Molecular full and abbreviated ionic form. Electrolyte solutions

When dissolved in water, not all substances have the ability to conduct electricity. Those compounds, water solutions which are capable of conducting electric current are called electrolytes. Electrolytes conduct current due to the so-called ionic conductivity, which many compounds with an ionic structure (salts, acids, bases) have. There are substances that have strongly polar bonds, but in solution they undergo incomplete ionization (for example, mercury chloride II) - these are weak electrolytes. Many organic compounds (carbohydrates, alcohols) dissolved in water do not decompose into ions, but retain their molecular structure. Such substances do not conduct electricity and are called non-electrolytes.

Here are some regularities, guided by which it is possible to determine whether one or another compound belongs to strong or weak electrolytes:

acids . Among the most common strong acids are HCl, HBr, HI, HNO 3 , H 2 SO 4 , HClO 4 . Almost all other acids are weak electrolytes.
Foundations. The most common strong bases are hydroxides of alkali and alkaline earth metals (excluding Be). Weak electrolyte - NH 3.
Salt. Most common salts - ionic compounds - are strong electrolytes. The exceptions are mainly salts of heavy metals.

Theory of electrolytic dissociation

Electrolytes, both strong and weak, and even very dilute ones, do not obey Raoult's law and . Having the ability to conduct electricity, the vapor pressure of the solvent and the melting point of electrolyte solutions will be lower, and the boiling point will be higher compared to the same values of a pure solvent. In 1887, S. Arrhenius, studying these deviations, came to the creation of a theory of electrolytic dissociation.

Electrolytic dissociation assumes that the electrolyte molecules in solution decompose into positively and negatively charged ions, which are called cations and anions, respectively.

The theory puts forward the following postulates:

In solutions, electrolytes decompose into ions, i.e. dissociate. The more dilute the electrolyte solution, the greater its degree of dissociation.
Dissociation is a reversible and equilibrium phenomenon.
Solvent molecules interact infinitely weakly (i.e., solutions are close to ideal).

Different electrolytes have different degrees of dissociation, which depends not only on the nature of the electrolyte itself, but also on the nature of the solvent, as well as electrolyte concentration and temperature.

Degree of dissociation α , shows how many molecules n decayed into ions, compared to the total number of dissolved molecules N:

α = n/N

In the absence of dissociation, α = 0, with complete dissociation of the electrolyte, α = 1.

From the point of view of the degree of dissociation, according to strength, electrolytes are divided into strong (α> 0.7), medium strength (0.3> α> 0.7), weak (α< 0,3).

More precisely, the process of electrolyte dissociation characterizes dissociation constant, independent of the concentration of the solution. If we present the process of electrolyte dissociation in a general form:

A a B b ↔ aA — + bB +

K = a b /

For weak electrolytes the concentration of each ion is equal to the product of α by the total concentration of electrolyte C, so the expression for the dissociation constant can be converted:

K = α 2 C/(1-α)

For dilute solutions(1-α) =1, then

K = α 2 C

From here it's easy to find degree of dissociation

Ionic–molecular equations

Consider an example of the neutralization of a strong acid by a strong base, for example:

HCl + NaOH = NaCl + HOH

The process is presented in the form molecular equation. It is known that both the starting materials and the reaction products are completely ionized in solution. Therefore, we represent the process in the form complete ionic equation:

H + + Cl - + Na + + OH - = Na + + Cl - + HOH

After the "reduction" of identical ions in the left and right parts of the equation, we obtain reduced ionic equation:

H + + OH - = HOH

We see that the process of neutralization comes down to the combination of H + and OH - and the formation of water.

When compiling ionic equations, it should be remembered that only strong electrolytes are written in ionic form. Weak electrolytes, solids and gases are written in their molecular form.

The precipitation process is reduced to the interaction of only Ag + and I - and the formation of water-insoluble AgI.

To find out whether the substance of interest to us is capable of solubility in water, it is necessary to use the insolubility table.

