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1 Water, pH and “p” notation

1 Water, pH and “p” notation

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17.1 | Water, pH, and “p” Notation


Often, for convenience, we omit the water molecule that carries the hydrogen ion and

write H+ in place of H3O+. The equilibrium equation for the autoionization of water then

simplifies to


H2O �

� H+ + OHThe equation for Kw based on this is likewise simplified.

[H+] [OH-] = Kw


In pure water, the concentrations of H+ and OH- produced by the

autoionization are equal because the ions are formed in equal numbers. It’s

been found that the concentrations have the following values at 25 °C.

Temperature (°C)



1.5 × 10-15


3.0 × 10-15


6.8 × 10-15


1.0 × 10-14


1.5 × 10-14


2.5 × 10-14


3.0 × 10-14


5.5 × 10-14


9.5 × 10-14

[H+] = [OH-] = 1.0 × 10-7 mol L-1

Therefore, at 25 °C,

Kw = (1.0 × 10-7) × (1.0 × 10-7)

Kw = 1.0 × 10-14


As with other equilibrium constants, the value of Kw varies with temperature (see Table 17.1). Unless stated otherwise we will deal with systems

at 25 °C.


Kw at Various Temperatures

Table 17.1


Effect of Solutes on [H ] and [OH ]


Water’s autoionization takes place in any aqueous solution, and because of

Normal body temperature

the effects of other solutes, the molar concentrations of H+ and OH- may

not be equal. Nevertheless, their product, Kw, is the same. Thus, although Equations

17.1–17.3 were derived for pure water, they also apply to dilute aqueous solutions. The

significance of this must be emphasized. In any aqueous solution, the product of [H+] and

[OH-] equals Kw, although the two molar concentrations may not actually equal each other.

Criteria for Acidic, Basic, and Neutral Solutions

One of the consequences of the autoionization of water is that in any aqueous solution,

there are always both H3O+ and OH- ions, regardless of what solutes are present. This means

that in a solution of the acid HCl there is some OH-, and in a solution of the base NaOH,

there is some H3O+. We call a solution acidic or basic depending on which ion has the

largest concentration.

A neutral solution is one in which the molar concentrations of H3O+ and OH- are

equal. An acidic solution is one in which some solute has made the molar concentration of

H3O+ greater than that of OH-. On the other hand, a basic solution exists when the molar

concentration of OH- exceeds that of H3O+.

Neutral solution [H3O+] = [OH-]

Acidic solution

[H3O+] > [OH-]

Basic solution

[H3O+] < [OH-]

n Because of the autoionization of

water, even the most acidic solution

has some OH-, and even the most

basic solution has some H3O+.

If we consider the autoionization constant of water, we can look at some quantitative

relationships of acids and bases. For instance, we can take a solution that has an easily

measurable 0.0020 M concentration of H3O+ and calculate an extremely small OH- concentration that exists in the same solution. Rearranging Equation 17.3, we get

[OH − ] =


1.0 × 10 −14


= 5.0 × 10 −12 M

[H3O+ ]

2.0 × 10 −3

A similar calculation allows us to determine the hydronium ion concentration if we know

the hydroxide ion concentration.

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776 Chapter 17 | Acid–Base Equilibria in Aqueous Solutions

Practice Exercises

17.1 | Commercial, concentrated, hydrochloric acid has a concentration of 12 moles per

liter. If the molarity of hydronium ions is assumed to also be 12 M, what is the concentration of hydroxide ions? (Hint: Use the relationships just discussed.)

17.2 | An aqueous solution of sodium bicarbonate, NaHCO3, has a molar concentration

of hydroxide ion of 7.8 × 10-6 M. What is the molar concentration of hydrogen ion? Is

the solution acidic, basic, or neutral?

The pH Concept

In most solutions of weak acids and bases, the molar concentrations of H+ and OH- are

very small, like those we just discussed. Writing and comparing two exponential values is not

always easy. A Danish chemist, S. P. L. Sørenson (1868–1939), suggested an easier approach.

To make comparisons of small values of [H+] easier, Sørenson defined a quantity that

he called the pH of the solution as follows.

pH = -log [H+]

Definition of pH

The properties of logarithms let us rearrange Equation 17.4 as follows.

n Calculating pH

pH = -log [H+]

= -log (3.6 × 10-4)

= -(-3.44)

= 3.44

Calculating [H+]

-log [H+] = pH

-log [H+] = 12.58

log [H+] = -12.58

[H+] = antilog (-12.58)

[H+] = 10-12.58

[H+] = 2.6 × 10-13 M

n If you still have questions

about these calculations, see

the mathematics review in

Appendix A.


