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Equilibrium – Acids and Bases 6

1. Strong Acids and Bases

Examples of strong acids are H2SO4, HNO3, HCl, HBr, and HI. (There are a few others, e.g. HClO4,

which we will not discuss.) Since strong acids dissociate completely in water, the H+ concentration (or

the H3O+ concentration) is equal to the concentration of acid. The exception is sulfuric acid which,

since it is diprotic, gives an H+ concentration which is almost twice the acid concentration (or would be

twice the concentration if the second proton came off completely).

Similarly strong bases, the hydroxides of Group IA and Group IIA metals, dissociate completely. So in

a solution of potassium hydroxide, the hydroxide concentration is equal to the concentration of

potassium hydroxide. In a solution of calcium hydroxide, Ca(OH)2, the hydroxide concentration is

twice the concentration of calcium hydroxide.

Because of their complete dissociation, calculating the pH of a strong acid or a strong base solution is

simple. Let us calculate, for example, the pH of a 2.0 10–4

M solution of HBr. The H+ concentration is

2.0 10–4

and the log of 2.0 10–4

is -3.70. (Recall that in a logarithm only the digits after the decimal

are significant.) Since pH is -log[H+], the pH of the solution is 3.70.

As an example of a strong base, suppose we have a solution which is 2.0 10–4

M in Ca(OH)2. This

leads to a hydroxide concentration of 4.0 10–4

. The log of 4.0 10–4

is -3.40. However since this is

based on hydroxide ion and not hydrogen ion, the log corresponds to something we call the pOH and

not the pH. Here the pOH will be 3.40. Because [H+][OH

–] = 1.00 10

–14 we know that the log of H

+

plus the log of OH– add up to 14 (i.e. pH + pOH = 14.0). Thus, pH = 14.0 - pOH = 14.00 - 3.40 = 10.60

2. Weak Acids and Weak Bases

Acids which are not listed among the strong acids (i.e. acids which are not H2SO4, HNO3, HCl, HBr, or

HI) are classed as weak acids. Bases which are not strong (i.e. bases other than the hydroxides of Group

IA and Group IIA metals) are weak. In contrast to a strong acid, which is completely dissociated in

aqueous solution, a weak acid dissociates only partly. That is, the dissociation of a weak acid or base is

an equilibrium reaction. In order to calculate the results of that equilibrium, we need to use the

equilibrium constant.

Let us consider the dissociation of a 0.50 M HF, a weak acid which reacts as below.

HF H+ + F

–

The dissociation has an equilibrium constant, referred to as "Ka", equal to 7.2 10–4

.

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The reaction starts with [HF] = 0.50 M and [H+] = [F

–] = 0. If the amount of HF which dissociates is x,

then [HF] = (0.50 - x) and [H+] = [F

–] = x. Therefore:

Clearing the fraction and combining like terms leads to a quadratic equation. Generally, however, it is

possible to make an approximation which allows us to avoid using the quadratic formula. If K is small,

x should be small. In fact it will probably be so small that (0.50 - x) will be equal to 0.50, and the x on

the bottom can be ignored. Neglecting x gives the following equation.

This is easily solved to give x = 0.019, which means that:

[H+] = [F

–] = 0.019 M and [HF] = 0.50 - 0.019 0.50 M

As before, we use the "5% rule" to determine whether neglecting x was valid. Since 0.019 is less than

5% of 0.50, the assumption is valid. Of course in some problems you know the value of x and are using

it to calculate K. In such a case you don't neglect x — even if it is less than 5%.

Weak base calculations typically involve ammonia or a similar compound (e.g. trimethylamine,

N(CH3)3). Let us calculate the pH of a 0.10 M ammonia solution, which reacts with water:

NH3 + H2O NH4+ + OH

–

The equilibrium constant, Kb, is equal to 1.8 10–5

. The equilibrium expression is, therefore:

Notice that water is in the reaction but not in the mass action expression. Substituting the

concentrations gives:

Neglecting the addition of x to 0.10, and multiplying gives:

x2 = 1.8 10

–6 M and x = [OH

–] = 1.34 10

–3 M

pOH = 2.87 and pH = 11.13

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3. Polyprotic Acids

A polyprotic acid has more than one removable proton. An example would be hydrogen sulfide:

H2S H+ + HS

– Ka1 = 1.1 10

–7

HS– H

+ + S

2– Ka2 = 1.2 10

–13

Why are the equilibrium constants for the removal of the two protons different? This is because the first

proton is removed from a neutral molecule while the second is removed from an anion.

