Acid-Base
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Readings for this section

Petrucci: Chapter 16

Acid Base Chemistry 19/07/2006

In discussing equilibrium chemical systems, we must consider as one of the most important of these topics reactions involving acids and bases. Most acid-base systems are equilibrium ones. Let's look at this field of chemistry, starting with some descriptive chemistry.

Acid properties are due to the hydrogen ion H⁺, more properly names the hydronium ion (H₃O⁺).

When we represent an acid in reaction, we often use simply HA (or HB) where the H is the proton that is released in reaction with bases and the A (or B) is the rest of the molecule. The text uses B exclusively but A is also commonly used.

A reaction between an acid and water can be represented as follows:

HB + H₂O ® H₃O⁺ + B^–

This representation of an acid-base reaction is a proton transfer reaction. This is the definition of acid/base systems called Brönsted-Lowry. The acid, HB, gives up a proton (H⁺) to become B^– while the base (water) accepts the proton to become H₃O⁺. Often, we ignore the water and simply write:

HB ® H⁺ + B^–.

It is important to remember in this short-hand system that the "rest of the acid" is not necessarily negatively charged. It merely is charged one negative more than the original acid. Take for example the acid NH₄⁺ ® H⁺ + NH₃. Here, B is neutral but the acid is +1 charged.

Previously, I said that most acids-base systems are equilibrium systems. Let's look at a few.

HCl (g) + H₂O (l) ® H₃O⁺ (aq) + Cl^– (aq) Complete reaction (100%)

HF (aq) + H₂O (l) H₃O⁺ (aq) + F^– (aq) not complete reaction (equilibrium)

Hydrogen chloride (When dissolved in water) is a strong acid. That means it reacts with water (and all other bases) completely. On the other hand, Hydrogen fluoride doesn't react completely with water. Some HF remains in solution undissociated. HF is a weak acid in water.

CH₃COOH (aq) + H₂O (l) H₃O⁺ + CH₃COO^– (aq)

Acetic acid, CH₃COOH, is an organic acid called a carboxylic acid with chemical structure shown in the reaction drawn below.

Acetic acid is also a weak acid. The hydrogen on the oxygen is the acidic hydrogen. It leaves behind a carboxylate ion (called the acetate ion in this case).

Actually, the vast majority of acids are weak acids. It would do you well to memorize the few acids that are strong and then you will always know that any other acid you encounter is a weak one.

The strong acids are:

• Hydrogen Halides (except HF), HCl, HBr, HI,
• Nitric acid, HNO₃
• Sulphuric Acid H₂SO₄ (in dilute solutions only)
• Perchloric acid HClO₄.
• Chloric acid HClO₃

Bases are proton acceptors according to the Brønsted-Lowry definition (B1923,L1923). We've seen that water can act as a base

HCl (g) + H₂O (l) ® H₃O⁺ (aq) + Cl^– (aq)

Ammonia is also a base

HCl (aq) + NH₃ (aq) ® NH₄⁺ (aq) + Cl^– (aq)

Actually, since HCl is a strong acid, it is not really a good representation of reality to show the HCl reacting with another solute in water. Since HCl dissociates (according to the previous reaction) 100% in water, we never actually have HCl (aq) existing as a species. We have instead, H₃O⁺ in the solution and it reacts with the ammonia

H₃O⁺ + NH₃ (aq) ® H₂O (l) + NH₄⁺ (aq).

We can also have water acting as an acid.

H₂O (l) + NH₃ (aq) OH^– (aq) + NH₄⁺.

Note that whereas the former reaction shows a 100% reaction this latter reaction is that of a weak base with water. Hence, an equilibrium is set up.

Alcohols can also act as an acid

R-O-H + :B^-  ® R-O^- + HB  

The anion RO^- is called the alcoxide ion (analogous to the hydroxide ion)

Like acids, The vase majority of bases are weak bases. The few strong bases include the alkali-metal hydroxides (NaOH, etc. and the oxide ion O^2–.

Brønsted-Lowry bases all react with water to produce the hydroxide ion (OH^–).

Brønsted-Lowry acids all react with water to produce the hydronium ion (H₃O⁺).

Lewis Acids:

Lewis gave a different definition of acids which involved the transfer of electron pairs, rather than the transfer of protons.  The two are compatible in that all Brønsted-Lowry acids are also Lewis acids but some Lewis acids are not Brønsted-Lowry acids.  The Lewis definition is more general than that of Brønsted-Lowry.

Lewis acids are electron pair acceptors [like a proton accepts an electron pair when it bonds with a base]

Lewis bases are electron pair donors. [like the electron pair on the oxygen of a hydroxide ion is shared into the empty 1s orbital on the hydrogen cation (proton)]

Consider BF₃ + F^- ® BF₄^-.  the fluoride has lone-pair electrons which it shares into the empty orbital on the boron to form a new covalent bond.  Hence, the fluoride is the electron pair donor (Lewis base) and the boron tri-fluoride is the electron pair acceptor (Lewis acid).  No protons were transferred in this reaction so this is not a Brønsted-Lowry acid base reaction.

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Acid-base Neutralization reactions

Consider

NaOH (aq) + HCl (aq) ® NaCl (aq) + H₂O (l)

Or

KOH (aq) + HBr (aq) ® KBr (aq) + H₂O (l)

Base + acid ® Salt + water.

In all cases, the net ionic reaction is simply

H⁺ + OH^– ® H₂O

The simplest type of chemical calculation involving acid-base reactions is a titration situation where the acid and base are exactly neutralized, i.e., they are both completely used up.

25.00 mL of a NaOH solution was neutralized by 32.72 mL of a 0.129 M HCl solution. What was the concentration of the original sodium hydroxide solution.

First, write a balanced chemical equation

NaOH + HCl ® NaCl + H₂O

We know that for perfect neutralization, all acid and base are exactly used up. The stoichiometry of this particular reaction shows a 1:1 ratio of acid and base, Therefore, the number of moles of acid added is equal to the number of moles of base in the original solution. Or, another way of saying this is:

Finally, the question asked for concentration.