Let us consider the third type of reactions, as a result of which a volatile compound is formed. These are reactions of interaction of carbonates, sulfites or sulfides with acids. For example,

When mixing some solutions of ionic compounds, the interaction between them may not occur, for example

So, to summarize, we note that chemical transformations occur when one of the following conditions is met:

Non-electrolyte formation. Water can act as a non-electrolyte.
Sediment formation.
Gas release.
The formation of a weak electrolyte, such as acetic acid.
Transfer of one or more electrons. This is realized in redox reactions.
The formation or rupture of one or more

Categories ,

Balance the complete molecular equation. Before writing the ionic equation, the original molecular equation must be balanced. To do this, it is necessary to place the appropriate coefficients in front of the compounds, so that the number of atoms of each element on the left side is equal to their number on the right side of the equation.

Write down the number of atoms of each element on both sides of the equation.
Add coefficients in front of the elements (except oxygen and hydrogen) so that the number of atoms of each element on the left and right side of the equation is the same.
Balance the hydrogen atoms.
Balance the oxygen atoms.
Count the number of atoms of each element on both sides of the equation and make sure it's the same.
For example, after balancing the equation Cr + NiCl 2 --> CrCl 3 + Ni, we get 2Cr + 3NiCl 2 --> 2CrCl 3 + 3Ni.

Determine the state of each substance that participates in the reaction. Often this can be judged by the condition of the problem. There are certain rules that help determine what state an element or connection is in.

Determine which compounds dissociate (separate into cations and anions) in solution. During dissociation, the compound decomposes into positive (cation) and negative (anion) components. These components will then enter the ionic equation of the chemical reaction.

Calculate the charge of each dissociated ion. When doing this, remember that metals form positively charged cations, and non-metal atoms turn into negative anions. Determine the charges of the elements according to the periodic table. It is also necessary to balance all charges in neutral compounds.

Rewrite the equation so that all soluble compounds are separated into individual ions. Anything that dissociates or ionizes (such as strong acids) will split into two separate ions. In this case, the substance will remain in a dissolved state ( rr). Check that the equation is balanced.

Solids, liquids, gases, weak acids and ionic compounds with low solubility will not change their state and will not separate into ions. Leave them as they are.
Molecular compounds will simply dissipate in solution, and their state will change to dissolved ( rr). There are three molecular compounds that not go to state ( rr), this is CH 4( G), C 3 H 8( G) and C 8 H 18( and) .
For the reaction under consideration, the complete ionic equation can be written in the following form: 2Cr ( tv) + 3Ni 2+ ( rr) + 6Cl - ( rr) --> 2Cr 3+ ( rr) + 6Cl - ( rr) + 3Ni ( tv) . If chlorine is not part of the compound, it breaks down into individual atoms, so we multiply the number of Cl ions by 6 on both sides of the equation.

Cancel the same ions on the left and right side of the equation. You can cross out only those ions that are completely identical on both sides of the equation (have the same charges, subscripts, and so on). Rewrite the equation without these ions.

In our example, both sides of the equation contain 6 Cl - ions, which can be crossed out. Thus, we obtain a short ionic equation: 2Cr ( tv) + 3Ni 2+ ( rr) --> 2Cr 3+ ( rr) + 3Ni ( tv) .
Check the result. The total charges of the left and right sides of the ionic equation must be equal.

When any strong acid is neutralized with any strong base, about heat is released for each mole of water formed:

This suggests that such reactions are reduced to one process. We will obtain the equation of this process if we consider in more detail one of the above reactions, for example, the first one. We rewrite its equation, writing strong electrolytes in ionic form, since they exist in solution in the form of ions, and weak electrolytes in molecular form, since they are in solution mainly in the form of molecules (water is a very weak electrolyte, see § 90):

Considering the resulting equation, we see that during the reaction, the ions and did not change. Therefore, we rewrite the equation again, excluding these ions from both sides of the equation. We get:

Thus, the reactions of neutralization of any strong acid with any strong base are reduced to the same process - to the formation of water molecules from hydrogen ions and hydroxide ions. It is clear that the thermal effects of these reactions must also be the same.

Strictly speaking, the reaction of formation of water from ions is reversible, which can be expressed by the equation

However, as we shall see below, water is a very weak electrolyte and dissociates only to a negligible degree. In other words, the equilibrium between water molecules and ions is strongly shifted towards the formation of molecules. Therefore, in practice, the reaction of neutralization of a strong acid with a strong base proceeds to the end.

