TOPIC 6.

CHEMICAL REACTIONS AND IONIC EQUATIONS.

Reactions involving ionic compounds.As discussed earlier, ionically bonded compounds consist of large aggregations ofcations and anions which pack together in crystal lattices in such a way that theelectrostatic attractions between oppositely charged ions are maximised andrepulsions between like charged ions minimised. When an ionic crystal is placedin water, in many cases the solid dissolves, releasing the component ions into theSOLVENT to form a SOLUTION. Such compounds are said to be SOLUBLEand the substance that dissolves is called the SOLUTE. A well known example ofa soluble ionic compound is table salt or sodium chloride. The process ofdissolving can be best represented by an equation which is slightly different fromthe formula equations used in Topic 5. Instead, an IONIC EQUATION is used toshow the ions released into the solution as follows.

NaCl(s) v Na + Cl+ –

The same rules apply as for formula equations in that all species shown on the leftmust also be present on the right hand side of the equation. In addition, notice thatthe electrical charge present on both sides of the equation also balances.Thus the equation for another ionic solid, barium chloride, dissolving in waterwould be as follows

2BaCl (s) v Ba + Cl + Cl2+ – –

which is usually written as

2BaCl (s) v Ba + 2Cl 2+ –

Because the Cl ions are separate individual species, they are represented as 2Cl– –

2and not as Cl , which would mean two Cl atoms bonded together and bearing a 22–

negative charge, (Cl!Cl) .2–

When ions are released into water solution, they all experience attractions to watermolecules which form spheres around them, as illustrated for NaCl below.

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The reason for this attractive force between water molecules and ions is the abilityof the oxygen atoms in water molecules to attract the electrons in their O–H bondsto a greater extent than do the hydrogen atoms. The O atom is said to be moreELECTRONEGATIVE than the H atom. This results in a slight negative chargeon the oxygen atom and a slight positive charge on each hydrogen atom. The O–Hbond is an example of a POLAR BOND and the water molecule, being angular inshape, has a non-symmetric distribution of charge and is a POLAR MOLECULE.

In the dissolution of ionic solids such as sodium chloride, the oxygen atoms ofwater molecules are attracted to the positive charge on cations (Na in this+

example) while its hydrogen atoms are attracted to the anions (Cl in this example). –

This results in considerable energy being released and the subsequent breaking upof the crystal lattice of the ionic solid to form individual ions surrounded by theattracted water molecules.

Consequently, ions in water solution are said to be AQUATED and sometimes thesuffix (aq) is used to emphasise this point. Thus the two equations above might alsobe written as

NaCl(s) v Na (aq) + Cl (aq)+ –

2BaCl (s) v Ba (aq) + 2Cl (aq)2+ –

Initially the (aq) suffix will be used here, but later it will be assumed that all ions inwater solution are aquated and the (aq) suffix will be omitted. Some otherexamples of ionic equations for ionic solids dissolving follow.

2 3 3K CO (s) v 2K (aq) + CO (aq)+ 2–

4 3 4 4 4 (NH ) PO (s) v 3NH (aq) + PO (aq)+ 3–

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Check your understanding of this section:Why do ionic compounds when soluble, dissolve best in water?Write an ionic equation for the dissolution of iron(III) chloride in water.Account for the water molecule being polar.What is the source of the energy required to break up an ionic crystal when itdissolves?

Many text books use the (aq) symbolism (incorrectly) to indicate a solution of asubstance in water by attaching (aq) to the formula of that solid. Clearly, therecan be no such species as an aquated ionic substance because if it has dissolved,it is totally present as ions. Therefore equations showing species such asNaCl(aq) are most misleading and should be ignored.

Not all ionic solids will dissolve in water to a significant extent. For example, theamount of the ionic solid silver chloride, AgCl, which will dissolve in water is sosmall that it is classed as insoluble. Insoluble ionic compounds of common metalsinclude three chlorides, about five sulfates, most carbonates, most phosphates andmost sulfides. How does one know if a given ionic compound is soluble?A table classifying the solubilities of ionic compounds in water is given at the endof this Topic. This table lists the ionic compounds of the common cations assoluble, insoluble or slightly soluble. This information probably will be needed tobe committed to memory at a later stage of your course, but for the present youshould consult it if needed for the ensuing exercises. The contents of the table aremore easily remembered in terms of the general situation for each anion. Forexample: all nitrates are soluble; most hydroxides are insoluble except for those ofthe first family members Na and K, plus Ba; most carbonates are insoluble exceptthose of the first family members Na and K. It is also useful to remember that allof the compounds of the elements Na and K from the first family are soluble.