As an example, we will calculate the ion concentration in 0.100 M H2S solution.

KA1

=[H+][HS-]

[H2S]

After neglecting the x in the denominator, we solve to get:

x = [H +] = [HS

-] = 1.0 10

–4 M

Additional hydrogen ions will be formed in the second ionization. However, the H+ formed in later

steps is ALWAYS neglected. Let us now determine the sulfide concentration:

KA2

=[H+][S-2]

[HS-]

Notice that we used "y" instead of "x" in order to lessen confusion. After neglecting the additive y's, the

problem becomes enormously easy, and we get:

y = [S -2

] = 1.2 10–13

M

Neglecting the y was clearly justified. Also note that "y" is equal to the H+ which comes from the

second dissociation. As stated above, y is always small and can be ignored in calculating H+.

4. Hydrolysis

Hydrolysis occurs when the conjugate base of a weak acid or the conjugate acid of a weak base is

dissolved in water. To calculate the pH of a sodium acetate solution, one calculates the hydroxide

concentration caused by the hydrolysis of the acetate.

C2H3O–2 + H2O HC2H3O2 + OH

–

You will notice that in this reaction acetate functions as a base. Therefore to solve the problem you

must use Kb for the acetate ion — which is Kw divided by Ka. Thus for 0.15 M acetate you calculate:

Setting x = [OH–] = [HC2H3O2]

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Neglecting the addition of x to 0.15 gives:

x2 = 8.32 10

–11 and x = [OH

–] = 9.12 10

–6 M

pOH = 5.04 and pH = 8.96

In hydrolysis only the conjugate base (or conjugate acid) is present. If the conjugate base is

accompanied by the acid — if acetate is accompanied by acetic acid — the solution is a buffer.

The following solutions should be treated as hydrolysis problems:

Sodium carbonate

Ammonium chloride

Equal moles of acetic acid and sodium hydroxide

A titration at its equivalence point

The following solutions should not be treated as hydrolysis problems:

Acetic acid

Ammonia

Unequal numbers of moles of acetic acid and sodium hydroxide

A titration anyplace except at its equivalence point

For qualitative questions — that is whether a given compound forms an acidic, basic or neutral solution

— you must know the following: The salt of a strong acid and a strong base is neutral; the salt of a

strong acid and a weak base is acidic; the salt of a weak acid and a strong base is basic; the salt of a

weak acid and a weak base is a nasty question which no one will ask you. Using this logic, you should

predict that ammonium nitrate is acidic, calcium chloride is neutral, potassium phosphate is basic and

aluminum sulfate is acidic.

5. Buffers

A buffer is an aqueous solution which contains roughly equal concentrations of a weak acid and its

conjugate base or a weak base and its conjugate acid. A buffer resists pH change because no matter

whether H+ or OH

– is added; there is something in the solution which will react with the added ion.

Consider a buffer made from the weak acid HF and its conjugate base F–. If an acid is added, it reacts

with the fluoride:

H+ + F

– HF

If a base is added, it will react with the hydrofluoric acid:

OH– + HF H2O + F

–

The pH of a buffer is easy to calculate. Consider a buffer based on HF, which has a KA = 7.2 10–4

. In

a solution where concentrations of both the weak acid and its conjugate base are equal to 0.10 M, the

pH can be calculated from the mass action expression.

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[H+] = 7.2 10

–4 M

pH = 3.14

If H+ is added to the buffer, it all reacts with F

– to form HF. (This is important!) So if we add (without

dilution) 0.01 M H+, we end up with 0.11 M HF and 0.90 M F

–.

Therefore:

[H+] = 8.8 10

–4 M and pH = 3.05

Now let us try another buffer calculation, based on hypochlorous acid, HOCl, for which

Ka = 3.5 10–8

. Suppose we start with 1 liter of 0.50 M HOCl and add, without diluting the solution,

0.20 moles of KOH. This is typical of how buffer problems start; so it is important you understand it.

The strong base (KOH) reacts completely with the weak acid (HOCl) to form H2O and OCl–.

HOCl + OH– OCl

– + H2O

This can be summarized in the ICE chart on the right:

The H+ is formed by "bounce back" and its concentration is

determined by the equilibrium expression:

KA

=[H+][OCl-]

[HOCl]

Therefor [H

+] = 5.25 10

–8 M and pH = 7.28

Now consider a buffer made by mixing 70.0 mL of 0.100 M ammonia with 30.0 mL of 0.100 M HCl.