The shorthand for an acid dissociation is HB Û H⁺ + B^–. In this system, we see that an equilibrium exists. On the left-hand side, the acid HB dissociates (in water). On the right-hand side, the base B reacts with H⁺ to reproduce HB. We call this pair of acid and base a conjugate pair since they are interconverted by exchange of a proton.

Some examples of conjugate acid-base pairs are:

• HB H⁺ + B^–
• HCl H⁺ + Cl^–
• H₂O H⁺ + OH^–
• NH₄⁺ H⁺ + NH₃
• H₃O⁺ H⁺ + H₂O
• Acid H⁺ + Base (conjugate pair)

Similarly

• B + H⁺ BH⁺
• H₂O + H⁺ H₃O⁺
• CH₃COO^– + H⁺ CH₃COOH
• F^– + H⁺ HF
• Base + H⁺ Acid (Conjugate pair)

Every Acid-base reaction actually involves two acid-base conjugate pairs

HA	+	B		A–	+	HB+
HF	+	H2O		F–	+	H3O+
H2O	+	CN–		OH–	+	HCN
H2O	+	NH3		OH–	+	NH4+
Acid 1		Base 2		Base 1		Acid 2

Acid 1 and Base 1 are a conjugate pair.

Acid 2 and Base 2 are a conjugate pair.

Notice that water is both an acid and a base. It can donate a proton and it can accept a proton. It is called amphiprotic. We'll see this later when we discuss the self-ionization of water.

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Strengths of acids and bases.

We already saw that HCl is a strong acid in water. That means that it dissociates 100% when dissolved in water. This carries two implications. First of all, we never find a species called HCl (aq) actually existing in water since it is instantly dissociated to its ions. Second, If we look at the conjugate base of HCl (Cl^–). We can immediately say that in any aqueous solution that contains the chloride ion, Cl^– will never react to form HCl. Thus, even though Cl^– is termed the conjugate "base" of HCl. It doesn't in fact act as a base. We call this a very weak base. A strong acid has a very weak base as its conjugate.

Consider the weak acid HF. If we dissolve some HF in water, some of it will dissociate to form the conjugate base F^–. However, if we dissolve some F^– (say, NaF, dissolved in water) we know that some of the fluoride will set up an equilibrium to form some HF. Thus, a weak acid has a weak base as it's conjugate.

Similarly, if we have a strong base (say, O^2–) we know that if we have a solution containing OH^–, it will never react as an acid to form the oxide ion. Thus, a strong base has a very weak acid as its conjugate.

Let's reconsider the strong acid HCl. It cannot exist in aqueous solution. Neither can any strong acid. They all react with water to form H₃O⁺. H₃O⁺ is the strongest acid that can actually exist in aqueous solution. All stronger acids are levelled to H₃O⁺.

Similarly, Strong bases are levelled to the strongest base that can exist in water, OH^–. Thus, any intrinsic differences in the strengths of strong acids (or bases) are not detectable in water since they can't exist in aqueous solution.

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Equilibria in acid-base systems

Consider the following reaction.

HCl + H₂O ® H₃O⁺ + Cl^–

If we were to write an equilibrium constant expression for this reaction, we might at first attempt create the following expression.

Recall that we write concentration here but actually an equilibrium constant is properly written using relative activities. Relative activities are written as:

In the case of the solute chemicals, standard concentration is 1 mol L^–1. Hence, we can use Molar concentrations directly in the equilibrium equations by simply dropping the units. However, the standard concentration of a pure liquid is it's normal molar concentration (in the case of water at 25ºC, 55.5 mol L^–1). If we're considering dilute aqueous solutions, we can easily assume that the concentration of the water (solvent) is almost identical to the pure water concentration. Hence, it is a reasonably good approximation to assume that the relative activity of the solvent in a dilute is 1.

Thus, we don't write the activities of solvents in the equilibrium expressions since they have no effect on the final value of K.

Our equilibrium for the reaction would thus be

The subscript a refers to the reaction type.  Specifically, this K value refers to a reaction where one mole of acid reacts with water.

Example,

What is the concentration of the species in a solution made up of 0.010 mol of HCl in 250 mL of water?

                   HCl + H2O   ---->  H3O+ + Cl–
   Initial        0.010 mol            0     0
                   250 mL
 Strong acid ( Change = 100%)
   Equil.             0             0.01 mol  0.01mol
                                    250 mL    250 mL  

Thus, [H⁺] = [Cl^–] = 0.01/.250 = 0.04 M.

Obviously, we did extra work here in this example. We could have simply said that the reaction was complete and that the stoichiometry gives the final answers because of the one-to-one molar ratio.

It's not quite that simple if the acid (or base) is a weak acid.

For example, Consider the following.

HF + H₂O H₃O⁺ + F^–

The subscript a merely indicates that the equilibrium is that of an acid reacting with water. Please don't think that there is anything else special about this relationship. In general, we can write an expression for any acid HB.

What is the concentration of H₃O⁺ in a 0.20 M solution of acetic acid?
[K_a (CH₃COOH) = 1.8 ×10^–5]

                    CH3COOH + H2O     H3O+ + CH3COO–
 Initial             0.20                   0      0
 CHANGE x reacts:=>    -x                  +x     +x
 Equil               0.20-x                 x      x   

Obviously, this is a quadratic equation (it has a term in X² and in X) and we should use the standard solution to the quadratic equation in order to solve this problem. However, in this case, there are two options.

1) Use the Quadratic formula

First, convert the K_a expression into the familiar quadratic form

after some algebra:

x = 1.9×10^–3M

2) since the molar concentration of the acid is more than 100 times it's equilibrium constant we can be fairly sure in making the assumption that the amount of reaction (x) is very small.

Assume 0.20 – x ~ 0.20. We can now rewrite the equilibrium constant expression by replacing the denominator as follows

Note here that this equation is easily generalizes as  [H+] =(Ca×Ka)½  where Ca is the initial concentration of acid. We'll use these notations in the "Summary of Calculation Types" section later.