How can a soluble ionic compound be obtained back from a solution?The aquated ions released into a solution when an ionic compound dissolves and itsionic bonds broken are free to move about throughout the solution. Ionic bondingis only present in the solid state. However, if the solution is boiled, the water isdriven off as a gas (VOLATILE) but the ions remain in the solution (NON-VOLATILE). Volatile substances have sufficiently weak forces of attractionbetween their constituent entities to allow them to escape to the vapour phase whenenough heat energy is supplied. In non-volatile substances, the attractive forcesoperating are much stronger and much higher temperatures would be required tovaporise them. When sufficient water has been evaporated, the cations and anionsare deprived of their surrounding water molecules and they can then recombine toform the ionically bonded solid again. In this way, any solution of an ioniccompound can be evaporated sufficiently for crystals to form. The equation for theevaporation process would simply be the reverse of the equation for the dissolution,for example

Na (aq) + Cl (aq) v NaCl(s)+ –

To emphasise that the solution is being evaporated, it may be helpful to write"evap" or similar over the arrow.

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Precipitation reactions.Silver chloride is seen from the solubility table to be insoluble in water. Silvernitrate and sodium chloride are both soluble compounds and in water-solutionwould release their component ions as shown in the following ionic equations.

3 3AgNO (s) v Ag (aq) + NO (aq)+ –

NaCl(s) v Na (aq) + Cl (aq)+ –

If a solution of silver nitrate were mixed with a solution of sodium chloride, thenthe Ag ions would react with the Cl ions to form a PRECIPITATE of the+ –

insoluble salt silver chloride, AgCl, as shown in the following ionic equation.

3 3Ag (aq) + NO (aq) + Na (aq) + Cl (aq) v AgCl(s) + NO (aq) + Na (aq)+ – + – – +

3Note that only the Ag and the Cl ions have reacted, leaving the Na and NO ions+ – + –

free in the solution. As these last two ions have not in fact entered into a reaction,they can be deleted from the equation in much the same way as common terms arecancelled from both sides of a mathematical equation. Such ions are calledSPECTATOR IONS. Initially it may be helpful to write down all the ions whichare being mixed together in order to establish whether any combination can forman insoluble salt, and then cancel out the spectator species. However, with practiceyou will be able to delete this step and write the final equation in one step. For thisreaction it would be

Ag (aq) + Cl (aq) v AgCl(s)+ –

Being a solid, the silver chloride could be obtained by filtering the mixture. Theprecipitate would be retained in the filter paper and the solution containing thespectator ions (called the FILTRATE) would pass through the filter paper. It

3would contain only the ions Na and NO , provided exactly equal amounts of the+ –

Ag and Cl ions were in the solutions which were mixed originally. By+ –

evaporating the water, crystals of sodium nitrate could be isolated from the filtrate.

evap3 3Na (aq) + NO (aq) v NaNO (s)+ –

From a knowledge of the solubilities of ionic compounds, one can predict whetherany combination of solutions of soluble compounds will lead to a precipitationreaction. The following examples show how to establish whether a precipitationreaction occurs when solutions are mixed.

Example 1. If a water solution of potassium chloride were added to a water solution ofcopper(II) nitrate, would any reaction occur and if so, what would be the product?

To answer this, consider all possible combinations of the four ions which are to be

3 3mixed - K , Cl , Cu and NO . The formulas of the possible products are KNO+ – 2+ –

2and CuCl . By consulting the solubility table given, it is seen that both of thesecompounds are soluble so there would be no reaction. If the solution formed bymixing this combination were to be evaporated, the resulting solid would be a

3 2 3 2mixture of all four possible compounds, KNO , CuCl , Cu(NO ) and KCl.

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Check your understanding of this section:Given that the formation of an ionic solid from its component ions involves therelease of energy, why is it necessary to supply heat to a solution of sodiumchloride in order to reclaim the solid from solution?What advantages are there in using an ionic equation to represent theprecipitation of silver chloride from a mixture of sodium chloride and silvernitrate solutions?In the previous reaction, which are the spectator ions?

Example 2. Water solutions of lead(II) nitrate and sodium sulfate are mixed. What, if any,reaction occurs?

The possible combinations of ions could produce the compounds of formulas

4 3PbSO and NaNO . Of these two, lead(II) sulfate is insoluble and sodium nitrate is

4soluble. Therefore, a precipitate of PbSO would form according to the equation

4 4Pb (aq) + SO (aq) v PbSO (s)2+ 2–

4In this example, only one of the possible products is insoluble (PbSO ) but if bothpossible products were insoluble, then a mixture of both compounds would form.

Example 3.Water solutions of sodium carbonate and barium chloride are mixed. What, if any,reaction occurs?

3The possible combinations of ions could produce the compounds NaCl and BaCO . Of these two, barium carbonate is insoluble while sodium chloride is soluble.

3Therefore, a precipitate of BaCO would form according to the equation

3 3Ba (aq) + CO (aq) v BaCO (s)2+ 2–

Example 4.Water solutions of iron(II) sulfate and potassium phosphate are mixed. What, ifany, reaction occurs?