[H+] = 10-pH


Equation 17.4 can be used to calculate the pH of a solution if its molar concentration of

H+ is known. On the other hand, if the pH is known and we wish to calculate the molar

concentration of hydrogen ions, we apply Equation 17.5. For example, if the hydrogen

ion concentration, [H+], is 3.6 × 10-4, we take the logarithm of this value (which is

-3.44) and change its sign so that the result is a pH of 3.44. With digital calculators, it’s

as simple as pressing the log key and then changing the sign. Determining the hydrogen

ion concentration using Equation 17.5 requires that we change the sign of the pH and

then take the antilogarithm. For instance, if we are given a pH of 12.58 we first change it

to -12.58. We then take the antilogarithm according to the instructions that came with

your calculator. In this case the result is 2.6 × 10-13 = [H+].

Acidic, Basic, and Neutral Solutions

One meaning attached to pH is that it is a measure of the acidity of a solution. Hence, we

may define acidic, basic, and neutral in terms of pH values. At 25 °C, in pure water, or in

any solution that is neutral,

[H+] = [OH-] = 1.0 × 10-7 M

n Neutral pH is 7.00 only for 25 °C.

At other temperatures a neutral

solution still has [H+] = [OH-], but

because Kw changes with temperature,

so do the values of [H+] and [OH-].

This causes the pH of the neutral

solution to differ from 7.00. For

example, at normal body temperature

(37 °C), a neutral solution has [H+] =

[OH-] = 1.6 × 10-7, so the

pH = 6.80.

Therefore, by Equation 17.4, the pH of a neutral solution at 25 °C is 7.00.2

An acidic solution is one in which [H+] is larger than 10-7 M and so has a pH less than

7.00. Thus, as a solution’s acidity increases, its pH decreases.

A basic solution is one in which the value of [H+] is less than 10-7 M and so has a pH

that is greater than 7.00. As a solution’s acidity decreases, its pH increases. These simple

relationships may be summarized as follows. At 25 °C:

pH = 7.00

Neutral solution

pH < 7.00

Acidic solution

pH > 7.00

Basic solution

The pH values for some common substances are given on a pH scale in Figure 17.1.


In this equation and similar ones, like Equation 17.6, the logarithm of only the numerical part of the

bracketed term is taken. The physical units must be mol L-1, but they are set aside for the calculation.


Recall that the rule for significant figures in logarithms is that the number of decimal places in the logarithm of

a number equals the number of significant figures in the number. For example, 3.2 × 10-5 has just two

significant figures. The logarithm of this number, displayed on a pocket calculator, is -4.494850022. We

write the logarithm, when correctly rounded, as -4.49.

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17.1 | Water, pH, and “p” Notation




0.1 M HCI


0.3 M citric acid

Lemon juice

0.1 M acetic acid

Pickles, sour


Orange juice

Carbonic acid



0.03 M boric acid





pH 7

“p” Notation


The logarithmic definition of pH, Equation 17.4, has proved to be so useful that it has

been adapted to quantities other than [H+]. Thus, for any quantity X,

we may define a term, pX, in the following way.


pX = - log X



Defining the

p-function, pX

pOH = - log [OH-]

Pure water


0.1 M NaHCO3

0.3 M borax


Milk of magnesia



For example, to express small concentrations of hydroxide ion,

we can define the pOH of a solution as

Soft drinks

One of the deceptive features of pH is how much the hydrogen ion concentration

changes with a relatively small change in pH. Thus, because of the logarithmic scale, a

change of one pH unit corresponds to a ten-fold change in the hydrogen ion concentration. For example, at pH 6 the hydrogen ion concentration is 1 × 10-6 mol L-1. At

pH 5, it’s 1 × 10-5 mol L-1, or ten times greater. That’s a major difference.

Solutions having hydrogen ion concentrations greater than 1 M can be prepared and

they should have negative pH values. However, at such high concentrations it is usually

easier to simply write the molarity of the hydrogen ions. For example, in a solution where

the value of [H+] is 2 M, the “2” is a simple enough number; we can’t simplify it further

by using its corresponding pH value. When [H+] is 2.00 M, the pH, calculated by

Equation 17.4, would be -log (2.00) or -0.301. There is nothing wrong with negative

pH values, but they offer no advantage over just writing the actual value of [H+].


