Exp. 12 -3

Data Sheet

Name

| | Chlorate | Perchlorate |
| :--- | :--- | :--- |
| 1. Wt of tube and contents | | |
| 2. Wt of tube empty | | |
| 3. Wt of sample | | |
| 4. Wt after heating | | |
| 5. Wt of oxygen evolved | | |
| 6. Wt of potassium chloride remaining | | |
| 7. Wt of oxygen associated with 1 g of potassium chloride | | |
| 8. Ratio* | | |
| 9. Wt of oxygen associated with 74.5 g of KCl (1 G.M./L.) | | |
| 10. Item 9 divided by 16 (at.wt. of oxygen) | | |

* Ratio = $\frac{\text{Wt of O}_2 \text{ associated with 1 g KCl in potassium chloride}}{\text{Wt of O}_2 \text{ associated with 1 g KCl in potassium perchlorate}}$

Lafayette College
Department of Chemistry

Experiment 13

TYPES OF CHEMICAL CHANGE

Object: To learn to recognize the types of chemical changes and to use the information thus gained.

Discussion:

In experiments previously performed you have observed many
chemical changes. In all of these cases you were interested in dis
covering the characteristics of a specific substance and very littl
attention was given to the nature of the change involved. In this
experiment you will recall some of these changes and classify them
according to the characteristics of the change itself.

The grouping of chemical changes into types, like any other
classification can be done on many different bases. First they will
be considered from the standpoint of the changes occurring in the
composition of the reactants and secondly from the standpoint of
the energy involved in the change.

The four types of chemical changes on the first basis mentione
above are: combination (synthesis), decomposition, displacement,
and double decomposition or metathesis. The first three of these
usually involve electron changes and have been illustrated by most
of the changes which you have studied. Metathesis never involves
electron changes but is merely a regrouping of ions which are alrea
present. Few of the changes which you have observed fall under this
latter class.

Combination: Characterized by involving fewer products than reactants. Electrons are transferred from the atoms of one element to those of another, or electrons are shared by the atoms of two elements.

$$
\mathrm{Na} + \mathrm{Cl} \longrightarrow \mathrm{Na}^{+} + \mathrm{Cl}^{-}
$$

$$
C + O_2 \rightarrow CO_2
$$

The above illustrations cover these two cases of combination. No-
tice that no molecules are formed in the first case. (1) Account
for the charge on the ions. Use diagramatic drawings. (2) Make a
drawing representing the carbon dioxide molecule. Use only the
outside electron rings of each atom.

Decomposition: Characterized by involving a larger number of products than reactants. This type of change is the reverse of combination and thus involves the regaining of electrons by ions or the breaking up of molecules into atoms in such a way that each atom retains its normal number of electrons. It is important to note that the products of a decomposition are not necessarily elements but may be simpler compounds.

Displacement: The number of reactants and products is the same. A free element replaces one of the elements of a compound. Changes of this type generally take place in solution and generally involve

Exp. 13 -2

the transfer of electrons from ions to atoms or vice versa. The following illustration will help in understanding this change.

$$Mg + 2H^+ + SO_4^{2-} \rightarrow Mg^{2+} + SO_4^{3-} + H_2$$

The magnesium atom has given up two electrons, one to each of the two hydrogen ions. Notice that the only action occurring is that between the magnesium atom and the hydrogen ions; the sulphate ion remains unchanged. (3) Explain, in terms of electron exchange, the ratio of ions and atoms involved. (4) What has been displaced? What displaced it?

Double decomposition or metathesis: This is another case in which the reactants and products are equal in number. Double decomposition reactions differ from displacement in that there is no transfer of electrons and thus no free elements are involved. The action takes place between the ions derived from two or more compounds. At least two of these ions combine to form a compound which differs from any of the original ones. The other ions may combine also, but in the majority of cases they remain in solution. The following is an example:

$$\begin{array}{c} \underline{\underline{\hspace{2cm}}} \\ \mathrm{Na}^+ + \mathrm{Cl}^- + \mathrm{Ag}^+ + \mathrm{NO}_3^- \rightarrow \mathrm{AgCl} + \mathrm{Na}^+ + \mathrm{NO}_3^- \\ \underline{\underline{\hspace{2cm}}} \quad \underline{\underline{\hspace{2cm}}} \quad \underline{\underline{\hspace{2cm}}} \quad \underline{\underline{\hspace{2cm}}} \\ 1 & 2 & 3 & 4 \end{array}$$

1. Ions derived from sodium chloride.

2. Ions derived from silver nitrate.

3. New compound formed by recombination of ions.

4. Ions remaining in solution.

Notice that the sodium and nitrate ions have undergone no change and, therefore, as far as the chemical change is concerned, could be omitted from the equation. The real chemical reaction took place between the silver ions and the chloride ions. This equation could be written in the following form:

$$\mathrm{Cl}^- + \mathrm{Ag}^+ \rightarrow \mathrm{AgCl}$$

One of the facts which you learned earlier in the course is that every chemical change is accompanied by a change in energy. On a basis of their energy changes then, chemical reactions are classified into two types: those that give off energy (exothermal) and those that take on or absorb energy (endothermal). This energy change is so characteristic of chemical changes that it was used as one of the criteria in determining whether or not a chemical change had occurred. The exothermic reaction is by far the most common in the layman's environment. Indeed, many of the reactions carried out in everyday life are useful because they are exothermal. It is the evolved energy which is the useful product and not the new form of matter. Thus coal is burned for the heat evolved, and not the ashes and smoke; the reaction which occurs in the automobile battery is useful because it evolves electrical energy and not because it produces lead sulphate.

Exp. 13 -3

A quantitative study of the energy changes which accompany chemical changes brings out some rather interesting relationships. Most of these quantitative relationships are too involved to be discussed fully here, but it is possible to make a few semi-quantitative generalizations on this topic:

a) The same amount of energy is evolved by the same reaction whether the change takes place slowly or rapidly. A ball of steel wool which is heated in the bunsen flame combines with oxygen, and enough energy is liberated to keep that material red hot for some time after it is removed from the flame. If the same weight of steel wool were exposed to the atmosphere and combined with oxygen to form the same compound as above, the same amount of energy would be evolved. (5) Explain why rusting iron does not feel hot.

b) When a gram molecular weight of a compound is formed from its elements, the quantity of energy involved in known as the heat of formation. The heat of formation is characteristic of the compound and may be positive or negative.

c) Those reactions which give off energy (exothermal) tend to take place, and the quantity of energy given off is indicative, but not necessarily a quantitative measure of this tendency.