We will assume the volumes are additive. After dilution, the concentrations are:

[NH3] = 0.070 M [HCl] = 0.030 M

The HCl (actually the H+ from the HCl) reacts with the NH3 as shown below:

H+ + NH3 NH4

+

This gives:

[NH4+] = 0.030 M [NH3] = 0.040 M

Plugging this in to the mass action expression gives us:

then [OH

–] = 2.4 10

–5 M

pOH = 4.62 and pH = 9.38

Remember that adding a strong

base to a mixture of HA and A-

both decreases the HA and

increases the A-.

HOCl OH- OCl

-

Initial 0.50 0.20 0.00

Change -0.20 -0.20 +0.20

Equilibrium 0.30 0.00 0.20

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Many textbooks explain buffers in terms of the Henderson-Hasselbalch Equation:

pH = pKA

+ log[base]

[acid]

This is equivalent to what we have been doing, and many students find Henderson-Hasselbalch makes

simple problems even simpler. However, it makes hard problems even more difficult. So we are not

using it here.

6. Indicators

How do people determine the pH of a solution? An electrochemical device called a pH meter can be

used. However, these are delicate and expensive and they need frequent calibration. A simpler method

involves the use of an organic acid or base which has different colors in its acidic and basic forms. Such

a material, an example of which is phenolphthalein, is called an indicator.

Phenolphthalein is colorless in its acid form (Figure 1) but

becomes pink when it loses a proton. The equilibrium constant

for the loss of a proton by this indicator (HIn H+ + In

–) is

Ka = 1 10–9

— which gives a pKA of 9. The pKA of an acid is

significant since it equals the pH at which half the material is in

the acid form and half is in the basic form. Thus the pK roughly

equals the pH at which color change occurs. Of course one can see

the pink color of the basic form long before the indicator is half

converted.

Consider what happens at pH 8.

Dropping the pH by one unit causes

the fraction of indicator in the basic

form to decrease from 50% to about

9%. However, a tinge of pink in an

otherwise colorless solution can

probably be seen at even less than

9%. For most indicators the color

changes over a range of about one

pH unit on either side of the pK.

Fortunately, this is narrow enough

for most purposes.

The strips of pH paper which most of you have used for measuring pH are based on indicators.

Typically a mixture of three indicators, which change color at different pH values, is used to give a

Indicator pH Range pKa Acid Form Base Form

Methyl violet 0.0 - 1.6 0.8 yellow blue

Methyl yellow 2.9 - 4.0 3.3 red yellow

Methyl orange 3.1 - 4.4 4.2 red yellow

Methyl red 4.2 - 6.2 5.0 red yellow

Chlorophenol red 4.8 - 6.4 6.0 yellow red

Bromothymol blue 6.0 - 7.6 7.1 yellow blue

Phenol red 6.4 - 8.0 7.4 yellow red

Cresol purple 7.4 - 9.0 8.3 yellow purple

Thymol blue 8.0 - 9.6 8.9 yellow blue

Phenolphthalein 8.0 - 9.8 9.0 colorless red

Figure 1. The acidic form

of phenolphthalein

Figure 2. Properties of Acid-Base Indicators at 25°C

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solution which changes color continuously over a wide pH range. Litmus is also an indicator, one

which turns red in acidic solution and blue in basic solution. (Remember bbblue for bbbasic.)

Indicators are widely used to determine the end point in a titration. What you need to know, how to

choose which indicator to use in a given titration, will be discussed in a later section.

7. Titrations

Titrations are used to determine the amount of acid or base in a sample. If the sample to be analyzed is

an acid, a base solution of known concentration (the titrant) is added. The point at which the moles of

acid and the moles of base are equal is known as the "end point" or "equivalence point." Calculations

are based on the volume of titrant needed to reach the end point. Thus one can calculate the moles of

acid in a sample:

MolesA = MB VB

If the acid is in a solution of known volume, its molarity can be calculated from:

MA VA = MB VB

Suppose a sample of organic acid weighing 0.255 g was found to require 35.6 mL of 0.0110 M KOH to

reach the equivalence point. What is the molecular weight of the unknown acid? We know the

molecular weight = grams/moles and that at the end point molesA = molesB. So:

What is the molarity of an acid solution if 5.0 mL are titrated by 28.8 mL of 0.0110 M KOH?