0.1 M ammonia

0.05 M Na2CO3

0.03 M Na3PO4




0.1 M KOH


1.0 M NaOH



Similarly, for Kw we can define pKw as follows.

pKw = - log Kw

The numerical value of pKw at 25 °C equals -log (1.0 × 10-14) or -(-14.00). Thus,

pKw = 14.00 (at 25 °C)

Figure 17.1 | The pH scale.

Common substances and their

approximate pH values.

A useful relationship among pH, pOH, and pKw can be derived by taking the

logarithm of all terms in Equation 17.1 or 17.2 and changing the sign to obtain

Equation 17.7:

pH + pOH = pKw = 14.00

(at 25 °C)


This tells us that in an aqueous solution of any solute at 25 °C, the sum of pH and pOH

is 14.00.

Relating pH and pOH to pKw

pH Calculations

Let’s now look at some pH calculations using Equations 17.4 and 17.7. These are calculations you will be performing often in the rest of this chapter.

If we know the [H+] of a solution—let’s use 6.3 × 10-5 M as an example—we can

calculate the pH by taking the logarithm and changing the sign. The pH we calculate

should be 4.20. We now have two ways to determine the pOH. Using Equation 17.2

allows us to determine [OH-], and then we can take the logarithm and change the sign

to calculate the pOH. An easier method is to use Equation 17.7 and simply subtract

the pH from 14.00. This subtraction tells us that the pOH must be 9.80. Equations

17.4 and 17.7 tell us that if we know one of the four terms, [H+], [OH-], pH, or

pOH, we can calculate the other three. Another way to look at this helps us with solving a problem. If we are given any one of these four values, it is the same as knowing

all of them. For instance, if you know the pOH but a problem requires the [H+], we

know how to easily calculate the needed value. The following practice exercises make

use of these calculations.

jespe_c17_773-829hr.indd 777

n Calculating pH and pOH

pH = -log (6.3 × 10-5)

= -(-4.20)

= 4.20

pKw = 14.00 = 4.20 + pOH

pOH = 14.00 - 4.20

= 9.80

11/18/10 4:18 PM

778 Chapter 17 | Acid–Base Equilibria in Aqueous Solutions

Practice Exercises

17.3 | Water draining from old coal and mineral mines often has pH values of 4.0 or

lower. The cause is thought to stem from reactions of ground water and oxygen with iron

pyrite, FeS. What are the pOH, [H+], and [OH-] in an acid mine drainage sample that

has a pH of 4.25? (Hint: Recall that pH, pOH, [H+], and [OH-] are all interrelated using

the Kw and pKw.)

17.4 | Because rain washes pollutants out of the air, the lakes in many parts of the world

have undergone pH changes. In a New England state, the water in one lake was found to

have a hydrogen ion concentration of 3.2 × 10-5 mol L-1. What are the calculated pH

and pOH values of the lake’s water? Is the water acidic or basic?

17.5 | Find the values of [H+] and [OH-] that correspond to each of the following values

of pH. State whether each solution is acidic or basic.

(a) 2.90 (the approximate pH of lemon juice)

(b) 3.85 (the approximate pH of sauerkraut)

(c) 10.81 (the pH of milk of magnesia, an antacid)

(d) 4.11 (the pH of orange juice, on the average)

(e) 11.61 (the pH of dilute, household ammonia)

pH Meters

One of the remarkable things about pH is that it can be easily measured using

an instrument called a pH meter (Figure 17.2). An electrode system sensitive to

the hydrogen ion concentration in a solution is first dipped into a standard solution of known pH to calibrate the instrument. Once calibrated, the apparatus

can then be used to measure the pH of any other solution simply by immersing

the electrodes into it. The accuracy of pH meter measurements is usually ± 0.02

pH units.

17.2 | pH of Strong Acid

Figure 17.2 | A pH meter. After the

instrument has been calibrated with a solution

of a known pH, the combination electrode is

dipped into the solution to be tested and the

pH is read from the meter. (Richard Megna/

Fundamental Photographs.)

and Base Solutions

Many solutes affect the pH of an aqueous solution. In this section we examine

how strong acids and bases behave and how to calculate the pH of their solutions. Weak acids and bases have similar effects, which we will discuss in the next


Strong Acids and Bases

n If necessary, review the list of

strong acids on page 170.

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In Section 16.3 we saw that strong acids and bases are considered to be 100% dissociated

in an aqueous solution. This makes calculating the concentrations of H+ and OH- in

their solutions a relatively simple task.