The last two generalizations are helpful in understanding the stability of compounds. If the reaction

$$2H_2 + O_2 \rightarrow 2H_2O$$

is carried out, it is found that 136,700 calories of heat are evolved when two gram molecular weights of water are formed, thus making this reaction a highly exothermal one. According to generalization (b), this reaction would take place readily. Now if water is to be decomposed, the reaction which must occur is the exact opposite of the above, and therefore it would absorb 136,700 calories. This would make the reaction highly endothermal. How would you classify water with regard to its stability (resistance to decomposition)?

### Directions:

(6) Review the directions of each experiment performed to date and write a word equation for each chemical change observed. Classify each change with regard to the four types mentioned above. Drop a piece of zinc into 5 ml of copper sulphate solution. Observe any change which occurs. Do you notice any signs of an energy change? The copper sulphate is composed of copper ions and sulphate ions. (7) Write an equation indicating the change which has occurred. Describe any electron change and name the particles which are formed as a result of such changes. (8) What type of reaction is this with regard to products; to energy?

Add about 2 ml of barium chloride solution to 2 ml of magnesium sulphate solution. The white product formed is insoluble barium sulphate. (9) What ions were present in each of the two solutions? What remains in the clear liquid after the barium sulphate settles? (10) Were electrons exchanged during this chemical

Exp. 13 -4

change? Classify it. Repeat the above directions substituting solutions of lead nitrate and potassium chromate for those formerly used. The yellow solid is lead chromate. (11) Apply questions 9 and 10 to this second change.

Questions:

I. From the information gained in experiment 6, what conclusions could you draw regarding the heat of formation of iron oxide, silicon dioxide (sand), manganese dioxide? What regarding barium peroxide, mercuric oxide? Explain.

II. In experiment 4, when you heated iron and sulphur together, you were told to observe the fact that the contents of the tube continued to glow for some time after the heat was removed. What is the significance of this observation with regard to the energy involved? With regard to the stability of ferrous sulphide?

III. Describe the electron changes which take place in any displacement reaction. Make a general statement regarding the tendencies of displaced and displacing elements to give up and take on electrons. Is there any correlation between this tendency and the arrangement in the electromotive series?

Experiment 14

RECRystALLIzATION AND EXTRACTION

Object: To learn to purify substances by fractional recrystallization and liquid-liquid extraction.

Discussion:

Many solids have a rather large temperature coefficient of solubility, i.e., their solubility increases markedly with an increase in temperature. This important property enables us to prepare a saturated solution at high temperature and then on cooling to obtain a large fraction of the dissolved solute as crystals. The solution obviously remains saturated all the time. If the solid in question has a small amount of an impurity in it, it will be possible in most cases to obtain a recrystallized material of higher purity. If the impurity is not present in sufficient quantity to form a saturated solution, it will not crystallize from the solution, only the pure material being obtained.

An example may serve to illustrate this point. Suppose you have a mixture of 10 g of A and 1 g of B and you wish to obtain pure A. Further let us suppose A dissolves to the extent of 20 g per 100 g of water at the boiling point and 5 g per 100 g of water at room temperature. If substance B is also soluble to the extent of 5 g per 100 g of water at room temperature, we have a favorable case for purification by recrystallization. The 11 g of mixture A + B is dissolved in about 55 or 60 g of water at the boiling point and evaporated slowly until crystals of A begin to form. At this point the solution contains 10 g of A, 1 g of B and 50 g of water. The solution now is cooled to room temperature. Since only 5 g of A is soluble in 100 g of water at this temperature, there will remain in the solution of 50 g of water only 2.5 g of A, 7.5 g having crystallized out. On the other hand, since B is soluble to the extent of 5 g per 100 g of water (which corresponds to 2.5 g in the solution of 50 g of water which we are considering none of compound B should crystallize out. One should obtain, after filtering, 7.5 g of pure A. Actually less than this amount would be recovered because after filtering it is necessary to wash the crystals of A with several small, successive portions of pure water in order to wash out any of the impure solution clinging to crystals of A. Of course, a little of A will be dissolved by this procedure.

It should be mentioned that we have omitted from this discussion secondary effects by which the solubility of one substance may be increased or decreased somewhat by the presence of other substances. Thus, in the above example, the presence of the large amount of A may modify slightly the solubility of B. However, for our purposes, such effects may be neglected.

You are familiar with the concept of electrolytic dissociation. When a salt is dissolved in water, it forms a solution of ions. Thus when potassium chlorate is dissolved, potassium ions and chlorate ions are present rather than molecules. Hence in a

Exp. 14 -3

potassium chlorate begin to form in the solution. Set the beaker aside to cool to room temperature. Filter and wash the crystals of potassium chlorate with 3 successive 8-10 ml portions of cold distilled water. Be sure to let the wash liquid drain thoroughly before adding the next portion. After washing, the crystals should show no more than a faint blue color, indicating the removal of practically all of the cupric nitrate.

Redissolve the crystals of potassium chlorate in as small an amount of hot distilled water as possible and evaporate and re-crystallize as before. Filter and wash this crop of crystals once with cold distilled water. Dry them between pieces of filter paper for at least 24 hours. Note that potassium chlorate is neither deliquescent nor efflorescent. After the pure potassium chlorate has dried, bottle it in a test tube, label with your name, material gross, tare and net weights and percentage recovery. Hand the product to your Instructor. You will be graded on purity of product, dryness, yield and neatness.

## B. Extraction

Add about 2 ml of carbon tetrachloride to 2-3 ml of distilled water and shake. (2) What do you observe? Water and carbon tetra-chloride are immiscible.

Dilute in a test tube 1 ml of bromine water with about 3 ml of water. (3) Note the distinctive color and odor of bromine water. Add 2 ml of carbon tetrachloride and shake. (4) What do you observe? Allow the two layers to separate and then pour the water layer into another test tube and repeat the extraction with a second 2 ml portion of carbon tetrachloride.