When mixing a solution of any silver salt with hydrochloric acid or with a solution of any of its salts, a characteristic white cheesy precipitate of silver chloride is always formed:

Similar reactions are also reduced to one process. In order to obtain its ionic-molecular equation, we rewrite, for example, the equation of the first reaction, writing strong electrolytes, as in the previous example, in ionic form, and the substance in the precipitate in molecular form:

As can be seen, the ions and do not undergo changes during the reaction. Therefore, we eliminate them and rewrite the equation again:

This is the ion-molecular equation of the process under consideration.

Here it must also be borne in mind that the silver chloride precipitate is in equilibrium with ions and in solution, so that the process expressed by the last equation is reversible:

However, due to the low solubility of silver chloride, this equilibrium is very strongly shifted to the right. Therefore, we can assume that the reaction of formation from ions practically comes to an end.

The formation of a precipitate will always be observed when ions and are in a significant concentration in one solution. Therefore, with the help of silver ions, it is possible to detect the presence of ions in a solution and, conversely, with the help of chloride ions, the presence of silver ions; an ion can serve as a reactant for an ion, and an ion as a reactant for an ion.

In the future, we will widely use the ion-molecular form of writing the equations of reactions involving electrolytes.

To draw up ion-molecular equations, you need to know which salts are soluble in water and which are practically insoluble. The general characteristics of the solubility in water of the most important salts are given in Table. fifteen.

Table 15. Solubility of the most important salts in water

Ionic-molecular equations help to understand the features of reactions between electrolytes. Consider, as an example, several reactions involving weak acids and bases.

As already mentioned, the neutralization of any strong acid by any strong base is accompanied by the same thermal effect, since it is reduced to the same process - the formation of water molecules from hydrogen ions and hydroxide ions.

However, when a strong acid is neutralized with a weak base, a weak acid with a strong or weak base, the thermal effects are different. Let us write the ion-molecular equations for such reactions.

Neutralization of a weak acid (acetic acid) with a strong base (sodium hydroxide):

Here, the strong electrolytes are sodium hydroxide and the resulting salt, and the weak ones are acid and water:

As can be seen, only sodium ions do not undergo changes during the reaction. Therefore, the ion-molecular equation has the form:

Neutralization of a strong acid (nitric acid) with a weak base (ammonium hydroxide):

Here, in the form of ions, we must write the acid and the resulting salt, and in the form of molecules, ammonium hydroxide and water:

Ions do not undergo changes. Omitting them, we obtain the ion-molecular equation:

Neutralization of a weak acid (acetic acid) with a weak base (ammonium hydroxide):

In this reaction, all substances, except for the resulting weak electrolytes. Therefore, the ion-molecular form of the equation has the form:

Comparing the obtained ion-molecular equations, we see that they are all different. Therefore, it is clear that the heats of the considered reactions are not the same.

As already mentioned, the reactions of neutralization of strong acids with strong bases, during which hydrogen ions and hydroxide ions combine into a water molecule, proceed almost to the end. Neutralization reactions, on the other hand, in which at least one of the starting substances is a weak electrolyte and in which molecules of weakly associated substances are present not only on the right, but also on the left side of the ion-molecular equation, do not proceed to the end.

They reach a state of equilibrium in which the salt coexists with the acid and base from which it is derived. Therefore, it is more correct to write the equations of such reactions as reversible reactions.

1. Write down the formulas of the substances that have reacted, put an "equal" sign and write down the formulas of the formed substances. Set up coefficients.

2. Using the solubility table, write down in ionic form the formulas of substances (salts, acids, bases) indicated in the solubility table by the letter “P” (highly soluble in water), the exception is calcium hydroxide, which, although indicated by the letter “M”, nevertheless, in an aqueous solution, it dissociates well into ions.

3. It must be remembered that metals, oxides of metals and non-metals, water, gaseous substances, water-insoluble compounds, indicated in the solubility table with the letter “H”, do not decompose into ions. The formulas of these substances are written in molecular form. Get the full ionic equation.

4. Reduce identical ions before and after the equal sign in the equation. Get the reduced ionic equation.

5. Remember!

P - soluble substance;

M - poorly soluble substance;

TP - table of solubility.

Algorithm for compiling ion exchange reactions (RIO)

in molecular, full and short ionic form

Examples of compiling ion exchange reactions

1. If as a result of the reaction a low-dissociating (md) substance is released - water.

In this case, the full ionic equation is the same as the reduced ionic equation.