MA VA = MB VB

MA 5.0 = 0.0110 28.8

MA = 0.0634 M

Since volume is stated in mL on both sides of the equation, the units cancel.

All this assumes a monoprotic acid. If the acid is diprotic, and assuming that both protons are titrated, it

is necessary to multiply the acid molarity by two. Thus:

2 MA VA = MB VB

8. Titration Curves

To understand what happens during a titration it is helpful to graph the way pH changes as the titration

proceeds. This graph is known as a titration curve.

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Figure 3. The titration of a

strong acid with a strong base.

Figure 4. The titration of a

weak acid with a strong base.

First look at what happens when a strong acid is titrated with a

strong base (Figure 3). The pH starts low. As the titration

proceeds (as base is added) the pH increases slowly. Suddenly, as

the titration nears the equivalence point, the slope increases and the

pH changes more rapidly. The equivalence point is, in fact, the

point at which the titration curve has its greatest slope. This is why

titrations work so well. At the equivalence point, the very point you

want to determine, a small amount of titrant will cause a large

change in pH. A single drop could cause the pH to change from 4

to 10, for example. This is fortunate, since indicators change color

gradually over a range of several pH units. This titration is often

done using phenolphthalein (pKA = 9) as an indicator.

The titration of a weak acid with a strong base gives a similar curve

(Figure 4), but there are differences. The initial pH is higher, since

the acid is weaker, and the pH at the equivalence point will be higher

as well. What this means is that the vertical portion of the titration

curve, the segment near the inflection point, is smaller than with a

strong acid. This means that determining the end point is more

difficult with a weak acid than with a strong acid, and choice of

indicator is more critical.

This is a good time to address a point which some students find

confusing. The simple comparison between strong and weak

acids given in the previous paragraph works only if the concentrations are the same. A concentrated

solution of a weak acid can be more acidic than a dilute solution of strong acid.

You must know what species are present at various points on the titration curve. Consider the titration

of an HF solution with a solution of KOH. At first the solution contains only HF. As the reaction

proceeds, the HF reacts with OH– to form H2O and F

–. Thus, the principal species are K

+, HF and F

–.

At the equivalence point, all the HF has been converted to F– and the principal species present are K

+

and F–. What happens when the volume of titrant is half of that needed to reach the equivalence point?

(This is called the half equivalence point). Here the concentrations of HF and F– are equal. If [HF] =

[F-], we get the following:

KA

=[H+][F-]

[HF] and K

A= [H+]

Therefore the pH is equal to the pK. This is not a calculation which a practicing chemist does often, but

it occurs frequently on the AP test. Watch for the half-equivalence point!

To calculate the pH at various points on the curve you should remember that the initial solution is a

weak acid, a calculation which you know well. The solution at the half-equivalence point (or at any

other point part way through the titration) is a buffer. Finally, at the equivalence point all the weak acid

and strong base have reacted. All the HF has been converted to F– and therefore the [H

+] and [OH

–]

concentrations are based on the reaction:

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F– + H2O HF + OH

–

The titration of a weak base with a strong acid is simply the reverse of this. It starts at a high pH (but not

as high a pH as it would if it were a strong base) and has its equivalence point below 7. Similarly the

titration of a strong base with a strong acid gives a curve which is the same as the titration of a strong

acid with a strong base. They both have their equivalence points at 7, but

in one the pH increases and in the other it decreases.

Students worry about what happens when you titrate a weak acid with a

weak base. Don't! The experimenter gets to choose the titrant and no one

would choose to titrate with a weak base.

The titration of a polyprotic acid is more complex. Consider a diprotic

acid with: KA1 = 10–3

and Ka2 = 10–10

. Since the acid has two K's, the

titration curve has two inflection points. When the K's of a polyprotic acid

are close, one deprotonation starts before the other has finished and

determining the end point is difficult.

9. Mixing Different Types of Acid/Base Problems

Acid/base problems are not difficult. Unfortunately students often don't know what category a problem

falls into and try to solve the wrong type. There are four types of acid/base problems. We talk about

acids, but everything here is equally applicable to bases.

Strong Acid -- nothing present except for a strong acid

Weak Acid -- nothing present except for the weak acid and its dissociation products. Finding the

pH of an acetic acid solution is a weak acid problem.

Buffer -- a weak acid and its conjugate base. Acetic acid and sodium acetate would make a buffer.