To the carbon tetrachloride used in the first extraction add an equal volume of water and shake. (5) What happens? How do you explain this behavior? (6) Is bromine more soluble in carbon tetrachloride or in water?

Repeat these experiments using iodine solution instead of bromine water.

Try the extraction also with carbon disulfide. Caution: Keep carbon disulfide away from flames. Do not pour it into sink but dispose of it in a special waste bottle in the hood.

## Questions and Problems:

I. Using the data given in the appendix, plot on coordinate paper the solubility of potassium chlorate between 0°-100°C and that of potassium nitrate between 0°-60°C. Indicate on this plot the course of potassium chlorate concentration during recrystallization.

II. Explain with aid of this plot and solubility data from the appendix why it is possible to recrystallize potassium chlorate effectively from a 1:1 admixture with cupric nitrate. Why is none of the other possible double decomposition products obtained?

III. Using your solubility graph, calculate the theoretical maximum percentage recovery of potassium chlorate which could be obtained in one recrystallization carried out by your procedure.

Experiment 15

FREEZING POINT DEPRESSION AND MOLECULAR WEIGHT

Object: To determine the molecular weight of a dissolved substance

Discussion:

One gram molecular weight (one mole) of any un-ionized solute dissolved in 1000 g of a solvent lowers the freezing point of that solvent a characteristic amount. The magnitude of this depression is not effected by the particular solute used, provided only that it does not ionize in the solvent. However, this depression, which is called the molal freezing point depression constant, varies from one solvent to another. In the case of water it is -1.86°C. Other solvents have different, but equally definite constants.

The fact that the amount which the freezing point of a particular solvent is lowered depends on the number of moles of solute dissolved only and not on the particular solute enables one to use freezing point depression measurements to determine the molecular weight of a substance when dissolved in a solvent. If one has a solution prepared by dissolving a known mass of solute in a known quantity of solvent, he can compute easily the number of grams of solute per 1000 g of solvent. If the freezing point of this solution has been measured and the molal freezing point depression constant for the solvent is known, it is a simple arithmetic exercise to compute the multiple or fraction of a mole of solute which is present in 1000 g of solvent. This multiple or fraction of a mole obviously weighs the same as the mass of solute dissolved in 1000 g of solvent. From this one computes by proportion the molecular weight.

It is important to have criteria by which the freezing point or melting point may be recognized. In the solid state the molecules are fixed more or less rigidly in equilibrium positions. As the temperature of a solid is raised, the molecules vibrate more and more violently until they acquire finally enough kinetic energy (energy of motion) to move about somewhat without being confined to fixed equilibrium positions. When this occurs, the solid is said to melt. It loses its rigid shape and takes on the shape of the confining vessel. In order for the molecules of the solid to acquire this kinetic energy of translation, heat must be supplied to the solid. While the solid is melting, heat is being added to it, but the temperature remains constant because this heat energy is being used to disrupt forces between molecules without increasing the average kinetic energy of the molecules. When the liquid freezes, it must give off the same amount of energy (in the form of heat) which the solid absorbed when it melted. This is a necessary condition according to the Law of the Conservation of Energy. Thus while the liquid is freezing, its temperature remains constant. It is evident that if a liquid is cooled toward its freezing point it will cool at about a constant rate until it begins to freeze. The temperature then will remain constant for some time, and, after the liquid has frozen, it will cool further. If one wishes to do so, he may think of this constant temperature at the freezing point as

Exp. 15 -2

caused by the heat which is given off maintaining the temperature at the freezing point value. We see that there are good criteria for determining the freezing point.

In this experiment you will determine the molecular weight of sulfur dissolved in naphthalene by observing the depression of the freezing point of such a solution.

### Directions:

Obtain a thermometer and stirrer from the stock-room. Weigh into a clean, dry 8" pyrex test tube about 10 g of naphthalene weighed to the nearest 0.1 g on a platform balance. Clamp the test tube in a vertical position in a 600 ml beaker of water. Insert the thermometer through one hole of a two-hole rubber stopper so that the bulb will be immersed in the naphthalene. The stirrer is placed in the test tube with its handle projecting through the other hole of the stopper. The loop of the stirrer is designed to fit around the thermometer. Stirring of the contents of the tube is carried out by raising and lowering the stirrer. Heat the beaker of water until the naphthalene has melted. Discontinue heating and stir while the naphthalene cools. Record the temperature every minute, reading the thermometer to the nearest 0.1°C. When the freezing point is reached, the temperature will remain constant for several minutes. If the naphthalene does not have a freezing point (melt-ing point) which agrees with the accepted value for pure naphtha-lene, it may be caused by either, or both, of the following. The naphtha-lene may contain impurities or the thermometer may be in error by a small amount. However neither of these factors will cause you to obtain erroneous results for the molecular weight of sulfur. (1) Explain why this will be true for both of these factors.

Weigh out on a small watch glass about one gram of powdered sulfur using the analytical balance, recording the mass to 0.01 g. Add this to the naphthalene after you have determined its freezing point as above. Be very careful not to lose either naphthalene or sulfur in these manipulations. It is essential that all of the sulfur shall dissolve in the naphthalene, none sticking to the wall of the test tube. Heat the beaker of water until the naphthalene has melted and all of the sulfur is dissolved. Then determine the freezing point of the solution of sulfur in naphthalene in the same way that you did for the naphthalene alone. Record all data immediately on your data sheets. (See problem I at end.)

The molal freezing point depression constant for naphthalene is -7.0°C. With this and the observed freezing point depression calculate the molecular weight of the dissolved sulfur.

### Questions and Problems:

I. Plot on co-ordinate paper your experimental cooling curves for both naphthalene and the sulfur solution of naphthalene. Plot the independent variable along the horizontal (x-) axis. You can determine the freezing point very nicely from these curves.

II. Calculate on the basis of your experimental results the freezing point of a solution of 13.48 g of sulfur in 321 g of naphthalene.

Exp. 15 -3

Data Sheet

Name ________________ Date ________________

Record your temperature - time cooling data in neat, tabular form on a separate sheet.