We should always check our assumptions.

0.20 – 0.0019 = 0.1982 (rounds to two sig figs) ~ 0.20. (no change) Actually, we might see a small change in the last digit. Since the limit for an assumption to be considered valid is 5%.

Thus, our assumption was good. We can normally consider that the assumption will be valid if
C(acid) > 100×K_a [or C(base) > 100*K_b].
For concentrations we meet regularly here in first-year chemistry, that means if the K constant is about 10^–5 or less, then the assumption will probably be good.

Example: What is [H₃O⁺] in 0.20 M HCN? K_a = 4.9×10^–10.

                       HCN  + H2O     H3O+  +  CN–
   Initial           0.20 M                 0      0
x reacts: CHANGE =>    -x                  +x     +x
   Equil             0.20-x                 x      x   

Thus, [H₃O⁺] = [CN^–] = 9.9×10^–6 M.

Just put the units on at the end since we dropped them so we could insert the numbers into the equilibrium expression.

The percent ionization is

(This means that out of 100,000 molecules, only 5 are ionized when they dissolve in water)

We can do all the same things for a weak base.

B + H₂O BH⁺ + OH^–

Subscript b merely means that this equilibrium constant expression is for one mole of base (B) reacting with water.

Example: What is [OH^–] in 0.10 M NH₃? K_b (NH₃)= 1.8*10^–5.

                       NH3(aq) + H2O (l)    NH4+  +  OH–
   Initial           0.10 M                       0      0
x reacts: CHANGE =>    -x                        +x     +x
   Equil             0.10-x                       x      x   

Thus, [NH₄⁺] = [OH^–] = 1.3×10^–3 M.

The percent ionization is

In this case, our assumption is good. The amount of reaction x is less than 5% of 0.10.

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Self-ionization of Water

We've seen that water can act as both an acid and a base. In fact, water can be simultaneously the acid and the base in a reaction called the self-ionization of water.

H₂O (l) + H₂O (l) H₃O⁺ (aq) + OH^– (aq)

One water molecule gives up a proton to another water forming one hydronium and one hydroxide ion. This is also called the autoprotolysis of water. The equilibrium constant for this reaction would be:

K_W = [H₃O⁺] [OH^–]

For pure water, the concentrations of the two ions are the same and are both 1.0× 10^–7. Hence,

K_W = (1.0× 10^–7)² = 1.0 × 10^–14 @ 25ºC

This equilibrium exists even in solutions of acids and bases and is the reason that there is always some H₃O⁺ and some OH^– ions present in solution even in strong acid and strong base solutions. Take for example, a solution of 0.010 M HCl.

HCl ® H⁺ + Cl^–

In solution, 0.010 M HCl will break up immediately and completely to 0.010 M H⁺ and 0.010 M Cl^–.We' ignore the Cl^– since it doesn't contribute to the acid-base system at all.

Now, in the self-ionization of water, we set up initial and final conditions.

                2H2O  H3O+  +  OH– 
Initial                    .010      0    (due to the HCl)
Change (x reacts)            +x     +x
Equilibrium               0.010+x    x
   

Now, we can fill this into the equation for K_W.

K_W = (0.010+x)(x) = 1.0×10^–14.

To simplify this calculation and avoid the quadratic formula, we'll assume that the amount of reaction x was very small. Hence, we can replace 0.010-x by ~0.010.

K_W = 0.010(x) = 1.0×10^–14.

x = 1.0×10^–12.

Obviously, x is indeed very much smaller than 0.01 and the assumption is valid.

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pH Scale

In water solution we can easily find H₃O⁺ and OH^– concentrations in the range between 1 and 10^–15. For reasons of convenience and other scientific reasons, concentrations are often expressed on a logarithmic scale. For this, the 'p' function is used. It was first proposed by Sørenson. (S1909) You will see this function used in numerous ways in Science. Don't get confused into thinking that each time you see it it is something new. In general

p X = –log X.

Whatever X is. In this case, we use pH and pOH to measure concentrations. Hence,

pH = –log [H⁺]    and    pOH = –log [OH^–]

We already know that in aqueous solution K_W = [H⁺][OH^–] = 1.0×10^–14. Because of a law in mathematics that says

log (x×y) = log x + log y

we can write

pK_w = p([H⁺][OH^–]) = pH + pOH = 14

We can also convert pH and pOH values back to [H⁺] and [OH^–] values by using the inverse of the p function

[H⁺] = antilog(–pH)       or       [H⁺] = 10^–pH

Let's set up a table of pH and pOH values

NOTES on pH (Theoretical definition):

pH has been more accurately defined as pH = -log a_H+ where a_H+ is the hydrogen ion activity. In solutions that contain other ions, activity and concentration are not the same. The activity is an effective concentration of hydrogen ions, rather than the true concentration; it accounts for the fact that other ions surrounding the hydrogen ions will shield them and affect their ability to participate in chemical reactions. These other ions effectively change the hydrogen ion concentration in any process that involves H⁺. In practice, Sørenson's original definition can still be used, because the instrument used to make the measurement can be calibrated with solutions of known [H⁺], with the concentration of background ions carefully controlled.

IUPAC has endorsed a pH scale based on comparison with a standard buffer of known pH using electrochemical measurements. The IUPAC pH scale is very slightly different from the theoretical definition, since it considers factors that are not included in the (thermodynamic) theoretical pH.

Now let's try some calculations,

What is the pH of a solution where:
a. [H₃O⁺] = 1×10^–3 M, b. [H₃O⁺] = 5×10^–3 M, c. [H₃O⁺] = 1 M.

1. pH = –log(1×10^–3) = 3
2. pH = –log(5×10^–3) = 2.3
3. pH = –log(1) = 0

That's relatively simple, now let's try some equilibrium calculations.

Example: What is the pH of a 0.20 M solution of acetic acid?