2. If a water-insoluble substance is released as a result of the reaction.

In this case, the full ionic reaction equation coincides with the reduced one. This reaction proceeds to the end, as evidenced by two facts at once: the formation of a substance insoluble in water, and the release of water.

3. If a gaseous substance is released as a result of the reaction.

COMPLETE THE TASKS ON THE TOPIC "ION EXCHANGE REACTIONS"

Task number 1.
Determine whether the interaction between solutions of the following substances can be carried out, write down the reactions in molecular, full, short ionic form:
potassium hydroxide and ammonium chloride.

Solution

We compose chemical formulas of substances by their names, using valences and write RIO in molecular form (we check the solubility of substances according to TR):

KOH + NH4 Cl = KCl + NH4 OH

since NH4 OH is an unstable substance and decomposes into water and gas NH3, the RIO equation will take the final form

KOH (p) + NH4 Cl (p) = KCl (p) + NH3 + H2 O

We compose the full RIO ionic equation using TR (do not forget to write down the charge of the ion in the upper right corner):

K+ + OH- + NH4 + + Cl- = K+ + Cl- + NH3 + H2 O

We compose a short RIO ionic equation, deleting the same ions before and after the reaction:

Oh - +NH 4 + =NH 3 + H2O

We conclude:
The interaction between solutions of the following substances can be carried out, since the products of this RIO are gas (NH3) and a low-dissociating substance water (H2 O).

Task number 2

Given scheme:

2H + + CO 3 2- = H2 O+CO2

Select substances, the interaction between which in aqueous solutions is expressed by the following abbreviated equations. Write the corresponding molecular and full ionic equations.

Using TR, we select reagents - water-soluble substances containing 2H ions + and CO3 2- .

For example, acid - H 3 PO4 (p) and salt -K2 CO3 (p).

We compose the RIO molecular equation:

2H 3 PO4 (p) +3 K2 CO3 (p) -> 2K3 PO4 (p) + 3H2 CO3 (p)

Since carbonic acid is an unstable substance, it decomposes into carbon dioxide CO 2 and water H2 O, the equation will take the final form:

2H 3 PO4 (p) +3 K2 CO3 (p) -> 2K3 PO4 (p) + 3CO2 + 3H2 O

We compose the full RIO ionic equation:

6H + +2PO4 3- + 6K+ + 3CO3 2- -> 6K+ + 2PO4 3- + 3CO2 + 3H2 O

We compose a short RIO ionic equation:

6H + +3CO3 2- = 3CO2 + 3H2 O

2H + +CO3 2- = CO2 + H2 O

We conclude:

In the end, we got the desired reduced ionic equation, therefore, the task was completed correctly.

Task number 3

Write down the exchange reaction between sodium oxide and phosphoric acid in molecular, full and short ionic form.

1. We compose a molecular equation, when compiling formulas, we take into account valencies (see TR)

3Na 2 O (ne) + 2H3 PO4 (p) -> 2Na3 PO4 (p) + 3H2 O (md)

where ne is a non-electrolyte, does not dissociate into ions,
md - a low-dissociating substance, we do not decompose into ions, water is a sign of the irreversibility of the reaction

2. We compose a complete ionic equation:

3Na 2 O+6H+ + 2PO4 3- -> 6Na+ + 2PO 4 3- + 3H2 O

3. We cancel the same ions and get a short ionic equation:

3Na 2 O+6H+ -> 6Na+ + 3H2 O
We reduce the coefficients by three and get:
Na2 O+2H+ -> 2Na+ + H2 O

This reaction is irreversible, i.e. goes to the end, since a low-dissociating substance water is formed in the products.

TASKS FOR INDEPENDENT WORK

Task number 1

Reaction between sodium carbonate and sulfuric acid

Write an equation for the reaction of ion exchange of sodium carbonate with sulfuric acid in molecular, full and short ionic form.

Task number 2

ZnF 2 + Ca(OH)2 ->
K2 S+H3 PO4 ->

Task number 3

Check out the following experiment

Precipitation of barium sulfate

Write an equation for the reaction of ion exchange of barium chloride with magnesium sulfate in molecular, full and short ionic form.

Task number 4

Complete the reaction equations in molecular, full and short ionic form:

Hg(NO 3 ) 2 + Na2 S ->
K2 SO3 + HCl ->

When completing the task, use the table of the solubility of substances in water. Remember about exceptions!