1. Mass of test tube and naphthalene ____________ g
2. " " " " empty ____________ g
3. " " naphthalene ____________ g
4. " " watch glass and sulphur ____________ g
5. " " " " empty ____________ g
6. " " sulphur ____________ g
7. Grams sulfur per 1000 g of naphthalene ____________ g
8. Freezing point of naphthalene ____________ °C
9. " " of sulfur solution ____________ °C
10. " " depression ____________ °C
11. " " " for naphthalene - 7.0 ____________ °C
12. Moles S per 1000 g naphthalene ____________
13. Molecular weight of sulfur in naphthalene ____________
14. Atoms of sulfur per molecule ____________
15. Molecular formula of sulfur in naphthalene ____________

Experiment 16

ACIDS AND BASES

**Object:** To study the characteristics of acids and bases and the reaction between them.

**Discussion:**

The majority of the compounds which you will study during this course can be classified as either acids, bases or salts. It is very important therefore that you understand the characteristics of each of these classes of compounds. In this experiment you will see the reactions which are typical of all acids, those typical of all bases, etc. The facts covered in this experiment then are very important because they apply to so many compounds.

According to the Bronsted theory, which is now the accepted one, acids are substances which give up protons or hydrogen atoms which carry a positive charge due to the loss of an electron. This theory states that in a water solution this positively charged particle, the proton, is accepted by a molecule of water to form the hydronium ion which therefore has a positive charge of one. (1) Use a diagram and account for the formation of the particle which is given up by acids. Show by means of an equation how the hydronium ion is formed. Since all water solutions of acids contain these hydronium ions in varying amounts, the properties, which are attributed to acids, are the properties of the ions only. In this exercise and those following the hydrogen ion will be used in place of the hydronium ion. This is done to save time in the writing of the ion and also to avoid some confusion.

Bases are proton acceptors and therefore are just the opposite of acids. The hydroxyl ion combines readily with the hydrogen ion and because of this pronounced tendency it is classed as one of the strongest bases. Here again, the concepts of acids and bases have changed much more rapidly than the usage of terms. It is still customary to speak of compounds as bases when, upon ionizing, they liberate hydroxyl ions. (2) What is a conjugate base? Explain the following statement: A strong conjugate base is formed when a weak acid ionizes.

If you go back to the older concept of acids and bases, that is, acids are compounds which form hydrogen ions, and bases that form hydroxyl ions, you still have a good working definition of these compounds, if not a strictly accurate one. The reaction which takes place between acids and bases will be studied here in order to find out as much as possible about how it takes place.

The change which takes place when a solution of an acid and one of a base are mixed is called neutralization, and is an example of a reaction that goes to completion. Just before mixing and momentarily at the time of mixing the solutions, all the ions which were derived from each of these substances are present. However, only minute concentrations of hydroxyl and hydrogen ions can exist together. The two, therefore, combine immediately to form a molecule of water. As an illustration, consider the reaction between dilute solutions of potassium hydroxide and nitric acid.

Exp. 16 -2

$$K^+ + OH^- + H^+ + NO_3^- \rightarrow H_2O + K^+ + NO_3^-$$

Since the hydrogen and hydroxyl ions combine almost complete the right side of the equation represents the composition of the solution at the end of the reaction. This complete shift is ex- plained by the fact that water is formed, and this substance has little tendency to ionize and produce those ions found on the lef side of the equation.

### Directions:

In this and all following experiments you are expected to re cord immediately all of your observations and experimental data i your notebook.

Pour one or two drops of dilute hydrochloric acid in half a test tube of water. Taste a little. Place a drop of this on a piece of litmus paper. Obtain about 5 ml of dilute hydrochloric acid in each of three test tubes. Place a small piece of zinc in the first, some iron filings in the second and a piece of copper in the third. Repeat the same test using dilute sulphuric acid. (3) What are the characteristics of acids? (4) Write an equation showing how the hydrogen ions reacted with the metals. (5) Is the true of all metals? Explain in terms of electron transfer. Repe the above tests using acetic acid. (6) Account for the differenc in results.

Put not more than two drops of dilute sodium hydroxide in ha a test tube of water. Make the taste and litmus test. Using the dilute solution from your desk, make about 5 ml of a 1:1 solution. Feel some of the solution between your fingers. (7) Wash your finge afterward.) Add a drop of phenolphthalein to about 5 ml of water in a test tube and then add one drop of sodium hydroxide. After you have noticed the change, add several drops of any dilute acid. (7) Why will any acid bring about this second change? (8) What properties of bases have you observed? (9) Why would you expect solutions of potassium hydroxide and barium hydroxide to act in the same manner?

Add 10 ml of distilled water to about 2 ml of sodium hydroxi and then pour half of this solution in a clean evaporating dish. Make a solution of hydrochloric acid using 2 ml of the dilute aci from your desk and 10 ml of water. (10) What ions are present in each of the solutions? Add one drop of phenolphthalein to the evaporating dish. Now add the dilute hydrochloric acid carefull and slowly with constant stirring until the color just disappears. At the end of this change the solution should be colorless, but c drop of the base should turn it pink and one of the acid should turn it colorless again. Set the dish over a low flame and evaporate to dryness just below its boiling point. (11) What wou be the objection to evaporating the solution by boiling vigorous Taste one crystal of the white solid remaining in the dish. (12) Write an ionic equation for this neutralization. (13) Assum that the final solution was exactly neutral, name all the particl (ions and molecules) present in this solution.

Exp. 16 -3

Questions:

I. Explain the fact that when one equivalent of a strong acid and one of a strong base react, the same quantity of heat is produced.

II. Acetic acid is ionized only slightly. Explain the fact that when a base is added, the reaction is complete; no molecules of acetic acid remain.

III. Define a solution of a salt from the standpoint of its composition. How does the solid salt differ from the solution?

Experiment 17

HYDROLYSIS

**Object:** To learn why solutions of salts are not/neutral and to learn to measure the acidity or basicity of a solution.