When the solute is a strong monoprotic acid, such as HCl or HNO3, the molarity of H+

will be the same as the molarity of the strong acid. Thus, a 0.010 M solution of HCl contains 0.010 mol L-1 of H+ and a 0.0020 M solution of HNO3 contains 0.0020 mol L-1

of H+.

To calculate the pH of a solution of a strong monoprotic acid, we use the molarity of

the H+ obtained from the stated molar concentration of the acid. Thus, the 0.010 M HCl

solution mentioned above has a pH of 2.00.

For strong bases, calculating the OH- concentration is similarly straightforward. A

0.050 M solution of NaOH contains 0.050 mol L-1 of OH- because the base is fully dissociated and each mole of NaOH releases one mole of OH- when it dissociates. For bases

11/18/10 4:18 PM

17.2 | pH of Strong Acid and Base Solutions


such as Ba(OH)2 we have to recognize that two moles of OH- are released by each mole

of the base.

Ba(OH)2(s) → Ba2+(aq) + 2OH-(aq)

Therefore, if a solution contained 0.010 mol Ba(OH)2 per liter, the concentration of

OH- would be 0.020 M. Of course, once we know the OH- concentration we can

calculate pOH, from which we can calculate the pH. Working through the following

practice exercises will help to solidify these methods.

17.6 | Calculate the [H+], pH, and pOH in 0.0050 M HNO3. (Hint: HNO3 is a strong


17.7 | Calculate the pOH, [H+], and pH in a solution made by weighing 1.20 g of KOH

and dissolving it in sufficient water to make 250.0 mL of solution.

17.8 | Rhododendrons are shrubs that produce beautiful flowers in the springtime. They

only grow well in soil that has a pH that is 5.5 or slightly lower. What is the hydrogen ion

concentration in the soil moisture if the pH is 5.5?

Practice Exercises

Effect of Solute on the Ionization of Water

In the preceding calculations, we have made a critical and correct assumption—namely,

that the autoionization of water contributes negligibly to the total [H+] in a solution of an

acid and to the total [OH-] in a solution of a base.3 Let’s take a closer look at this.

In a solution of an acid, there are actually two sources of H+. One is from the ionization

of the acid itself and the other is from the autoionization of water. Thus,

[H+]total = [H+]from solute + [H+]from H2O


Except in very dilute solutions of acids, the amount of H contributed by the water

([H+]from H2O) is small compared to the amount of H+ contributed by the acid ([H+]from solute).

For instance, in 0.020 M HCl the molarity of OH- is 5.0 × 10-13 M. The only source of

OH- in this acidic solution is from the autoionization of water, and the amounts of OHand H+ formed by the autoionization of water must be equal. Therefore, [H+]from H2O

also equals 5.0 × 10-13 M. If we now look at the total [H+] for this solution, we have

n A similar equation applies to the

equilibrium OH- concentration in a

solution of a base.

[OH-]equilib =

[OH-]from solute + [OH-]from H2O

[H+]total = 0.020 M + 5.0 × 10-13 M

(from HCl)

(from H2O)

= 0.020 M (rounded correctly)

In any solution of an acid, the autoionization of water is suppressed by the H+ furnished by the solute. It’s simply an example of Le Châtelier’s principle. If we look at

the autoionization reaction, we can see that if some H+ is provided by an external source

(an acidic solute, for example), the position of equilibrium will be shifted to the left.


H+ + OH−

Adding H+ from a solute

causes the position of

equilibrium to shift to the left.

As the results of our calculation have shown, the concentrations of H+ and OH- from the

autoionization reaction are reduced well below their values in a neutral solution (1.0 ×

10-7 M). Therefore, except for very dilute solutions (10-6 M or less), we will assume that

all of the H+ in the solution of an acid comes from the solute. Similarly, we’ll assume that

in a solution of a base, all the OH- comes from the dissociation of the solute.


The autoionization of water cannot be neglected when working with very dilute solutions of acids and bases.

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780 Chapter 17 | Acid–Base Equilibria in Aqueous Solutions

17.3 | Ionization Constants, Ka and Kb

As you’ve learned in Chapter 16, weak acids and bases are incompletely ionized in water

and exist in solution as molecules that are in equilibria with the ions formed by their

reactions with water. To deal quantitatively with these equilibria it is essential that you

be able to write correct chemical equations for the equilibrium reactions, from which

you can then obtain the corresponding correct equilibrium laws. Fortunately, there is a

pattern that applies to the way these substances react. Once you’ve learned how to write

the correct equation for one acid or base, you can write the correct equation for any other.