                     CH3COOH + H2O     H3O+ + CH3COO–
   Initial           0.20 M                 0      0
x reacts: CHANGE =>    -x                  +x     +x
   Equil             0.20-x                 x      x   

Assume 0.20 – x ~ 0.20. We can now rewrite the equilibrium constant expression by replacing the denominator as follows

We should always check our assumptions.

0.20 – 0.0019 = 0.1982 (rounds to two sig figs) ~ 0.20. (no change) Actually, we might see a small change in the last digit. Since the limit for an assumption to be considered valid is 5%.

Thus, [H⁺] = x = 1.9×10^–3. Hence, pH = –log(1.9×10^–3) = 2.72

Example: What is the pH of a solution of 25 mL of 0.20 M NaOH added to 40.0 mL of 0.15 M HCl?

Let's first calculate the number of moles of each species.

25 mL NaOH × 0.20 mol/L = 5.0×10^–3 mol NaOH (this is the limiting reagent)

40 mL HCl × 0.15 mol/L = 6.0×10^–3 mol HCl (this is in excess)

Whenever we add an acid to a base, even a weak acid and a weak base, we react them with each other 100% (limiting reagent calculation) and then allow the excess react with water to find the final pH.

              HCl   +   NaOH   ®   NaCl   +   H2O
Initial      6.0×10–3    5.0×10–3      0
NaOH is used up
Equil        1.0×10–3    0          5.0×10–3

So Now, we see that the HCl is excess, we allow it to react with water.

              HCl + H2O ® H3O+ + Cl– (100 % reaction) 
Initial      1.0 mmoles    0      0
Change (100%, i.e., x=1 mmole)
Final          0          1.0     1.0  (mmoles)

#mol H⁺ = 1.0×10^–3 mol.  Now, divide this final number of moles by the total volume of solution to get the concentration [H⁺].

Hence, pH = –log(0.015) = 1.82.

Examples:

Calculate the pH of the following solutions: a) 0.0100 M NaOH, b) 0.134 M NaOH

1. Since NaOH is a strong base, [OH^–] = 0.0100 M.
   K_W = [H₃O⁺][OH^–] = 1.0×10^–14
   [H₃O⁺] = 1.0×10^–14 / 0.0100 = 1.00×10^–12.
   pH = –log(1.00×10^–12) = 12.
   
   Alternatively, we could have calculated pOH first and converted to pH
   
   pOH = –log(0.01) = 2
   pH = 14 – 2 = 12
2. [H₃O⁺] = 1.0×10^–14 / 0.134 = 7.46×10^–14
   pH = –log(7.46×10^–14) = 13.13
   
   or
   
   pOH = –log(0.134) = 0.87
   pH = 14 – 0.87 = 13.13

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Distribution diagrams

It is useful to look at a graphical technique of viewing acid-base equilibria. In this technique, we plot the relative amount of the components as a function of pH. Using this type of diagram, we can tell at a glance what the major constituents of a solution will be for any given pH (or pOH). The equations for creating these plots are found in the text book. We will not be expecting you to memorize them. However, those equations can be used to program an excel spreadsheet so that the plots can be created easily. A sample spread-sheet file is available here for your use.

Let's look at a distribution diagram for Acetic acid (K_a = 1.8×10^–5).

CH₃COOH (aq) Û H⁺ + CH₃COO^– (aq)

Here, we see two curves. These represent the concentrations of the acetic acid (red) and the acetate ion (blue). The K_a value of 1.8×10^–5 gives a pK_a = 4.74. You can see that in a solution of exactly pH=4.74, half of the original acetic acid would be ionized to acetate ion (and H⁺). At pH lower than 4.74, the proportion of acetic acid is larger and at pH higher than 4.74, more acetate ion is present.

Recall the example in the pH discussion above where we calculated the pH of a 0.20 M solution of acetic acid. We assumed that the amount of dissociation of acetic acid (x) was small compared to the initial amount. Once we had calculated the value for [H⁺] (and hence, the pH) we found that our assumption was true. The calculated pH was 2.72. Checking our chart, at a pH of 2.72 we can easily see that almost all of the original acetic acid is still present (relative amount ~ 1). Hence, by using this distribution chart, we can see visually that our assumption is good.

It is important to think of distribution diagrams in the correct terms. Remember, pH is the x axis. That means, this is a plot of how the acid-base system will behave in a solution of given pH. It is not usually used to tell us what pH a solution will attain by addition of some amount of acetic acid. For that, you need to do the calculations as we did above.

In yet another example, we looked at the equilibrium reached in 0.20 M HCN? K_a = 4.9×10^–10. We calculated the value of [H⁺] to be 9.9×10^–6 M. This gives a pH of 5. Let's look at the distribution curve for HCN to see if pH 5 really does have mostly HCN present.

HCN + H₂O CN^– + H₃O⁺

Since pKa = 9.3, we see that the cross-over point (where [HCN]=[CH^–]) occurs there. The value pH =5 is well below the region where any significant dissociation of HCN occurs. Hence, it is safe to approximate the amount of HCN in solution as equal to the initial amount (to an accuracy of a few digits, anyway).

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Relationship between K_a and K_b for conjugate acid/base pairs

Consider the reaction between the weak acid HF and water.

HF (aq) + H₂O H₃O⁺ (aq) + F^– (aq).

HF and F^– are conjugate acid and base respectively. We could say HF is the conjugate acid of F^– or equally well, we could say that F^– is the conjugate base of HF.

The K_a (HF) can be written as:

We could also have written the equilibrium as that of the weak base F^– reacting with water along with the expression for K_b(F^–).

F^– (aq) + H₂O HF (aq) + H⁺ (aq).

Now, if we add the two chemical equations together, we get the reaction for the self-ionization of water.

2 H₂O H₃O⁺ (aq) + OH^– (aq).

Thus, the equilibrium constant K_W can be represented as

K_W = K_a × K_b

Or

This is a general relation, good for any conjugate weak acid/base pair.