**Discussion:** always

In one of the previous experiments you found that an acid is substance which gives off hydrogen ions and that it is these ions which account for the properties of acids. Similarly, using the older concept, a base is a substance which ionizes to form hydroxyl ions, and the properties of bases can be attributed to this ion. Since normal salts in the ionized state supply neither hydrogen nor hydroxyl ions, it is necessary to find some explanation for the fact that their solutions may be acidic or basic.

The solutions which you will study in this experiment are all water solutions of salts, and it is well known to you by now that water is a neutral substance, giving neither an acidic nor basic reaction with litmus. It is also well known that although water is generally classed as a non electrolyte, it is not completely un-ionized but dissociates to a slight degree according to the following equation:

$$H_2O \rightleftharpoons H^+ + OH^-$$

It is important to note then that water is neutral because it has an **equal number** of hydrogen and hydroxyl ions and not because these ions are absent.

It was found from Experiment 16 that salts are formed when acids react with bases. While this is not the only method of preparing salts, they all may be considered as derivatives of acids and bases. Thus all sodium salts are derivatives of the base, sodium hydroxide, and likewise all nitrates are derivatives of nitric acid. (1) Name the acid and base from which each of the following is derived: Copper sulphate, sodium chloride, potassium phosphate, calcium nitrate, potassium acetate, magnesium carbonate.

The reaction which takes place when some some salts are dissolved in water can be illustrated by the following ionic, horizontal, vertical equations. This reaction is called hydrolysis.

$$ \begin{array}{ccc} & \overset{++}{\mathrm{Cu}} & + \mathrm{SO_4}^{--} \\ \mathrm{CuSO_4} & \underset{\underset{2 \mathrm{OH}^-}{\uparrow}}{\rightleftharpoons} & \mathrm{SO_4}^{+} \\ 2 \mathrm{HOH} & \underset{\underset{\mathrm{Cu(OH)_2}}{\uparrow}}{\rightleftharpoons} & 2 \mathrm{H} \end{array} $$

An examination of the above equation shows the following points:

a. Copper sulfate is ionized highly, as are most salts.

Exp. 17 - 2

b. Water is very slightly ionized but forms an equal number of the two ions.

c. Copper hydroxide is a weak base (slightly ionized), hence only a small percent of the hydroxyl ions remain. Most of them combine with the cupric ions to form slightly ionized cupric hydroxide.

d. Sulfuric acid is a strong acid (highly ionized). Therefore hydrogen ions remain in the solution and _do not_ combine with another ion to form a neutral molecule.

e. The original balance between the hydroxyl and hydrogen ions formed from the water has been displaced and there are now more hydrogen ions than hydroxyl ions in the solution. Hence the solution is acidic.

(2) Write similar equations for the hydrolysis of sodium acetate. Would a solution of this salt give an acidic or a basic reaction? Explain. Note that acetic acid is a weak acid.

In the two cases mentioned above there was a very delicate balance existing among the several reversible reactions in each system. Consider the copper sulfate solution for example. The salt itself is in equilibrium with cupric and sulfate ions. These same cupric ions are in equilibrium with the copper hydroxide molecules and the other ion, in this case the hydroxyl ion, which in turn is in equilibrium with the water molecules and hydrogen ions. There are four equilibria in the system and they are so inter-related that changing the concentration of any one ion would shift all four of them. (3) If it were possible to remove copper hydroxide molecules from the solution what effect would this have on each of the four equilibria?

While copper hydroxide molecules are _not removed_ in the hydrolysis of copper sulfate, this type of action does occur in the case of certain salts. The following equation represents the action which takes place in the case of antimony trichloride.

$$ \underset{\text{Sb}}{\overset{\text{+++}}{\text{Sb}} + \text{3 Cl} + \text{3 H}_2\text{O} \longrightarrow} \underset{\substack{\text{Sb}^- \\ \text{Cl}}}{\overset{\text{OH}^-}{\text{Sb}} + \text{2 H} + \text{2 Cl}} $$
$$ \xrightarrow[\text{white}]{\text{insoluble}} \text{SbOCl} $$
$$ \text{H}_2\text{O} $$

(4) Explain why the hydrolysis of antimony trichloride is much more nearly complete than that of copper sulfate.

You have seen that in the hydrolysis of a salt the system reaches an equilibrium in which the hydrogen ions out-number the hydroxyl ions or in which the hydroxyl ions out-number the hydrogen ions. Two salts, both of whose solutions are acidic may reach equilibrium in solutions when different amounts of hydrogen ions have been produced. Thus one solution will more acidic than

Exp. 17 -3

the other. It is convenient to be able to measure this acidity, and the simplest method of doing so is to use indicators. Of course indicators can be used to measure acidity whether it is due to the presence of hydrolyzed salts or the acids themselves.

Acidity of a solution is measured in terms of the concentrati of hydrogen ions and while the topic cannot be discussed completel at this time, it is possible to give you some understanding of its use. Pure water ionizes very slightly. This can be expressed quantitatively as follows: Conc. of hydrogen ion x conc. of hydroxyl ion = $10^{-14}$ when concentrations are expressed in moles per liter. In other words,

$$[H^+] [OH^-] = 10^{-14}$$

In pure water, of course, the concentration of the two ions is equ and $[H^+] = [OH^-] = 10^{-7}$ moles/liter. This need not be true of water solutions; the concentration of one of the ions may be great than that of the other, but the product of the two is $10^{-14}$. The is, therefore, a definite relationship between the concentration of the two ions, as one increases the other decreases. Therefore, either the acidity or basicity of a solution may be expressed in terms of its hydrogen ion concentration. (5) What would be the hy droxyl ion concentration of a solution whose hydrogen ion con centration was $10^{-5}$? Would the solution be acidic or basic? A solution whose $[H^+] = 10^{-9}$, A more convenient way of expressing the acidity of a solution is in terms of pH. ($pH = -\log [H^+]$) The following will show you the relationship between the two values

<table><tr><td>[H<sup>+</sup>]</td><td>10<sup>-1</sup></td><td>10<sup>-3</sup></td><td>10<sup>-6</sup></td><td>10<sup>-8</sup></td><td>10<sup>-10</sup></td></tr><tr><td>pH</td><td>1</td><td>3</td><td>6</td><td>8</td><td>10</td></tr></table>

Notice that the pH increases as the hydrogen ion concentration de creases. (6) What would be the pH of a neutral solution? A solu tion whose $[OH^-] = 10^{-5}$?