2 4The possible combinations of ions could produce the compounds K SO and

3 4 2Fe (PO ) . Of these two, iron(II) phosphate is insoluble while potassium sulfate is

3 4 2soluble. Therefore, a precipitate of Fe (PO ) would form according to the equation

4 3 4 23Fe (aq) + 2PO (aq) v Fe (PO ) (s)2+ 3–

Reactions of acids.4Apart from the cations from metals and the polyatomic cation NH , another cation+

frequently encountered in reactions is the hydrogen ion, H , which is also+

sometimes called the oxonium ion or the hydronium ion. Recall from Topic 2 thatthe H atom consists of just one proton for the nucleus, surrounded by a single orbiting electron. If this electron were removed, the H ion would be formed and it+

would be just a free proton. A proton is extremely small and the charge density on

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it would therefore be very concentrated. Consequently, the H ion does not have an+

independent existence in solution. Instead it associates with water molecules byjoining on to them using one of the lone pairs of electrons on the O atom, and is

3more correctly represented as H (aq), or frequently as H O . These are all equally+ +

acceptable ways of representing the hydrogen ion. Hydrogen ions are supplied inwater solution by compounds called ACIDS. The three most common acidsencountered are:

3 3nitric acid, HNO : contains H (aq) and NO (aq) ions+ –

2 4 4 4sulfuric acid, H SO : which contains H (aq), HSO (aq) and SO (aq) ions + – 2–

hydrochloric acid: a water solution of the gas hydrogen chloride, HCl(g),which ionizes to H (aq) and Cl (aq) ions in the water. + –

Acids such as these provide H ions as well as the anion from the acid in solution. +

Both the H ions and the anions can participate in reactions. Thus addition of+

hydrochloric acid to silver nitrate solution produces silver chloride precipitate justlike the previous example when solutions of silver nitrate and sodium chloride were

3mixed. Remaining in solution would be the spectator ions, H and NO , which is a+ –

solution of nitric acid. The ionic equation for the reaction is

3 3H (aq) + Cl (aq) + Ag (aq) + NO (aq) v AgCl(s) + H (aq) + NO (aq)+ – + – + –

When the spectator ions are deleted, it reduces to

Cl (aq) + Ag (aq) v AgCl(s) – +

which is identical to the ionic equation given previously for the reaction betweensilver nitrate solution and sodium chloride solution.From this example, it can be seen that any soluble silver salt and any solublechloride or hydrochloric acid, would react when mixed together in a water solutionto precipitate silver chloride. One advantage of writing ionic equations is that onlythe actual species reacting are shown.

Reactions involving the H ion.+

In the precipitation reactions involving acids discussed above, it is the anion fromthe acid which is reacting and the H is the spectator ion. However, there are many+

reactions of acids where the important ion is H while the anion from the acid is+

merely a spectator ion.All acids supply hydrogen ions in solution. This ion can participate in a number ofcommon reaction types, recognition of which makes the writing of chemicalequations much simpler. These reaction types are:

1. Acids with reactive metals.Many metals react with acids to form the cation of the metal and produce hydrogen

2gas, H , from the H ions. These metals are sometimes called "reactive metals", as+

distinct from the "coinage metals" such as silver, gold, copper, and platinum whichare inert to most acids. In particular, hydrochloric acid and sulfuric acid behave inthis way with reactive metals, and in the process, will leave the chloride or sulfate ions in the solution. Evaporation of the water from the resulting solution willproduce the ionic compound of the metal and the acid used. Ionic compounds are

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an acid + a reactive metal forms a salt and hydrogen gas

an acid + an oxide forms a salt and water.

frequently called "SALTS" which is a general term and, as pointed out earlier, isnot restricted to common table salt, NaCl. For example, hydrochloric acid added to magnesium metal produces hydrogenwhich is evolved as a gas and leaves in solution magnesium ions, Mg and2+

chloride ions, Cl . Evaporation of this solution results in isolation of the salt,–

magnesium chloride.

2 2Formula equation: Mg + 2HCl v MgCl + H

2Ionic equation: Mg(s) + 2H (aq) + 2Cl (aq) v Mg (aq) + H (g) + 2Cl (aq)+ – 2+ –

or, deleting the Cl spectator ions,–

2 Mg(s) + 2H (aq) v Mg (aq) + H (g)+ 2+

2. Acids with oxides of metals.All ionic oxide compounds, regardless of whether they are water soluble, react withacids to release the metal ion from the oxide into the solution and also to formwater. As an example, consider the equation for the reaction of hydrochloric acid withcalcium oxide.

2 2Formula equation: CaO + 2HCl v CaCl + H O

2Ionic equation: CaO(s) + 2H (aq) + 2Cl (aq) v Ca (aq) + 2Cl (aq) + H O(l)+ – 2+ –

If the water were evaporated from the solution, the Ca and Cl ions would2+ –

crystallise to form the solid compound, calcium chloride.