Reaction of a Weak Acid with Water

In aqueous solutions, all weak acids behave the same way. They are Brønsted acids and,

therefore, proton donors. Some examples are HC2H3O2, HSO4-, and NH4+. In water,

these participate in the following equilibria.


HC2H3O2 + H2O �

� H3O+ + C2H3O2���

HSO4- + H2O �

� H3O+ + SO42���

NH4+ + H2O �

� H3O+ + NH3

Notice that in each case, the acid reacts with water to give H3O+ and the corresponding

conjugate base. We can represent these reactions in a general way using HA to represent

the formula of the weak acid and A- to represent the anion.

General equation for the

ionization of a weak acid

n Some call Ka the acid dissociation



HA + H2O �

� H3O+ + A-


As you can see from the equations above, HA does not have to be electrically neutral; it

can be a molecule such as HC2H3O2, a negative ion such as HSO4- or a positive ion such

as NH4+. (Of course, the actual charge on the conjugate base will then depend on the

charge on the parent acid.)

Following the procedure developed in Chapter 15, we can also write a general equation for the equilibrium law, omitting the concentration of liquid water as usual, for

Equation 17.8,

[H3O+ ][[ A− ]

= Ka

[HA ]


The new constant Ka is called an acid ionization constant. Abbreviating H3O+ as H+, the

equation for the ionization of the acid can be simplified as


HA �

� H+ + Afrom which the expression for Ka is obtained directly.

[H+ ][[ A− ]

[HA ]

Ka =


You should learn how to write the chemical equation for the ionization of a weak acid

and be able to write the equilibrium law corresponding to its Ka. For example, if we have

a weak acid such as HNO2 we can write the ionization reaction as


HNO2 �

� H+ + NO2and the corresponding equilibrium law is

The meat products shown here

contain nitrite ion as a preservative. (Andy Washnik)

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Ka =

[H+ ][[NO2− ]

[HNO2 ]

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17.3 | Ionization Constants, Ka and Kb 781

17.9 | For each of the following acids, write the equation for its ionization in water and

the appropriate expression for Ka: (a) HC2H3O2, (b) (CH3)3NH+, (c) H3PO4. (Hint:

Determine the conjugate base for each of these acids.)

Practice Exercises

17.10 | For each of the following acids, write the equation for its ionization in water and

the appropriate expression for Ka: (a) HCHO2, (b) (CH3)2NH2+, (c) H2PO4-.

For weak acids, values of Ka are usually quite small and can be conveniently represented

in a logarithmic form similar to pH. Thus, we can define the pKa of an acid as

pKa = -log Ka

n Also, Ka = antilog (-pKa) or

The strength of a weak acid is determined by its value of Ka; the larger the Ka, the stronger

and more fully ionized the acid. Because of the negative sign in the defining equation for

pKa, the stronger the acid, the smaller is its value of pKa. The values of Ka and pKa for some

typical weak acids are given in Table 17.2. A more complete list is located in Appendix C.8.

17.11 | Use Table 17.2 to find all the acids that are stronger than acetic acid and weaker

than formic acid. (Hint: It may be easier to focus on the Ka values since they are directly

related to acid strength.)

Ka =10-pKa

n The values of Ka for strong acids are

very large and are not tabulated. For

many strong acids, Ka values have not

been measured.

Practice Exercises

17.12 | Two acids, HA and HB, have pKa values of 3.16 and 4.14, respectively. Which is

the stronger acid? What are the Ka values for these acids?

Table 17.2

Ka and pKa Values for Weak Monoprotic Acids at 25 °C

Name of Acid

Iodic acid

Chloroacetic acid

Nitrous acid

Hydrofluoric acid

Cyanic acid

Formic acid

Barbituric acid

Hydrazoic acid

Acetic acid

Butanoic acid

Propanoic acid

Hypochlorous acid

Hydrocyanic acid


Hydrogen peroxide













































1.3 × 10



2.4 × 10



1.7 × 10

1.4 × 10

4.6 × 10

3.5 × 10

2 × 10

1.8 × 10

9.8 × 10

2.5 × 10

1.8 × 10

1.5 × 10

1.3 × 10

3.0 × 10

4.9 × 10

Reaction of a Weak Base with Water

As with weak acids, all weak bases behave in a similar manner in water. They are weak

Brønsted bases and are therefore proton acceptors. Examples are ammonia, NH3, and

acetate ion, C2H3O2-. Their reactions with water are


NH3 + H2O �

� NH4+ + OH���

C2H3O2- + H2O �

� HC2H3O2 + OH-

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1 Water, pH and “p” notation

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