Indicators are compounds having characteristic colors in sol tions with different pH values. For example, one of the best kno v indicators, litmus, is red in any solution whose pH is 6 or less but is blue in any which is 8 or more. Other indicators have different pH ranges. Phenolphthalein, for example, is colorless at 8 but red at 10. By use of several indicators with different ranges it will be possible for you to determine the pH of a solut See the accompanying table of indicators

### Directions:

A. Make a solution of each of the following salts by adding abo l g of the salt to half a test tube of distilled water: potassium carbonate, aluminum sulphate, sodium chloride, and ammonium acetate. Test each solution with litmus and (7) record the result and their interpretation in tabular form. (8) Write horizontal, vertical ionic equations which explain the behavior of each salt solution. (9) Write generalizations concerning the acid and base from which a salt is derived and the reaction of its solution to litmus. (10) Name one salt whose solution will have a reaction which is similar to each of the above salts. (11) What is the product of the hydrogen and hydroxyl ion concentrations in each of

Exp. 17 -4

the solutions? Reason? (12) Which solutions have a pH greater than 7; which less than 7?

Add about 5 ml of water to 1 ml of antimony trichloride solution. Notice the white product. (13) Show by means of an ionic equation how it is formed. Why is this hydrolysis? Add conc. hydrochloric acid to the tube until the solid disappears. (14) What effect did the addition of hydrochloric acid have on the equilibrium? Is the system still in equilibrium? Explain fully.

Hydrolysis is often used to supply hydrogen or hydroxyl ions needed in a reaction. Add several drops of dilute hydrochloric acid to a small amount of sodium bicarbonate on a watch glass. The gas evolved is carbon dioxide, and its formation is accounted for by the following:

$$ \underset{\text{Na}^+ + \text{HCO}_3^- \rightarrow \text{H}_2\text{CO}_3}{\text{L}} \quad \text{Na}^+ + \text{Cl}^- \rightarrow \text{H}_2\text{O} + \text{CO}_2 $$

Repeat the experiment using dilute sulphuric acid in place of the hydrochloric acid. (15) Write an ionic equation using only those ions that undergo any change. (16) Would nitric or acetic acid have the same effect? Explain fully. If in doubt, make the tests using dilute acids in each case. (17) Why doesn't water have this effect?

Add several drops of aluminum sulphate solution to some sodium bicarbonate. (18) Explain the results, using ionic equations throughout. What was the source of the ion which reacted with the bicarbonate ion? This reaction is a very important one and is used in certain types of baking powders and fire extinguishers.

B. In this part of the experiment you will learn to use indicators for determining the hydrogen ion concentration of a solution. First it is necessary to make up standards (solutions of known pH values in order to determine the point at which each indicator changes color. You will need twenty-one clean, dry test tubes.

Obtain a test tube of 1 N hydrochloric acid and after labeling it "pH 0", set it in your rack. Wash a burette twice with water, rinse with about 5 ml of 1 N HCl and then fill slightly above the 40 ml mark with this same acid. After filling the tip of the burette exactly 5 ml of the acid into a clean, dry beaker. Add 45 ml of distilled water from your graduated cylinder and stir with a clean rod. The new solution is 0.1 N; what is the hydrogen ion concentration? The pH? Fill a test tube with the acid, label with correct pH, and set in rack. Do not discard the acid in the beaker.

Empty the burette, wash with water, rinse with a small amount of the 0.1 N acid remaining in your beaker and repeat the above process using the 0.1 N acid remaining in the beaker in place of the 1 N HCl. (19) What is the normality of the new solution? The [H⁺] in the pH?

Experiment 18

TITRATION AND EQUIVALENT WEIGHT OF AN ACID

**Object:** To standardize a base and use it in titration of acids to determine concentration and equivalent weights.

**Discussion:**

The process of measuring accurately the amount of solution required to complete a reaction is called titration. In a titration some indicating device is used which will show a decided change in one or more of its properties when very nearly an equivalent amount of the titrating solution has been added to the material being titrated. The point in the titration where this change in property occurs is called the 'end-point.' Often an indicator is used which undergoes a sharp color change at the end-point.

It is convenient for the chemist to express concentrations of solutions used in titrating in terms of the number of gram equivalent weights per liter of solution. An equivalent weight, it will be recalled, is the weight in grams of a substance which will react completely with, or displace, 1.008 g of hydrogen or its equivalent. A solution which contains one equivalent weight per liter of solution is termed a one normal solution. It is evident that equal volumes of equi-normal solutions will be exactly equivalent to one another. A simple relationship thus exists between volumes required for complete reaction and the normality of these solutions. If the concentration of one of the solutions is known, the concentration of the other may be calculated from the volumes of the two solutions which were used in the titration.

The titration of acids and bases, which involves neutralization reactions, is a very common procedure in the analytical laboratory. It has many important applications, for instance to determine the strength of some acid or base or perhaps to evaluate the equivalent weight of some substance. Acid base indicators with which you are familiar may be used, for example, phenolphthalein or methyl orange.

**Directions:**

Obtain a burette, a 25 ml pipette and a 10 ml pipette from the stock-room. (NOTE: These must be cleaned and returned to the stock room at the end of the laboratory period.) In the laboratory you will find several bottles containing approximately 0.25 N hydrochloric acid (S.R.). The exact normality of the acid is marked on each bottle. This hydrochloric acid is to be used as a standard. You will determine the concentration of a solution of sodium hydroxide by titrating it against this acid. Be sure that in all of your experiments you use the same acid and the same base. You must work carefully in standardizing the sodium hydroxide because it in turn will be used to titrate other acids.