Note that in the equation above, the chloride ion is a spectator ion. This ion shouldbe omitted, to give for the correct equation

2CaO(s) + 2H (aq) v Ca (aq) + H O(l)+ 2+

3. Acids with hydroxides of metals.In the same way as the H ions of acids react with metal oxides to form a salt and+

water, so too can acids react with hydroxide compounds of metals, regardless ofwhether they are water soluble, to also form a salt and water. As in the case ofoxides, if the salt formed is soluble in water, then the resulting solution must beevaporated in order to isolate the solid compound.For example, nitric acid reacts with copper(II) hydroxide to form water andcopper(II) nitrate in solution.

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an acid + a hydroxide forms a salt and water

an acid + an oxide or hydroxide forms a salt and water.

2 3 3 2 2Formula equation: Cu(OH) + 2HNO v Cu(NO ) + 2H O

2 3 2 3Ionic: Cu(OH) (s) + 2H (aq) + 2NO (aq) v 2H O(l) + Cu (aq) + 2NO (aq)+ – 2+ –

3Again, the NO (aq) ion is a spectator ion and should be deleted from both sides of –

the equation, as the H (aq) is the only component of the acid which reacts with the+

2Cu(OH) (s).

2 2 Cu(OH) (s) + 2H (aq) v Cu (aq) + 2H O(l)+ 2+

3 2Evaporation of this solution would produce the salt, Cu(NO ) , according to theequation

evap3 3 2 Cu (aq) + 2NO (aq) v Cu(NO ) (s)2+ –

If the hydroxide reacting with the acid is soluble in water and a solution of thathydroxide is specified, then the cation of the hydroxide is also a spectator ion andshould be deleted along with the anion from the acid. For example, hydrochloricacid reacting with a solution of sodium hydroxide would be represented by theionic equation

2 H (aq) + OH (aq) v H O(l)+ –

The salt NaCl(s) could be obtained by evaporating the solution.

or, combining this with the previous section,

4. Acids with carbonates.Another common reaction of an acid leads to the evolution of a gas when the acidis mixed with the metal salts of certain anions. This type of reaction is typified bythe well-known case of acids reacting with carbonates to produce carbon dioxidegas, water and a salt. Again, the reaction occurs regardless of whether thecompound reacting with the acid is soluble or insoluble in water. For example,sulfuric acid reacts with solid magnesium carbonate to produce carbon dioxide gasand magnesium sulfate in solution as follows:

3 2 4 4 2 2Formula equation: MgCO + H SO v MgSO + CO + H O

3 2 2Ionic: MgCO (s) + 2H (aq) v Mg (aq) + CO (g) + H O(l)+ 2+

4Note that the spectator ion, SO (aq) has been deleted from both sides of the ionic2–

equation. If the carbonate compound used were soluble and was already in solution, then theionic equation would not include the metal ion because it would be a spectator ion,

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an acid + a carbonate forms carbon dioxide gas, a salt and water

Check your understanding of this section:By what criterion would you decide if a species is an acid?What is the expected reaction when an acid is mixed with a reactive metal?What happens when an acid is placed with an insoluble oxide or hydroxide?When an acid is placed on a compound such as calcium carbonate, what isobserved?

as in the following example of adding nitric acid to a solution of sodium carbonate:

3 32Na (aq) + CO (aq) + 2H (aq) + 2NO (aq) v + 2– + –

3 2 2 2Na (aq) + 2NO (aq) + H O(l) + CO (g)+ –

which simply becomes, after deleting spectator ions,

3 2 2 CO (aq) + 2H (aq) v CO (g) + H O(l)2– +

Objectives of this Topic.When you have completed this Topic, including the tutorial questions, you shouldhave achieved the following goals:

1. Know the meaning of the terms solvent; solute; solution; aquated ions;volatile; non-volatile; spectator ions; precipitate; filtrate; electronegativeatom; polar bond; polar molecule; non-polar bond; non-polar molecule.

2. Understand the concept of polarity in covalent molecules, especially as it relates to the role of water as a solvent for ionic compounds.

3. Be able to write ionic equations for any dissolution process or anycrystallisation of a salt.

4. Know how to find out if any given ionic compound is soluble.

5. Be able to write ionic equations for any precipitation of a salt.

6. By use of solubility tables, be able to predict whether any given combinationof solutions of salts will lead to formation of a precipitate.

7. Recognise acids as a source of hydrogen ions.

8. Know the names and formulas for several common acids.

9. Recognise acids as being able to precipitate insoluble salts by supplying the relevant anion.

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10. Know the reactions of acids with (i) reactive metals; (ii) oxides andhydroxides; (iii) carbonates.

