CuSO4)=64+32+(4×16)=160 M_r(\text{CuSO}_4) = 64 + 32 + (4 \times 16) = 160 Mr​(CuSO4​)=64+32+(4×16)=160
  • Add the mass of the five water molecules.
Mr(5H2O)=5×[(2×1)+16]=90 M_r(5\text{H}_2\text{O}) = 5 \times \left[(2 \times 1) + 16\right] = 90 Mr​(5H2​O)=5×[(2×1)+16]=90 Mr(CuSO4⋅5H2O)=160+90=250 M_r(\text{CuSO}_4{\cdot}5\text{H}_2\text{O}) = 160 + 90 = 250 Mr​(CuSO4​⋅5H2​O)=160+90=250
  • The percentage by mass of water in the crystals follows from these two values.
percentage of water=90250×100=36.0% \text{percentage of water} = \frac{90}{250} \times 100 = 36.0\% percentage of water=25090​×100=36.0%
  • This means that 36.0%36.0\%36.0% of the mass of the blue crystals is lost as steam when they are heated until no more water is given off.
Example

Number of water molecules from heating data

  • A 5.00 g5.00\ \text{g}5.00 g sample of hydrated magnesium sulfate, MgSO4⋅xH2O\text{MgSO}_4{\cdot}x\text{H}_2\text{O}MgSO4​⋅xH2​O, is heated until its mass stops changing, leaving 2.44 g2.44\ \text{g}2.44 g of anhydrous MgSO4\text{MgSO}_4MgSO4​.
  • The mass of water lost is the difference between the two masses.
5.00−2.44=2.56 g 5.00 - 2.44 = 2.56\ \text{g} 5.00−2.44=2.56 g
  • Divide each mass by its molar mass, using Mr(MgSO4)=120M_r(\text{MgSO}_4) = 120Mr​(MgSO4​)=120 and Mr(H2O)=18M_r(\text{H}_2\text{O}) = 18Mr​(H2​O)=18.
moles of MgSO4=2.44120=0.0203 mol \text{moles of MgSO}_4 = \frac{2.44}{120} = 0.0203\ \text{mol} moles of MgSO4​=1202.44​=0.0203 mol moles of H2O=2.5618=0.142 mol \text{moles of H}_2\text{O} = \frac{2.56}{18} = 0.142\ \text{mol} moles of H2​O=182.56​=0.142 mol
  • Divide by the smaller number of moles to find the ratio.
0.1420.0203=7.0 \frac{0.142}{0.0203} = 7.0 0.02030.142​=7.0
  • The value of xxx is 7, so the formula is MgSO4⋅7H2O\text{MgSO}_4{\cdot}7\text{H}_2\text{O}MgSO4​⋅7H2​O.
Common Mistake
  • Do not keep the filtrate when making an insoluble salt, because the product is the residue left on the filter paper.
  • Do not choose an insoluble reactant such as barium carbonate for a precipitation, because its ions are not free in solution to react.
  • Do not confuse water of crystallisation with water that simply makes crystals damp, because water of crystallisation is chemically bound.
  • Do not leave out the water of crystallisation when calculating the MrM_rMr​ of a hydrated salt.
Exam technique

Describing a precipitation method

  • Name two soluble reactants, one containing each ion of the insoluble salt.
  • Give the steps in order: mix the solutions, filter, wash the residue with distilled water, then dry it.
  • Explain that washing removes the soluble product, which would otherwise contaminate the salt.
  • Write the ionic equation with only the two reacting ions and the state symbol (s) on the precipitate.
Self review
  • Why can an insoluble salt not be made by crystallising it from solution?
  • Name two solutions that could be mixed to make barium sulfate.
  • Why is the precipitate washed with distilled water?
  • What is water of crystallisation?
  • What is the MrM_rMr​ of CuSO4⋅5H2O\text{CuSO}_4{\cdot}5\text{H}_2\text{O}CuSO4​⋅5H2​O?
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7.3.2 Insoluble salts and water of crystallisation Revision Guide

  1. IGCSE
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Revision notes for CIE IGCSE Chemistry 7.3.2 Insoluble salts and water of crystallisation: explanations and worked examples.

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