Acetic acid and sodium hydroxide would also make a buffer, because the hydroxide would

react with the acetic acid to form acetate. However hydrochloric acid and sodium chloride

would not form a buffer, because hydrochloric acid is strong.

Hydrolysis -- only the conjugate base of a weak acid. Thus, sodium acetate solution is a hydrolysis

problem. However, if you add an acid to this solution, it becomes a buffer.

Some examples:

1.0 M SODIUM FLUORIDE. Forms sodium ions and fluoride ions in solution. The fluoride ions react:

F– + H2O HF + OH

–, which is hydrolysis.

1.0 M SODIUM FLUORIDE + 0.5 M HYDROFLUORIC ACID. These dissolve to form F– and HF. They

don't react; they form a buffered equilibrium.

1.0 M HYDROFLUORIC ACID + 1.0 M SODIUM HYDROXIDE. They form HF and OH-, which react

according to HF + OH– F

– + H2O. Because these are present in equal concentrations, you

end up with 1.0 M fluoride which then hydrolyzes.

1.0 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. The solution contains 1.0 M F– and 1.0 M

H+.

They react to form 1.0 M HF which then dissociates in a typical weak acid reaction.

Figure 5. Titration Curve

for a diprotic acid.

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0.5 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. These form H+ and F

– ions. All the

fluoride is protonated by 0.5 M of the H+ and the remaining 0.5 M H

+ is left to act as a strong

acid.

1.0 M HYDROFLUORIC ACID + 0.5 M SODIUM HYDROXIDE. These form HF and OH-. All of the

hydroxide reacts with 0.5 M HF to form 0.5 M F-. This mixture, which now contains 0.5 M

fluoride and 0.5 M HF, acts as a buffer.

1.0 M SODIUM FLUORIDE AND 0.5 M HYDROCHLORIC ACID. These dissolve to give a solution

containing H+ and F

– ions. All the H

+ from the hydrochloric acid is used to protonate 0.5 M

of the fluoride forming 0.5 M HF. This leaves a mixture containing 0.5 M HF and the

remaining 0.5 M F-, which is a buffer.

10. Brønsted and Lewis Models

After all we have learned about acids and bases and their properties, it is strange to realize that the

definition of acids and bases which we have been using is only one of several. The definition which we

have used, that an acid supplies H+ ions and a base gives OH

– ions, is the original definition as put forth

by the Swedish chemist Svante Arrhenius. It is, therefore, called the Arrhenius model or the Arrhenius

definition. It is very useful, but there are others.

The Brønsted Model, or more properly the Brønsted-Lowry Model, defines an acid as a proton donor

and a base as a proton acceptor. Consider the reaction of hydrofluoric acid with ammonia

HF + NH3 F– + NH4

+

The hydrofluoric acid donates a proton to the ammonia, making the hydrofluoric acid an acid and the

ammonia a base. Fluoride, which results from the loss of a proton by hydrofluoric acid, is considered a

base since in the reverse reaction it gains a proton. It is said to be the "conjugate base" of hydrofluoric

acid. Similarly the ammonium ion, which loses a proton in the reverse reaction, is said to be the

"conjugate acid" of ammonia.

Although this is less obvious, the dissolving of hydrogen chloride in water can also be considered an

acid base reaction.

HCl + H2O H3O+ + Cl

–

Here HCl is the acid, water is the base, hydronium ion (H3O+, essentially another way of writing H

+) is

the conjugate acid and chloride is the conjugate base.

Water is a base. How strange! Now look at what happens when ammonia is added to water:

NH3 + H2O NH4+ + OH

–

Ammonia is a base, water is an acid, ammonium ion is the conjugate acid and hydroxide ion is the

conjugate base. So water is either an acid or base, depending on the circumstances.

In the Brønsted model acid/base reactions can be considered to be a competition between the base and

the conjugate as to which one wants the proton more.

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Imagine a creature on some distant

planet where lakes are made of ammonia

instead of water. Acid/base chemistry in

ammonia would be based on the

reaction:

NH3 + NH3 NH4+ + NH2

–

Chemistry students on this planet might

think it strange that acid/base

neutralization reactions could occur in

liquid water. Yet they could easily

understand aqueous chemistry in terms

of the Brønsted model.

Another way of defining acids and bases is seen in the Lewis model. Lewis acids are electron pair

acceptors and Lewis bases are electron pair donors. This is consistent with the Arrhenius model, since a

proton is itself an excellent acceptor of electrons while hydroxide is a good electron donor. Thus a

neutralization reaction is water is still an acid base reaction.