11. Be able to write ionic equations for each of the above reaction types.

SOLUBILITY TABLE OF SOME COMMON SALTS

ANIONBEHAVIOUR WITH COMMON CATIONS

USUAL EXCEPTIONALSLIGHTLYSOLUBLE

F– soluble except Mg , Ca , Sr , Ba , Mn , Pb Al2+ 2+ 2+ 2+ 2+ 2+ 3+

Cl– 2soluble except Ag , Hg Pb+ 2+ 2+

Br– 2soluble except Ag , Hg Hg , Pb+ 2+ 2+ 2+

I– 2soluble except Ag , Hg , Hg , Pb+ 2+ 2+ 2+

4SO 2– 2soluble except Sr , Ba , Hg , Pb Ca , Ag2+ 2+ 2+ 2+ 2+ +

3NO – soluble

3 2CH CO – 2soluble Ag , Hg+ 2+

OH– insoluble except Na , K , Ba Ca , Sr+ + 2+ 2+ 2+

3SO 2– 4insoluble except Na , K , NH , Mg+ + + 2+

4PO 3– 4insoluble except Na , K , NH+ + +

3CO 2– 4insoluble except Na , K , NH+ + +

2 4C O 2– 4insoluble except Na , K NH+ + +

S2– 4insoluble except Na , K , NH , Mg , Ca , Sr , Ba+ + + 2+ 2+ 2+ 2+

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SUMMARY

Ionic compounds consist of very large aggregates of cations and anions packed intoa crystal lattice so as to maximise the attractive forces between cations and anionsand minimise the repulsive cation/cation and anion/anion forces. Despite thestrength of ionic bonds, many ionic compounds can dissolve in solvents such aswater to form a solution, a process best represented by an ionic equation. Ionicequations show the species involved in the actual form in which they are present, soions in solution are shown as such. Ionic equations require not only that the massshould balance on both sides but also the charge. Otherwise, ionic equations followthe same rules as for formula equations. The reason why ionic compounds may dissolve in water is that water is a polarmolecule in which the oxygen atoms attract more than their share of the bondingelectrons (said to be more electronegative) than the hydrogen atoms, leaving a slightnegative charge on the O atoms and a slight positive charge on the H atoms. Thisuneven charge distribution allows water molecules to be attracted to both cationsand anions in the ionic crystal, releasing much energy in the process. If enoughenergy is released through these attractions, the ions may leave the crystal latticeand go into solution as aquated ions, each surrounded by a sheath of watermolecules. These aquated ions are free to move around throughout the solution. Once dissolved, ionic bonds cease to exist as this form of bonding is really onlypresent in the solid state. Not all ionic compounds are soluble and solubility tableslisting this data are available. If the solvent is boiled off from a solution of an ionic compound, the ions, strippedof their surrounding water molecules, are non-volatile and remain to recombine ascrystals of the solid compound.If cations and anions from two different salts are mixed in the same solution andany combination of them corresponds to an insoluble salt, then that compound willbe precipitated from the solution. Other cations and anions present will remain insolution and are termed spectator ions. The precipitated solid can be collected byfiltration and the remaining solution (the filtrate) can be boiled down to collect anysoluble compounds. This type of reaction is best represented by an ionic equationin which only the reacting ions are shown and the spectator ions are deleted as theyare present on both sides of the reaction equation. Acids are compounds that release the H ion in solution. This ion would strictly be+

just a proton with no electron, but this species would have too large a charge on toosmall a volume to be stable, so it associates with water molecules via a lone pair of

3electrons on the O atom and thus can also be represented as H (aq) or H O . Apart+ +

from the anion from the acid in solution, reactions of the H ion are conveniently+

grouped into several classes.Acids on reactive metals: form a salt and hydrogen gas.Acids on oxides of metals: form a salt and water.Acids on hydroxides of metals: form a salt and water.Acids on a carbonate: form a salt and water and carbon dioxide gas.In each case, the anion in the salt formed is the anion that belonged to the acid, forexample, chloride from hydrochloric acid or nitrate from nitric acid.Each of these reaction types are conveniently represented by ionic equations whichmore clearly show the process which is taking place than overall formula equations.

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TUTORIAL QUESTIONS - TOPIC 6.

1. Explain the meaning of the following terms: solvent; solution; precipitate; filtrate; formula equation; ionic equation; volatile substance.

2. Give the missing formula or name for each entry in the following table. FORMULA NAME

2CuBr

4BaSO

4NH I

3 2Mn(NO )

AgF

NaCl

3 4 2Fe (PO )

4FePO

3ZnCO

2 3K CO

3 2 2Ca(CH CO )

3 2Zn(HCO )

LiOH

4CuSO

3 2Pb(NO )

2 2Pb(NO )

2 3Li SO

cobalt(II) chloride

magnesium sulfate

lead(II) carbonate

strontium sulfite

aluminium nitrate

silver oxalate

tin(II) acetate

iron(II) bromide

sodium nitrite

iron(III) bromide

nickel(II) iodide

lithium carbonate

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3. Write equations for the dissolution of the following ionic compounds in water.

cobalt(II) sulfate

ammonium nitrate

iron(III) chloride

sodium phosphate

aluminium nitrate

potassiumcarbonate

iron(II) bromide

4. To be done as a group: Write ionic equations for any reaction that occurs whenthe following pairs of substances are mixed. If no reaction occurs, write "noreaction".