Quite often, schoolchildren and students have to make up the so-called. ionic reaction equations. In particular, problem 31, proposed at the Unified State Examination in Chemistry, is devoted to this topic. In this article, we will discuss in detail the algorithm for writing short and complete ionic equations, we will analyze many examples of different levels of complexity.

Why ionic equations are needed

Let me remind you that when many substances are dissolved in water (and not only in water!) A process of dissociation occurs - substances break up into ions. For example, HCl molecules in an aqueous medium dissociate into hydrogen cations (H + , more precisely, H 3 O +) and chlorine anions (Cl -). Sodium bromide (NaBr) is in an aqueous solution not in the form of molecules, but in the form of hydrated Na + and Br - ions (by the way, ions are also present in solid sodium bromide).

When writing the "ordinary" (molecular) equations, we do not take into account that not molecules enter into the reaction, but ions. Here, for example, is the equation for the reaction between hydrochloric acid and sodium hydroxide:

HCl + NaOH = NaCl + H 2 O. (1)

Of course, this diagram does not quite correctly describe the process. As we have already said, there are practically no HCl molecules in an aqueous solution, but there are H + and Cl - ions. The same is true for NaOH. It would be better to write the following:

H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O. (2)

That's what it is complete ionic equation. Instead of "virtual" molecules, we see particles that are actually present in the solution (cations and anions). We will not dwell on the question why we have written H 2 O in molecular form. This will be explained a little later. As you can see, there is nothing complicated: we have replaced the molecules with ions, which are formed during their dissociation.

However, even the complete ionic equation is not perfect. Indeed, take a closer look: both in the left and in the right parts of equation (2) there are identical particles - Na + cations and Cl - anions. These ions do not change during the reaction. Why then are they needed at all? Let's remove them and get short ionic equation:

H + + OH - = H 2 O. (3)

As you can see, it all comes down to the interaction of H + and OH - ions with the formation of water (neutralization reaction).

All complete and short ionic equations are written down. If we solved problem 31 at the exam in chemistry, we would get the maximum mark for it - 2 points.

So, once again about the terminology:

HCl + NaOH = NaCl + H 2 O - molecular equation ("usual" equation, schematically reflecting the essence of the reaction);
H + + Cl - + Na + + OH - = Na + + Cl - + H 2 O - complete ionic equation (real particles in solution are visible);
H + + OH - = H 2 O - a short ionic equation (we removed all the "garbage" - particles that do not participate in the process).

Algorithm for writing ionic equations

We compose the molecular equation of the reaction.
All particles that dissociate in solution to a noticeable degree are written as ions; substances that are not prone to dissociation, we leave "in the form of molecules."
We remove from the two parts of the equation the so-called. observer ions, i.e., particles that do not participate in the process.
We check the coefficients and get the final answer - a short ionic equation.

Example 1. Write a complete and short ionic equation describing the interaction of aqueous solutions of barium chloride and sodium sulfate.

Solution. We will act in accordance with the proposed algorithm. Let's set up the molecular equation first. Barium chloride and sodium sulfate are two salts. Let's look at the section of the reference book "Properties of inorganic compounds". We see that salts can interact with each other if a precipitate forms during the reaction. Let's check:

Exercise 2. Complete the equations for the following reactions:

KOH + H 2 SO 4 \u003d
H 3 PO 4 + Na 2 O \u003d
Ba(OH) 2 + CO 2 =
NaOH + CuBr 2 =
K 2 S + Hg (NO 3) 2 \u003d
Zn + FeCl 2 =

Exercise 3. Write the molecular equations for the reactions (in aqueous solution) between: a) sodium carbonate and nitric acid, b) nickel (II) chloride and sodium hydroxide, c) orthophosphoric acid and calcium hydroxide, d) silver nitrate and potassium chloride, e) phosphorus oxide (V) and potassium hydroxide.

I sincerely hope that you had no problems completing these three tasks. If this is not so, it is necessary to return to the topic "Chemical properties of the main classes of inorganic compounds".

How to turn a molecular equation into a complete ionic equation

The most interesting begins. We must understand which substances should be written as ions and which should be left in "molecular form". You have to remember the following.