It is possible to use the p function on the K values just as we did for pH and pOH. Thus, we can write

pK_W = pK_a + pK_b. = 14

where the K_a and the K_b refer to a conjugate acid/base pair.

Example: What is the base dissociation constant for the F^– ion? Given K_a = 3.5×10^–4.

Since HF and F^– are a conjugate pair, we can write

K_a (HF) × K_b (F^–) = 1.0×10^–14

What if we had been given a value for pKa rather than K_a? Say, pK_a = 3.46.

We could convert pK_a to K_a and then use the relationship above. Or, easier, we convert pK_a to pK_b and then to K_b.

pK_a(HF) + pK_b(F^–) = 14

pK_b(F^–) = 14 – 3.46 = 10.54

K_b = antilog(–10.54) = 10^–10.54 = 2.9×10^–11.

Notice that the reverse of function p ( –log X = Y) is p^–1 [X = antilog(–Y) or X = 10^–Y].

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Buffer Solutions

In many chemical processes, both industrial and biological reactions can only take place in a narrow range of [H₃O⁺] concentrations (narrow pH range). To ensure that the desired reactions take place, buffers are used which hold the pH constant (within limits).

Examples of such chemical processes are:

• Sewage treatment: Bacteria only survive in a narrow pH range
• Photographic developing
• Electroplating
• Most metabolic processes

• Stomach pH ~ 1.5
• Saliva pH ~ 6.8
• Blood pH ~ 7.4

Buffers are used to maintain pH at some desired range. A Buffer is made up of approximately equimolar amounts of a weak acid and its conjugate base. To make these solutions, we could, for example add a weak acid (like Acetic acid) and the salt of the acid (like Sodium acetate) in equal proportions.

Video demonstration of the Buffer effect: (Broadband) (Dialup)

Let's consider the solution made up of CH₃COOH and CH₃COO^– in approximately equal amounts.

CH₃COO^– + H₃O⁺ CH₃COOH + H₂O

According to La Châtelier's principal If we add a small amount of acid, the acetate ion will react with it and reduce the effective change in pH. If we add a small amount of base, the acetic acid will react with it and again reduce the effective change in pH. The amount of acid and base added must be smaller than the amount of buffer or it will simply use up all the buffer and then proceed to change the pH.

To restate what we said in more mathematical terms:

Here, we see that the equilibrium constant for the reaction of the acetate ion with acid is very large. Hence, any acid added to the buffer solution will be immediately used up in the buffer equilibrium. Similarly, the K for the reaction of the acetic acid with a base is large and the base will be used up.

Example: What is the pH of a solution of 0.050 mol acetic acid and 0.050 mol of acetate ion in 1 L of solution?

             CH3COOH (aq) + H2O  CH3COO– (aq) + H3O+ (aq) 
Initial:        0.05                  0.05           0
Change;           -x                    +x          +x
Equilibrium:   .05-x                 .05+x           x  

Here, we assume that x is small and that we can replace 0.050+x by 0.050 and 0.050–x by 0.050.

(Yes, x is small compared to 0.050)

[H₃O⁺] = x = 1.8×10^–5

pH = –log(1.8×10^–5) = 4.74

NOW, let's add 0.001 mol HCl to the solution (Pretend no volume change just for simplicity in this example).

The HCl will react immediately with the acetate to use 0.001 mole of acetate and produce the same amount of acetic acid.

                CH3COOH (aq) + H2O  CH3COO- (aq) + H3O+ (aq) 
Initial:        0.051                   0.049           0
Change;           -x                      +x           +x
Equilibrium:   .051-x                  .049+x           x   

(Yes, x is small compared to 0.049)

pH = –log(1.9×10^–5) = 4.72

We see in the above example, that the pH has changed from 4.74 to 4.72, a very small change, considering that if we had added the HCl to 1 L of pure water, the resulting pH would have been 3.

What if we had added .01 mol HCl rather than .001 mol. That's ten times as much.

Our initial values would have changed by 0.01 rather than by 0.001.

                CH3COOH (aq) + H2O  CH3COO– (aq) + H3O+ (aq) 
Initial:            0.06                 0.04            0
Change;               -x                   +x           +x
Equilibrium:       .06-x                .04+x            x  

(Yes, x is small compared to 0.050)

[H₃O⁺] = x = 2.7×10^–5

pH = –log(2.7×10^–5) = 4.57

Even adding 0.01 mol of HCl only changed the pH by about .16 units. Had we added the HCl to pure water, the pH would have changed from 7 to 2 (5 units). Thus, the buffer was effective in holding the pH steady.

Let's look at the distribution diagram for acetic acid/acetate again to see this effect visually.

CH₃COOH (aq) H⁺ + CH₃COO^– (aq)

NOTE: NH₄Cl is a salt that dissolves completely to form ammonium ion and chloride ion.

                  NH3 (aq) + H2O  NH4+ (aq) + OH– (aq) 
Initial:          1.00                0.40         0
C                   -x                  +x        +x
Equilibrium:      1.00–x             0.04+x        x  

assume x is small:

Hence, [OH^–] = x = 4.5×10^–5

pOH = 4.34

pH = 9.66

To see if this is a buffer, we can look at the ratio of acid to base:

This ratio is well within the range 10 – .1 so it is a buffer solution.

Looking at the distribution diagram for ammonia (pK_b = 4.74 so pK_a = 9.26) we can also see that pH = 9.66 is in the buffer zone (slope of the line is steep).

NH₃ + H₂O NH₄⁺ + OH^–

We can derive a general equation to deal with calculations of pH of buffer solutions.

HA + H₂O A^– + H₃O⁺

We are using initial concentrations because if we are truly in a buffer zone the amount of change in the relative concentrations of the acid and conjugate base is small.

This equation is called the Henderson-Hasselbalch equation. And it is a very useful shortcut to calculating the pH of a buffer solution once we have established the initial concentrations like we did in the examples above.