(a) a water solution of sodium sulfate and a water solution of barium chloride

(b) a water solution of potassium hydroxide and a water solution of copper(II) nitrate

(c) a water solution of silver nitrate and a water solution of sodium chloride

(d) a water solution of silver nitrate and a water solution of barium chloride

(e) a water solution of silver nitrate and dilute hydrochloric acid

(f) a water solution of lead(II) nitrate and a water solution of potassium chloride

(g) a water solution of copper(II) chloride and a water solution of lead(II) nitrate

(h) a water solution of iron(III) sulfate and a water solution of sodium hydroxide

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(i) a water solution of cobalt(II) sulfate and a water solution of ammonium carbonate

(j) a water solution of sodium phosphate and a water solution of magnesium chloride

(k) a water solution of magnesium bromide and a water solution of lead(II) nitrate

(l) a water solution of copper(II) nitrate and a water solution iron(III) sulfate

(m) a water solution of potassium iodide and a water solution of sodium chloride

(n) a water solution of aluminium sulfate and a water solution of potassium phosphate

5. To be done as a group: Write an ionic equation for any reaction that occurswhen the following pairs are mixed. If no reaction occurs, write "no reaction".

(a) solid sodium carbonate and dilute nitric acid

(b) copper(II) carbonate and dilute sulfuric acid

(c) copper(II) carbonate and dilute hydrochloric acid

(d) silver carbonate and dilute nitric acid

(e) iron(II) hydroxide and dilute hydriodic acid

(f) solid sodium hydroxide and dilute nitric acid

(g) a water solution of sodium hydroxide and dilute sulfuric acid

(h) a water solution of sodium carbonate and dilute nitric acid

(i) solid sodium hydroxide and dilute hydrochloric acid

(j) magnesium oxide and dilute hydrochloric acid

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6. Write ionic equations for the reaction of magnesium carbonate with each of theacids - hydrochloric acid, nitric acid, sulfuric acid, hydrobromic acid, hydriodicacid. What do you notice about the equations? Explain your observation.

7. What is a "salt"? Which of the following compounds are salts and how did youdecide in each case?

3 4 3 2 3 3 2 2 2 4 2 4 2 3Na PO , SO , BaBr , NH , Mg(NO ) , CO , H O, AgCl, CCl , H SO , H S, HNO

8. To be done as a group: Write ionic equations for any reaction which occurs inthe following cases:

(a) Solutions of barium chloride and copper(II) sulfate are mixed.

(b) Solutions of sodium carbonate and calcium nitrate are mixed.

(c) Solutions of hydrochloric acid and lead(II) nitrate are mixed.

(d) Solutions of nitric acid and iron(III) chloride are mixed.

(e) Solutions of sodium phosphate and zinc nitrate are mixed.

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(f) Solutions of potassium iodide and silver(I) nitrate are mixed.

(g) Solutions of barium hydroxide and strontium chloride are mixed.

(h) Solutions of sodium phosphate and iron(III) sulfate are mixed.

(i) Solutions of potassium carbonate and nickel(II) bromide are mixed.

(j) A solution of nitric acid is added to calcium carbonate. [NB calcium carbonate is a solid.]

(k) A solution of hydrochloric acid is added to magnesium metal.

(l) A solution of sulfuric acid is added to zinc oxide.

(m) A solution of nitric acid is added to calcium hydroxide.

(n) A solution of hydrobromic acid is added to a solution of silver(I) nitrate.

(o) A solution of sulfuric acid is added to a solution of lead(II) nitrate.

(p) A solution of sulfuric acid is added to a solution of potassium carbonate.

(q) A solution of hydrochloric acid is added to solid potassium carbonate.

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ANSWERS TO TUTORIAL TOPIC 6

1. Solvent: Substance in which another substance (solute) dissolves, thecombination being termed a solution. Solvents are usually but not necessarilyliquids.

Solution: A homogeneous combination of two or more substances which retaintheir separate identities.

Precipitate: A solid formed in a chemical reaction.

Filtrate: The liquid phase that remains after any solids previously present havebeen removed, for example by filtering.

Formula equation: A representation of a chemical reaction in which all thereactants and products are represented by their entire formulas.

Ionic equation: A representation of a chemical reaction in which any specieswhich are present as ions are shown as such. Only the species actually reactingare shown and any other ions are called spectator ions because they play no partin the actual reaction.

Volatile substance: Element or compound that can be readily converted to thegaseous state.

2 4 3 3 3 3 2 2 42. Formulas: CoCl ; MgSO ; PbCO ; SrSO ; Al(NO ) ; Ag C O ;

3 2 2 2 2 3 2 2 3Sn(CH CO ) ; FeBr ; NaNO ; FeBr ; NiI ; Li CO .

Names: copper(II) bromide; barium sulfate; ammonium iodide;

manganese(II) nitrate; silver(I) fluoride; sodium chloride;

iron(II) phosphate; iron(III) phosphate; zinc carbonate;

potassium carbonate; calcium acetate; zinc hydrogencarbonate;

lithium hydroxide; copper(II) sulfate; lead(II) nitrate;

lead(II) nitrite; lithium sulfite.