Take about 60 ml of hydrochloric acid (S.R. approx. 0.25 N) in a small, dry beaker to your desk. Rinse the 25 ml pipette with a few ml of this acid. Then withdraw 25 ml of the acid with your pipette and deliver it into a 250 ml erlenmeyer flask. Dilute with

Exp. 18 -2

about 40-50 ml of distilled water. The exact volume of water added is not important. (1) Explain. Add 2 drops of phenolphthalein indicator. Clamp your burette in a vertical position to a ring stand with a burette clamp. Rinse the burette with 5 - 8 ml of sodium hydroxide solution (S.R. approx. 0.25 N) and then fill it with sodium hydroxide and read the volume to the nearest 0.01 ml. Remember to read the bottom of the meniscus. It is not necessary to set the level of the solution in the burette exactly at the zero. (2) Why not? Now titrate the hydrochloric acid in the erlenmeyer flask adding the sodium hydroxide slowly with swirling until a very faint pink color is obtained which persists in the solution for about 15 seconds. This is the end point. Care must be exercised to prevent running over the end point. Read the burette to the nearest 0.01 ml and obtain the volume of solution used by difference of final and initial volumes read on the burette. Calculate the normality of your solution of sodium hydroxide. Repeat the titration with a second 25 ml portion of hydrochloric acid and compare your results. They should agree to within about ± 1% of each other.

Obtain a sample of vinegar from your Instructor. This is to be analyzed by you to determine the amount of acid in it. After rinsing the 10 ml pipette with a little of your sample of vinegar, deliver 10 ml of it into a 250 ml erlenmeyer flask and dilute it with about 40 ml of distilled water. Add 2 drops of phenolphthalein indicator and titrate with the sodium hydroxide which you have standardized. From the determined normality of your base and the volumes involved compute the normality of the vinegar as an acid. Assuming that the vinegar contains acetic acid (CH3COOH) and that the specific gravity of the vinegar is 1.03, calculate the percent of acetic acid in the vinegar sample. (NOTE: Acetic Acid produces one H per molecule only.)

Obtain from your Instructor an unknown solid acid of which you are to determine the equivalent weight. Using the analytical balance and weighing to the nearest 0.001 g, weigh out on a small watch glass an amount of this acid which is within the limits given you by your Instructor. Dissolve this in about 75 ml of distilled water and titrate it with your standardized base, using 2 drops of phenolphthalein as indicator. Be very careful not to lose any of the acid after you have weighed it, because to do so would introduce a considerable error. Calculate the number of equivalents of base from the ml used and the normality. This is also the number of equivalents of acid which were present in the mass of solid unknown which you used. (3) Explain this. From this calculate the equivalent weight of the unknown acid. Repeat with a second portion of the acid. The two results should agree to within ± 1%. Report in neat form on a sheet of paper both values and the average and deviation to your Instructor. You will be graded on both concordance of your values and accuracy of the average.

Questions and Problems:

1. 27.53 ml of 0.2472 N base neutralizes 28.19 ml of acid. 0.2673 g of a certain base is neutralized by 29.89 ml of this acid. What is the equivalent weight of the base? If each molecule of this base liberates two hydroxyl ions what is its molecular weight? If the acid used was H2SO4, calculate number of grams of it per liter of solution.

Experiment 19

THE HALOGENS

**Object:** To study the chemical characteristics of the halogens and to observe the relationship between the members of this family.

**Discussion:**

In your study of the various chemical elements you have found that there is a great similarity between certain of them. You are familiar with the fact that when the elements are arranged in order of their atomic numbers, as in the periodic table, those elements which are most similar occur in such order as to form a vertical column which is called a group. Each of these groups contains two families or sub-groups of elements. While the elements within a family resemble each other closely, they are generally quite different from the members of the other family making up this group. The above is called to your attention at this time because the elements which you will study in this experiment form a family of the periodic table whose members show a high degree of similarity. Observe the relationship between the various elements carefully, for by understanding these relationships and formulating the generalizations which grow out of them you can eliminate much of the "pure memory work" of chemistry.

It would be well to point out at this time that the properties of the elements of a family are not exactly alike but exhibit certain similarities. The different elements do show variation and this variation, in the case of most properties, is an orderly one. Thus most properties of elements within a family show progressive changes beginning with the top member in the table and ending with that member which is at the bottom of the vertical column. (1) Explain clearly how color of the element in the case of the halogens illustrates the variation of properties within a family. What color would you predict for the element 85?

The elements of this family are found in Group VII of the periodic table and show strong non-metallic properties; that is, they accept electrons readily. (2) Explain the fact that they all have a negative valence of one. In addition to this negative valence they have several different positive valences when they form oxides and oxy-compounds. The highest of these positive valences is 7. Can you explain why this is to be expected? The ions which are formed when the elements accept electrons are active reducing agents in acid solution but all differ in the degree of this activity. After carrying out this experiment you should be able to state some generalization regarding the way in which this reducing property of the halide (negative halogen) ion varies within the family.

You are familiar with the fact that certain free metals will displace other metals from solutions of their compounds. The action in this case takes place between the free metal and the

Exp. 19--3

the results. The taper is made up of compounds which consist entirely of hydrogen and carbon. (6) In view of this fact how would you interpret your results? What new gas would you expect to find in the bottle? (7) Is this an oxidation-reduction reaction? Justify your answer. Mark a piece of filter paper with a pencil and with ordinary ink. Moisten this and add it to the third bottle of chlorine. To this same bottle add moistened calico and moistened red and blue litmus. (8) Describe and explain the results. Add a piece of dry calico to the fourth bottle and observe the results after one or two minutes. (If the bottle has any moisture in it the test will not be valid.) How do you account for the difference in the action of chlorine on wet and dry calico. How do you account for the difference in the case of the ink and pencil mark? (9) State two factors which determine whether or not chlorine will act as a bleaching agent. The fourth bottle, containing the dry calico, may now be used to test the solubility of chlorine in water. Add about 15 ml of water to the bottle, place your hand over the mouth quickly and shake. (10) What do you notice when you lift your hand? What is the meaning of this?

## B. Preparation of one of the oxy-compounds of chlorine.

Dissolve 3 grams of solid potassium hydroxide in 7 ml of water in a small test tube. Dissolving of potassium hydroxide in water is strongly exothermal; hence the solution will be come quite warm. While it is still warm pass a stream of chlorine from your generator through it until crystals begin to form. Cool the test tube and filter off the crystals using a special piece of filter paper obtained from your instructor. Dry the crystals between filter paper, heat them in a small pyrex test tube and determine whether or not oxygen is given off. (11) Explain your results and write equations for all reactions encountered in this section. Point out any oxidations and reductions which take place. Show the valence changes involved.