We see that the primary contribution to the pH of a buffer solution is the pK_a of the acid/base pair. And modifications to the pH are affected by adjusting the relative concentrations of the acid and base. Thus, by carefully choosing the correct acid/base pair, we can create a buffer of any pH we wish.

A buffer is most effective at resisting change in pH when [base] = [acid]

pH = pK_a + log(1) = pK_a.

A one-to-one molar ratio is not always desired. What if we need to achieve a specific pH value, not quite equal to pKa for the buffer system we are using?

Example: Blood has a pH of 7.4. and is buffered by a carbonic acid/hydrogen carbonate buffer system. What is the ratio of acid to base required to keep the pH at 7.4, given pK_a(H₂CO₃) = 6.37

We need a ratio of 10.7 base to acid to maintain a pH of 7.4. This is right on the edge of the buffer region for the buffer system we used here. In reality, the buffer system used by our blood is much more complex system than we used here and so thankfully, we are not quite on the edge, so to speak.

Well, we've learned how to look at distribution diagrams and how to do equilibrium calculations. In this next section, we'll review a bit of both and show more how these two ideas can be combined to help in the understanding of acid-base equilibria.

We've seen that for a monoprotic acid/base system (one proton is exchanged) the distribution diagram consists of two curves that cross at a value pH = pK_a. (or pOH = pK_b). We've also seen that the width of the curved space is about 2 pH units. This means that we can always draw a quick distribution diagram for any acid/base conjugate system knowing only the K_a (or pK_a), or the K_b (pK_b) of the weak acid or weak base respectively.

Let's take for example, the acetic acid – acetate system. We know that K_a (HAc) = 1.8×10^–5. So pKa = –log(1.8×10^–5) = 4.74. Thus, we can quickly draw a distribution diagram by drawing a rectangle to represent the plot area. The axes are:

• Y: fraction, range = 0 – 1.
• X: pH, range= 0 – 14.

The cross-over of the two curves occurs at about 4.7 and the width of the cross-over area is 1 pH unit to either side of 4.7 (about 3.7 to 5.7). The resulting sketch of the distribution diagram is shown below.

Here, we see the cross-over at 4.74 and we know the width of the area is 2 units (NOTE). Our drawing doesn't have to be perfect. The information is still there.

Summary of calculation types.

Now, lets look again at the example we did above. In that example, we asked for the pH of a 0.20 M solution of acetic acid. After some work and assumptions, we worked out that [H⁺] = (C_aK_a)^1/2 = 1.9×10^–3. And therefore that pH = 2.72.

We could have reached this assumptions more directly by taking the negative log of the previous equation.
-log[H⁺] = -log(C_aK_a)^1/2 
or, rearranged:
pH = 1/2{pC_a + pK_a}.

In words, the pH is the average of the pC_a and pK_a.

This is easily visualized in the diagram where the pC_a and pH are indicated. pH is half-way between pK_a and pC_a.

We could do the same in the case of a solution of weak base where the pOH is half-way between pK_b and pC_b or [OH^-] = (C_bK_b)^1/2. Thus, the pOH of a (C_b = 0.30 M; pC_b = -log 0.30 = 0.52) solution of ammonia (pK_b = 4.75) is (4.75+0.52)/2 = 2.64.  The pH can easily be calculated to be 14 - 2.64 = 11.36.

In both of these cases, the concentration C is larger than K by a factor of at least 100. C > 100 * K. This is called the concentrated limit. In the case where the concentration is not so high (100K > C_a > .1K), the assumption that the reaction is small is not valid and we must use the quadratic formula.

Take acetic acid again, if C_a were to be, say, .0002 we would find that the amount of dissociation is no longer negligible (x is not small) and we would need to use the quadratic formula.  Thus,

[H⁺] = {-K_a + (K_a²+4K_aC_a)^1/2}/2

[H⁺] = {-1.8*10^-5 + ([1.8*10^-5]²+4[1.8*10^-5]*0.0002)^1/2}/2 = 7.0*10^-5_.

Notice that in this last example, the amount of dissociation (0.00007) is a sizable fraction of the initial amount (0.0002)

0.00007/0.0002 = 35%, definitely not negligible.  In this case, we would lie on the distribution curve in the region (Near pK_a) where the slope of the curves is very steep and we cannot make any assumptions.

In an extremely dilute weak acid, we can always assume that it dissociates 100% just like a strong acid. This is often called the strong acid limit because of that.

In this case, pH = pC_a is a reasonable approximation unless C_a is so small that pH approaches 7. In this case, the self-ionization of water takes over and we must include that in any calculations. Thus, a concentration of acetic acid of C_a ~ 1*10^-6 gives a pH of about 6 but, for example, a  1*10^-8 M solution would have a pH of about 7, not 8.

We summarize the shortcuts we've just discussed in the following table. The table is given for weak acids and so the subscripts are all a and we calculate [H⁺] and pH. In a weak base case, we calculate using the exact same equations [OH^-] and pOH.

These equations are all shortcuts that you can use only when you're dealing with the addition of a weak acid to water with nothing else added.  The can be used for weak base added to water by changing all references to acid into references to base, i.e.,

 [H⁺] ==> [OH^-], pH ==> pOH, C_a ==> C_b and K_a ==> K_b.

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Acid-base properties of salts.

In an earlier part of this section on acid-base reactions, we said that we can say the the reaction between an acid and a base results in a salt and water. The questions arises, Is this solution neutral?. To know the answer, we must look at the acid/base properties of the two ions that make up the salt. If we don't know the answer, we can sometimes figure out the answer by checking the conjugates of the ions. There are four possible cases:

1) Salt of strong acid and strong base

We gave as an example, hydrochloric acid and sodium hydroxide. If the two compounds are added to a solution such that they exactly neutralize each other the resulting solution will be a NaCl solution.

HCl + NaOH --> NaCl + H₂O

In this case, the solution is the result of the neutralization of a strong acid with a strong base and the pH is exactly 7. In other words, the salt NaCl is neither acidic nor basic. The Na⁺ is the conjugate acid of a strong base, NaOH and is therefore a very weak acid, in other words, it is not an acid. The Cl^- is the conjugate base of a strong acid, HCl, and is therefore a very weak base. In other words, it is not an acid. So NaCl is a salt made up of a cation that is not an acid and an anion that is not a base. It is neutral.