Note the use of Roman numerals in conjunction with those cations that canoccur with more than one cationic charge, e.g. copper

4 43. CoSO (s) v Co (aq) + SO (aq)2+ 2–

4 3 4 3NH NO (s) v NH (aq) + NO (aq)+ –

3FeCl (s) v Fe (aq) + 3Cl (aq) (note 3Cl required to balance)3+ – –

3 4 4Na PO (s) v 3Na (aq) + PO (aq) (note 3Na required to balance)+ 3– +

3 3 3 3Al(NO ) (s) v Al (aq) + 3NO (aq) (note 3NO required to balance)3+ – –

2 3 3K CO (s) v 2K (aq) + CO (aq) (note 2K required to balance)+ 2– +

2FeBr (s) v Fe (aq) + 2Br (aq) (note 2Br required to balance)2+ – –

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4. Make sure that you have given the correct number of each ion on both sides ofthe equation for any compound that contains more than one of a given ion - allequations must balance. Pay special attention to the correct balancing of theequation in parts (b), (f), (g), (h), (j) and (k) - the same numbers of each of theions must appear on both sides of the equation.

4 4(a) Ba (aq) + SO (aq) v BaSO (s)2+ 2–

2(b) Cu (aq) + 2OH (aq) v Cu(OH) (s)2+ –

(c) Ag (aq) + Cl (aq) v AgCl(s)+ –

(d) Ag (aq) + Cl (aq) v AgCl(s)+ –

(e) Ag (aq) + Cl (aq) v AgCl(s)+ –

2(f) Pb (aq) + 2Cl (aq) v PbCl (s)2+ –

2(g) Pb (aq) + 2Cl (aq) v PbCl (s) 2+ –

3(h) Fe (aq) + 3OH (aq) v Fe(OH) (s)3+ –

3 3(i) Co (aq) + CO (aq) v CoCO (s)2+ 2–

4 3 4 2(j) 3Mg (aq) + 2PO (aq) v Mg (PO ) (s)2+ 3–

2(k) Pb (aq) + 2Br (aq) v PbBr (s)2+ –

(l) No reaction

(m) No reaction

4 4(n) Al (aq) + PO (aq) v AlPO (s)3+ 3–

2 3 2 25. (a) Na CO (s) + 2H (aq) v 2Na (aq) + CO (g) + H O+ +

3 2 2(b) CuCO (s) + 2H (aq) v Cu (aq) + CO (g) + H O+ 2+

3 2 2(c) CuCO (s) + 2H (aq) v Cu (aq) + CO (g) + H O+ 2+

2 3 2 2(d) Ag CO (s) + 2H (aq) v 2Ag (aq) + CO (g) + H O+ +

2 2(e) Fe(OH) (s) + 2H (aq) v Fe (aq) + 2H O+ 2+

2(f) NaOH(s) + H (aq) v Na (aq) + H O+ +

2(g) OH (aq) + H (aq) v H O– +

3 2 2(h) CO (aq) + 2H (aq) v CO (g) + H O2– +

2(i) NaOH(s) + H (aq) v Na (aq) + H O+ +

2(j) MgO(s) + 2H (aq) v Mg (aq) + H O+ 2+

Particularly check that you have the correct balance in parts (a) - (f) and (j).

VI - 19

3 2 26. MgCO (s) + 2H (aq) v Mg (aq) + CO (g) + H O(l)+ 2+

3 2 2MgCO (s) + 2H (aq) v Mg (aq) + CO (g) + H O(l)+ 2+

3 2 2MgCO (s) + 2H (aq) v Mg (aq) + CO (g) + H O(l)+ 2+

3 2 2MgCO (s) + 2H (aq) v Mg (aq) + CO (g) + H O(l)+ 2+

3 2 2MgCO (s) + 2H (aq) v Mg (aq) + CO (g) + H O(l)+ 2+

The equations are identical because the reaction does not involve the chloride,nitrate, sulfate, bromide or iodide ions which are merely spectator ions in eachcase.

7. A salt is any ionic compound in which the cation is not H or the anion is not+

3 4 2 3 2OH . The compounds Na PO , BaBr , Mg(NO ) , AgCl are all salts–

because they meet these criteria. The other compounds are either covalently

3 3 2 2 4 2bonded compounds involving only non-metals (SO , NH , CO , H O, CCl , H S)

3 2 4or acids (HNO , H SO ).

8. (a) Consider the formulas of the two possible compounds that might form, 4 2BaSO and CuCl . Only barium sulfate is insoluble, so it will precipitate while

copper(II) ions and chloride ions remain in solution.

4 4Ba (aq) + SO (aq) v BaSO (s)2+ 2–

3 3(b) The possible compounds have the formulas CaCO and NaNO . Onlycalcium carbonate is insoluble, so it precipitates while the sodium and nitrateions remain in solution.