In the form of ions write:

soluble salts (I emphasize that only salts are highly soluble in water);
alkalis (let me remind you that water-soluble bases are called alkalis, but not NH 4 OH);
strong acids (H 2 SO 4 , HNO 3 , HCl, HBr, HI, HClO 4 , HClO 3 , H 2 SeO 4 , ...).

As you can see, this list is not difficult to remember: it includes strong acids and bases and all soluble salts. By the way, to especially vigilant young chemists who may be outraged by the fact that strong electrolytes (insoluble salts) are not included in this list, I can tell you the following: NOT including insoluble salts in this list does not at all reject the fact that they are strong electrolytes.

All other substances must be present in the ionic equations in the form of molecules. For those demanding readers who are not satisfied with the vague term "all other substances", and who, following the example of the hero of a famous film, demand "announce the full list", I give the following information.

In the form of molecules, write:

all insoluble salts;
all weak bases (including insoluble hydroxides, NH 4 OH and similar substances);
all weak acids (H 2 CO 3 , HNO 2 , H 2 S, H 2 SiO 3 , HCN, HClO, almost all organic acids ...);
in general, all weak electrolytes (including water!!!);
oxides (all types);
all gaseous compounds (in particular H 2 , CO 2 , SO 2 , H 2 S, CO);
simple substances (metals and non-metals);
almost all organic compounds (with the exception of water-soluble salts of organic acids).

Phew, I don't think I forgot anything! Although it is easier, in my opinion, to remember list No. 1. Of the fundamentally important in list No. 2, I will once again note the water.

Let's train!

Example 2. Make a complete ionic equation describing the interaction of copper (II) hydroxide and hydrochloric acid.

Solution. Let's start, of course, with the molecular equation. Copper (II) hydroxide is an insoluble base. All insoluble bases react with strong acids to form a salt and water:

Cu(OH) 2 + 2HCl = CuCl 2 + 2H 2 O.

And now we find out which substances to write in the form of ions, and which - in the form of molecules. The lists above will help us. Copper (II) hydroxide is an insoluble base (see solubility table), a weak electrolyte. Insoluble bases are written in molecular form. HCl is a strong acid, in solution it almost completely dissociates into ions. CuCl 2 is a soluble salt. We write in ionic form. Water - only in the form of molecules! We get the full ionic equation:

Cu (OH) 2 + 2H + + 2Cl - \u003d Cu 2+ + 2Cl - + 2H 2 O.

Example 3. Write a complete ionic equation for the reaction of carbon dioxide with an aqueous solution of NaOH.

Solution. Carbon dioxide is a typical acidic oxide, NaOH is an alkali. When acidic oxides interact with aqueous solutions of alkalis, salt and water are formed. We compose the molecular reaction equation (do not forget, by the way, about the coefficients):

CO 2 + 2NaOH \u003d Na 2 CO 3 + H 2 O.

CO 2 - oxide, gaseous compound; keep the molecular shape. NaOH - strong base (alkali); written in the form of ions. Na 2 CO 3 - soluble salt; write in the form of ions. Water is a weak electrolyte, practically does not dissociate; leave it in molecular form. We get the following:

CO 2 + 2Na + + 2OH - \u003d Na 2+ + CO 3 2- + H 2 O.

Example 4. Sodium sulfide in aqueous solution reacts with zinc chloride to form a precipitate. Write the complete ionic equation for this reaction.

Solution. Sodium sulfide and zinc chloride are salts. When these salts interact, zinc sulfide precipitates:

Na 2 S + ZnCl 2 \u003d ZnS ↓ + 2NaCl.

I will immediately write down the full ionic equation, and you will analyze it yourself:

2Na + + S 2- + Zn 2+ + 2Cl - = ZnS↓ + 2Na + + 2Cl - .

I offer you several tasks for independent work and a small test.

Exercise 4. Write the molecular and full ionic equations for the following reactions:

NaOH + HNO3 =
H 2 SO 4 + MgO =
Ca(NO 3) 2 + Na 3 PO 4 =
CoBr 2 + Ca(OH) 2 =

Exercise 5. Write complete ionic equations describing the interaction of: a) nitric oxide (V) with an aqueous solution of barium hydroxide, b) a solution of cesium hydroxide with hydroiodic acid, c) aqueous solutions of copper sulfate and potassium sulfide, d) calcium hydroxide and an aqueous solution of iron nitrate ( III).