Such is not the case if the original acid and base are not both strong.

2) Salt of strong acid and weak base

Take, for example, the reaction between  Hydrochloric acid and ammonia, a weak base.

HCl + NH₃ --> NH₄Cl

Here, we still have the anion Cl^- that we know to be not a base. the cation, NH₄⁺, however is a weak acid. It's the conjugate acid of the weak base NH₃, since the conjugate of a weak base is always a weak acid. Thus, the salt ammonium chloride is made up of a cation ammonium which is a weak acid and an anion Cl^- that is not a base. Thus, the solution of ammonium chloride is acidic.

3) Salt of weak acid and strong base

What about a solution of sodium acetate, NaCH₃COO? Is it acidic or basic?

The cation is Na⁺ and is obviously neutral. The anion is acetate CH₃COO^-, the conjugate base of the weak acid acetic acid CH₃COOH. Since acetic acid is weak, acetate is also weak. So, sodium acetate is made up of a neutral cation and a basic anion. This salt is basic.

4) Salt of weak acid and weak base

Now for the hard one. What is both the anion and the cation are non-neutral?

Take, for example a salt ammonium acetate, NH₄CH₃COO. Is it acidic or basic?

We've seen that the ammonium ion NH₄⁺ is acidic and the acetate ion CH₃COO^- is basic. The answer to the questions lies in a comparison of the relative strengths of the acid and the base. The stronger one wins. In other words, if K_a > K_b the solution will be acidic. If K_a < K_b then the solution is basic.   It turns out, that in this case, the two happen to be almost equal (to about 2 sig. figs) so the salt sodium acetate is almost neutral.

We now have a way of determining the acid-base properties of any salt solution. In each of these cases, we could have said that the stronger ion wins, in the first three cases, at least one of the ions was very weak and we could easily discard it. In the fourth case both ions were weak and we had to compare directly the two to determine the answer.

What if you were asked to calculate the pH of a salt solution. In the first three cases above, the calculations are relatively simple.

In case 1, the pH is 7. That's simple.

In case 2 and 3, one of the ions is neutral so we can simply do a calculation on the other ion as we have done before.

What is the pH of a 0.1 M solution o sodium cyanide?

We know that the sodium ion is a very weak acid (i.e., not an acid) and we will not consider it further. It's a spectator ion. The cyanide ion CN^-, however is the conjugate of the weak acid hydrogen cyanide HCN and is a weak base. We do the calculation of a weak base in water

                CN- + H2O  HCN + OH-
initial        0.1                0   0
x reacts
equilibrium    .1-x               x   x
Kb = [HCN][OH-] 
       [CN-]
Kb = 2.0*10-5.  
In this case, Cb > 100Kb so we can assume x is small.
Kb = x2/.1 or x = [OH-] = (Kb * .1)1/2 = 1.4*10-3.
pOH =-log(1.4*10-3) = 2.85 so pH = 14 - 3.85 = 11.15
                        

In this case, we could have used the shortcut we discussed above to get to the pH quicker.

pOH = (pKb + pCb)/2 = (4.70 + 1)/2 = 2.85
pH = 14 - 2.85 = 11.15        

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Polyprotic acids.

So far, in our discussion of acids and bases, we have focused on acids that donate only one proton (and bases that accept only one). This type of acid are called monoprotic acids. Other acids can have more than one proton to donate. The general class of these is called polyprotic acids.

Polyprotic acids can be categorized according to how many protons they donate:

• Diprotic acids donate two protons. Examples include H₂SO₄, H₂SO₃, H₂CO₃ and (COOH)₂
• Triprotic acids donate three protons. An example of this class of acids is H₃PO₄

In the case of strong acids, we can merely consider that the ratio of acid to base in a stoichiometric calculation is not 1:1, for example, in dilute solutions H₂SO₄ can be considered to be a strong acid. And especially in a reaction with a base, we can write,

H₂SO₄ (aq) + 2 B ® SO₄^2– + 2 HB⁺.

Here we can plainly see that the ratio of acid to base is 1:2, not 1:1 as would be the case if a monoprotic acid had reacted with the same base.

If the acid is not a strong acid, then we must consider the equilibria that occur in the solution.

The generalized reaction scheme for reaction of a diprotic acid with water can be written as follows. We take the protons off the acid one at a time and set up an equilibrium for each step.

1. H₂B + H₂O Û H₃O⁺ + HB^–
   
   
2. HB^– + H₂O Û H₃O⁺ + B^2–
   
   

These two equilibria exist simultaneously in the solution. For the sake of calculations, there are three different sets of conditions we should consider.

1. The K_a values are separated by at least 4 orders of magnitude (pK_a values are separated by at least 4). In this case, we can consider the equilibria separate from each other. One does not affect the other. It allows for certain assumptions to be made in the calculations.
2. The K_a values are less than 4 but greater than 1 order(s) of magnitude apart (1< DpK_a < 4) we are in an intermediate area where the equilibrium of step 1 is never completely separated from that of step 2 but where some separation is still visible.
3. The K_a values are only one order of magnitude apart (1 pH unit separation of pK_as). In this case, it is nearly impossible to distinguish the two equilibria using experimental techniques. Calculations can still determine the amounts of all reactants.

Lets look at the distribution diagrams for cases like these. (Spreadsheet)

Case 1: pK_a separation is 5.

Here, the curves are colour coded: Red represents H₂B, Green is HB^– and Blue is B^2–. We see that the red-green crossover area is completely separated from the green-blue crossover area. This means that the acid H₂B is completely converted to HB^– before any significant amount of HB^– dissociates to form B^2–.