3 3Ca (aq) + CO (aq) v CaCO (s)2+ 2–

(c) The hydrochloric acid provides H (aq) and Cl (aq) ions in the solution. The+ –

combination of Pb and Cl ions results in the formation of lead(II) chloride2+ –

precipitate. The H (aq) and nitrate ions remain in solution.+

2Pb (aq) + 2Cl (aq) v PbCl (s)2+ –

3(d) The ions in the mixture are H (aq), NO , Fe and Cl . No combination of+ – 3+ –

these ions would lead to an insoluble compound or any other reaction.

No reaction

(e) The combination of zinc ions and phosphate ions results in an insolublecompound, zinc phosphate, being precipitated while the sodium and nitrate ionsremain in solution.

4 3 4 23Zn (aq) + 2PO (aq) v Zn (PO ) (s)2+ 3–

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(f) The combination of silver(I) ions and iodide ions results in an insolublecompound, silver(I) iodide, being precipitated while the potassium and nitrateions remain in solution.

Ag (aq) + I (aq) v AgI(s)+ –

(g) The combination of strontium ions and hydroxide ions results in an insoluble compound, strontium hydroxide, beingprecipitated while the barium and chloride ions remain in solution.

2 Sr (aq) + 2OH (aq) v Sr(OH) (s)2+ –

(h) The combination of iron(III) ions and phosphate ions results in an insolublecompound, iron(III) phosphate, being precipitated while the sodium and sulfateions remain in solution.

4 4 Fe (aq) + PO (aq) v FePO (s)3+ 3–

(i) The combination of nickel(II) ions and carbonate ions results in an insolublecompound, nickel(II) carbonate, being precipitated while the potassium andbromide ions remain in solution.

3 3 Ni (aq) + CO (aq) v NiCO (s)2+ 2–

(j) An acid reacts with a carbonate to form carbon dioxide gas, water and a saltconsisting of the cation from the carbonate compound (Ca in this case) and the2+

3anion from the acid used (NO in this case). The salt in this example will–

remain in solution as separate ions because calcium nitrate is soluble.

3 2 2 CaCO (s) + 2H (aq) v Ca (aq) + CO (g) + H O(l)+ 2+

(k) An acid reacts with reactive metals to form hydrogen gas and a saltconsisting of the cation from the metal element (Mg in this case) and the anion2+

from the acid used (Cl in this case). The salt in this example will remain in–

solution as separate ions because magnesium chloride is soluble.

2Mg(s) + 2H (aq) v Mg (aq) + H (g)+ 2+

(l) An acid reacts with an oxide to form water and a salt consisting of the cation

4from the oxide (Zn in this case) and the anion from the acid used (SO in this2+ 2–

case). The salt in this example will remain in solution as separate ions becausezinc sulfate is soluble.

2ZnO(s) + 2H (aq) v Zn (aq) + H O(l)+ 2+

VI - 21

(m) An acid reacts with a hydroxide to form water and a salt consisting of thecation from the hydroxide (Ca in this case) and the anion from the acid used2+

3(NO in this case). The salt in this example will remain in solution as separate–

ions because calcium nitrate is soluble.

2 2 Ca(OH) (s) + 2H (aq) v Ca (aq) + 2H O(l)+ 2+

(n) The H (aq) from the hydrobromic acid won’t react with the nitrate ion, but+

the anion, Br , will combine with the silver(I) ions from silver(I) nitrate to–

3precipitate the insoluble compound silver(I) bromide. The H (aq) and NO ions+ –

remain in solution as nitric acid.

Ag (aq) + Br (aq) v AgBr(s)+ –

(o) The H (aq) from the sulfuric acid won’t react with the nitrate ion, but the+

4anion, SO , will combine with the lead(II) ions from lead(II) nitrate to2–

3precipitate the insoluble compound lead(II) sulfate. The H (aq) and NO ions+ –

remain in solution as nitric acid.

4 4Pb (aq) + SO (aq) v PbSO (s)2+ 2–

(p) An acid reacts with a carbonate to form carbon dioxide gas, water and a saltconsisting of the cation from the carbonate compound (K in this case) and the+

4anion from the acid used (SO in this case). The salt resulting in this example2–

will remain in solution as separate ions because potassium sulfate is soluble. Note that as the potassium carbonate reactant was specified as a solution, it is

2 3therefore initially present already as separate ions and so the solid, K CO , is not

3used in the equation. Instead, just the CO ion is shown without the spectator2–

ion, K .+

3 2 22H (aq) + CO (aq) v CO (g) + H O(l)+ 2–

(q) An acid reacts with a carbonate to form carbon dioxide gas, water and a saltconsisting of the cation from the carbonate compound (K in this case) and the+

anion from the acid used (Cl in this case). The salt in this example will remain–

in solution as separate ions because potassium chloride is soluble. Compare thiswith the previous example - here the potassium carbonate is specified as beingthe solid so it will be written as such in the equation.

2 3 2 2K CO (s) + 2H (aq) v 2K (aq) + CO (g) + H O(l)+ +