H+ + OH

– H2O

However, other reactions which we do not think of as acid/base reactions are acid/base reactions

according to the Lewis model. Consider the reaction of trimethylamine, which we know to have an

unshared pair of electrons and to thus be a Lewis

base. Suppose it reacts with BF3, a molecule

which lacks an octet and is, therefore, capable of

acting as an electron acceptor. The molecule

which results, BF3-N(CH3)3, is referred to as a

"Lewis acid/Lewis base adduct."

11. Relation of Structure to Acidity

A compound which contains a hydrogen atom will often give it up when dissolved, forming an acidic

solution. However many compounds, for example organic compounds other than carboxylic acids, do

not lose protons in water. The question of why some proton-containing compounds ionize easily

(strong acids), others ionize with difficulty (weak acids) and still others don't ionize at all (non-acids)

has to do with bond strength. Why bonds strengths vary as they do is complex.

The simplest example of a structure/acidity relationship

occurs in the hydrogen halides — HF, HCl, HBr and HI.

Since HF contains the most electronegative, and therefore

the most non-metallic, of the halogens, it seems that it should form the

most acidic hydride. However that's not the way it works. Being the

most electronegative element means it will form the strongest bond

with hydrogen, which means it will form the weakest acid.

Acid Bond

Strength

(kJ/mol)

Acid

strength

(in Water)

HF 565 weak

HCl 427 strong

HBr 363 strong

HI 295 strong

Figure 6. A Lewis acid – Lewis base reaction

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The next trend is among the oxy-acids, for example HClO4, HClO3, HClO2 and HClO. The trend is that

the more oxygens there are, the stronger the acid. Thus the trend in acid strength is: HClO4 > HClO3 >

HClO2 > HClO. To understand this you must first realize that the formulas written above give a

misleading picture of the structures. In almost all oxy-acids the hydrogen is bound to an oxygen and not

to the central atom. Thus, HClO2 is actually H-O-Cl-O. (That's why HCl, a strong acid, doesn't fit the

trend. The hydrogen in HCl is bound directly to chlorine.)

Why do additional oxygen atoms make an oxy-acid stronger? The easiest way to view this (and there

are several legitimate ways) is to say that oxygen, being very electronegative, withdraws electron

density from the chlorine. Since losing a proton (a hydrogen cation) will leave behind a negative

charge, something which withdraws negative charge will make the ion more stable. The acidity of other

oxy-acids follows the same trend. Thus: H2SO4 > H2SO3 and HNO3 > HNO2.

Another trend you should know is that metal hydroxides are basic while non-metal hydroxides are

acidic. Thus ClOH is an acid (usually written as HOCl) and KOH is a base. Similarly metal oxides

generally react with water to form basic solutions, for example:

CaO + H2O Ca2+

+ 2 OH–

Non-metal oxides, on the other hand, react with water to form acidic solutions, for example:

SO2 + H2O H2SO3

Why is this? The answer is that an electronegative element, such as chlorine, forms a covalent bond

with oxygen. This bond, like other covalent bonds, does not break easily in water. Thus the only bond

which can break is the H-O bond. In metal hydroxides, however, the central atom is electropositive and

thus forms an ionic bond. The ionic bond between the metal ion and the hydroxide ion is strong, but

unlike the covalent bond it comes apart easily in a polar solvent such as water. (As you may recall,

polar water molecules surround the ions and stabilize the charges.)

The intermediate case, in which the central atom is neither strongly electronegative nor strongly

electropositive, leads to a situation where both the X-O bond and the O-H bond are capable of breaking.

This leads to a phenomenon known as "amphoterism," in which the oxide or hydroxide is capable of

acting as either an acid or a base. It is difficult to predict whether a transition metal oxide will be acidic,

basic or amphoteric. In general you will not have to do that. However, when we get to the reaction

section of this course, you will learn that aluminum and zinc form amphoteric oxides. But that's for

another day.

12. "How many acids" problems

One good way of testing your knowledge of acid-base chemistry is to ask how many compounds in a

list form acidic (or basic) solutions in water. Consider the following list of compounds.

CaBr2, NH4Cl, SO3, AlCl3, KCN If you answered that "two form acidic solutions," you are wrong. The correct answer is "three,"

specifically NH4Cl, SO3, and AlCl3. Make sure you understand this.