When you have finished with the chlorine generator, rinse the contents into the large waste jar under the hood. **Do not pour this into the sink.**

## C. Displacement series of the halogens.

Before performing this part of the experiment review part B of Experiment 14 on liquid-liquid extraction.

Observe a few ml. of chlorine water in a test tube. Is there an appreciable color? In the tests to be performed in this section the water takes no part in the reaction and therefore chlorine water may be considered merely as a solution of the element chlorine and its formula would be $Cl_2$. In another test tube obtain about 3 ml. of potassium iodide solution. (12) What ions are present in this solution? What color are these ions? Now add the chlorine water to the potassium iodide solution. Notice the color change. (13) What molecule is responsible for this color in the solution? Write the ionic equation which accounts for its formation. The change observed so far might be used as a test for

Exp. 19--4

this ion but the test becomes much more reliable if the molecules
which are formed are extracted from the water by means of carbon
tetrachloride in which they are much more soluble. Add about
2 ml. of carbon tetrachloride to the brown solution in the test
tube and shake. Allow the liquids to separate. (14) What change
do you notice in the carbon tetrachloride layer; in the water
layer? Account for this change. It should be emphasized that the
carbon tetrachloride did not enter into any of the chemical
changes involved in the test. (15) Explain clearly the part
played by the carbon tetrachloride and the advantage of using it
in this test.

Repeat the above test using chlorine water and potassium bromide which is then extracted with carbon tetrachloride. (16) What do the results indicate regarding the relative displacing power of chlorine and the other two halogens. Repeat the test using bromine water on potassium iodide. (17) Results? Ionic equation. Arrange the three halogens tested in the order of their displacing powers. Why is it not necessary to try bromine water and the chloride ion?

D. The hydrogen halides and the reducing power of the halide ions.

Add about 2 ml of concentrated sulphuric acid to about a gram of sodium chloride in a test tube. (Caution: the tube should be held with a test tube holder and in such a position that the mouth of the tube is directed away from your face.) Hold a piece of wet blue litmus near the mouth of the tube. Blow your breath across the mouth of the tube. What gas is being evolved? (18) Write an equation for its formation. Is there an oxidation-reduction involved? What would be formed if this gas were dissolved in water? Bring the mouth of an ammonium hydroxide bottle up to the mouth of the tube. Ammonia, NH₃, is being given off by the ammonium hydroxide. (19) Account for the results by means of an equation. Add a pinch of powdered manganese dioxide to the test tube containing the sulphuric acid and sodium chloride. (20) Account for the results. Warm a little sodium chloride with concentrated phosphoric acid. Make the litmus, ammonia, and breath tests on the gas evolved.

Add 2 ml of concentrated sulphuric acid to about a gram of potassium bromide. Make the litmus, ammonia, and breath tests. Is any hydrogen bromide formed? What is the heavy red vapor? Sulphur dioxide is also being given off. (21) Account for the formation of the letter two substances by means of an equation. Why did you not get similar results from sodium chloride and sulphuric acid? Try some concentrated phosphoric acid on potassium bromide. (22) How do sulphuric acid and phosphoric acid compare as oxidizing agents?

Repeat all the tests in the above paragraph substituting potassium iodide for potassium bromide. When the iodide ion reacts with sulphuric acid, hydrogen sulphide is produced. (23) Write equations for the reactions with both sulphuric acid and

Exp. 19--5

phosphoric acid. (24) List the halide ions in the order of their reducing power. Justify the position of each.

## E. Tests for the halide ions.

In testing for an ion you may use any reaction or series of reactions which is characteristic of that particular ion only. Such reactions are not numerous in the case of any ion but you have studied one in this experiment which might be used as a test for the bromide and iodide ions. (25) Describe such a test.

The most widely used and one of the most reliable tests for the halide ions is their reaction with silver nitrate. Obtain in three separate test tubes solutions containing chloride ions, bromide ions, and iodide ions. Add about 3 ml of silver nitrate solution to each of the test tubes. Notice the colors and nature of the precipitates which are formed. (26) Account for each precipitate by means of an ionic equation. If these three ions were the only ones which formed a precipitate with silver ions, the test as performed would be sufficient. There are many other ions, however, which react in this way and consequently it is necessary to carry this test further in order to distinguish these other precipitates from the silver halides. Shake the tube containing the silver chloride until a suspension is formed and quickly pour half of the contents into another test tube. Add dilute nitric acid to one of the tubes. Most precipitates which are formed with silver ions are soluble in dilute nitric acid. Would this test distinguish silver chloride from these? Add ammonium hydroxide to the other tube. Note results. Repeat these tests with the tube containing silver bromide and then with the one containing silver iodide. (27) Record your results in tabular form:

<table><thead><tr><th></th><th>AgCl</th><th>AgBr</th><th>AgI</th></tr></thead><tbody><tr><td>HNO<sub>3</sub></td><td></td><td></td><td></td></tr><tr><td>NH<sub>4</sub>OH</td><td></td><td></td><td></td></tr></tbody></table>

## F. Analysis of an unknown.

Obtain an unknown compound from your instructor. This will be a solid substance in a corked test tube. Record the number which is on the test tube: (Unknown No. __). Dissolve about half of the solid in a test tube of distilled water and using small portions of this solution perform tests which will identify the halide ion which is present. Report your results and the number of your unknown to your instructor. If your conclusion is incorrect, you may perform further tests using the remainder of the solid. Of course, in this case you will not receive as high a grade on your unknown as you would have, had your first conclusion been correct.

Exp. 19--6

Questions:

I. Calculate the density of chlorine in gram/liters at 25°C. and 750 mm.

II. Explain why it is not possible to prepare fluorine by the action of manganese dioxide on hydrofluoric acid. Why will you get pure hydrofluoric acid from sodium fluoride and concentrated sulphuric acid?

III. How many grams of sodium iodide would be needed to make 40g. of iodine? How many liters of hydrogen iodide, measured at 27°C and 747 mm, could be made from this same weight of sodium iodide?