Case 2: pK_a separation is 2:

Here, we wee that the green curve never completely reaches mole fraction of 1. In this case, the second equilibrium takes effect and uses up the HB^–. In the intermediate region between two crossover points, significant concentrations of all three species H₂B, HB^– and B^2– are present simultaneously. Obviously, calculations in this region will not be as simple as in case 1.

Case 3: pK_a separation is .5

Here, we see that the amount of HB^– is always quite low and experimentally, it would be very hard to observe the two different equilibria. We have what looks like a distorted crossing zone between the red and blue curves with the green poking through as a slight perturbation on the curves. We might think of the HB^– as an intermediate species whose concentration doesn't build up very much. In these cases, except under extremely carefully controlled pH situations, the reaction of H₂B with a base occurs in "nearly" a single step.

Let's look at some calculations involving the triprotic Phosphoric acid, H₃PO₄.

We will look at each equilibrium in turn:

1. H₃PO₄ (aq) + H₂O H₃O⁺ (aq) + H₂PO₄^– (aq)
   
   
   
2. H₂PO₄^– (aq) + H₂O H₃O⁺ (aq) + HPO₄^2– (aq)
   
   
   
3. HPO₄^2– (aq) + H₂O H₃O⁺ (aq) + PO₄^3– (aq)
   
   

We see that the K_a values are all separated by about 5 orders of magnitude. That means that the equilibria can all be treated independent of each other.   We can see this even more clearly if we look at the distribution diagram for H₃PO₄.

Here, the colored lines represent the following species:

• RED => H₃PO₄,
• GREEN =>  H₂PO₄^–,
• BLUE =>  HPO₄^2–,
• VIOLET =>  PO₄^3–,

It is clear to see in the distribution diagram that the three crossover regions corresponding to the three K_a values are completely separate from each other.

Using the three K_a relationships, we can now calculate the concentration of all species involved in the equilibria.

Calculate the concentration of H₃PO₄, H₂PO₄^–, HPO₄^2–, PO₄^3– and H₃O⁺ in a 0.100 M solution of H₃PO₄.

We cannot assume that x<<0.100 since C_a is less than 100×K_a. Must use quadratic equation. (NOTE)

x² + 6.9×10^–3x – 6.9×10^–4 = 0

recall: so, after some algebra, we find

x = 0.023 º [H⁺] = [H₂PO₄^–] (in mol L^–1).

[H₃PO₄] = 0.100 – 0.023 = 0.077 M

However, some of the H₂PO₄^– will dissociate in the second equilibrium. Will it affect the concentration found here significantly? Probably not since the value of K_a2 is about 10⁵ smaller than K_a1. Let's do the calculation to prove it.

This time, C_a is greater than 100K_a2 and we can assume x<<0.023.

x = 6.2×10^–8 º [HPO₄^2–] (in mol L^–1)

NOTE that the concentrations of H⁺ and of H₂PO₄^– calculated above are not significantly changed because x is so small.

Finally, in a like manner, we can use the third equilibrium to calculate the concentration of the PO₄^3– using the initial concentrations we've calculated here for HPO₄^2– and H⁺.

x = 1.2×10^–18 º [PO₄^3–] (in mol L^–1).

Again, we see that the concentrations of H⁺ and HPO₄^2– are not significantly affected by this equilibrium.

We've now calculated the concentrations of all the species in the 0.100 M solution of H₃PO₄.

 In cases where the K_a values lie closer than about four orders of magnitude, we get into a situation where assuming that the two equilibria are independent of each other (as we were able to do for H₃PO₄) leads to increasingly erroneous results. We won't delve into this further in this level. That's for a future course in analytical chemistry.

In the distribution diagrams of polyprotic acids, there are several special points that we should consider. We'll look at the distribution diagram for phthalic acid.

Here we see the crossover points representing the two equilibria (color coded)

(Dihydrogen phthalate or phthalic acid) H₂P H⁺ + HP^–; (hydrogen phthalate, cf., Potassium Hydrogen Phthalate, KHP)
K_a1 = 1.30×10^–3; pK_a1 = 2.89

(hydrogen phthalate) HP^– H⁺ + P^2– (phthalate ion)                                        
K_a2 = 3.1×10^–6; pK_a2 = 5.51

First off, we see that the two equilibria overlap somewhat and there are three species present in solution. Now let's look at the regions of this diagram try to understand them. (see numbers on diagram)

1. Low pH range (pH < K_a1 – 2). In this range, the ion H₂P exists mostly undissociated, just as we would expect in a monoprotic acid.
2. pH = pK_a1. At this point, the concentration of H₂P and HP^– are equal. We are in the centre of the first buffer region.
3. pH = (pK_a1 + pK_a2)/2. The point half-way between the two crossover regions is where the solution can be considered to be made up of the HP^– ion (and some H₂P and P^2–). This is the pH we would get if we made a solution by adding the salt of the HP^– ion to water.
4. pH = pK_a2. At this point, the concentration of HP^– and P^2– are equal. This is the centre of the second buffer region.
5. pH > pK_a2 + 2. In this region, both equilibria are completed and the solution is one of the pure base P^2– in water.

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Titrations.

Titrations are used to determine amounts (concentrations) of acids and bases in solutions of unknown composition. You performed a titration in your first lab where you determined the concentration of an unknown acid by reacting it with a carefully measured volume of standardized base (from the burette). Knowing the volume and concentration of the base, and the stoichiometry of the acid-base reaction involved, we were able to calculate the amount of acid in the reaction vessel (beaker) and hence, its concentration.

In a titration, the object is to measure exactly the amount of titrant it takes to react completely with the sample. This point, where the acid and base have exactly neutralized each other is called the equivalence point. Let's explore the process of a titration so we can understand the system. Then, we can devise a method to experimentally determine equivalence points and perform the experiments.

For our first example we'll be titrating 25 cm³ of a 0.10 M strong acid (in the beaker) with a 0.10 M strong base (in the burette). The discussion following is valid for any strong acid and any strong base. The titration curve depicted below will be identical in all cases.

Since both the strong acid and the strong base are 100% dissociated upon addition to water, we'll not consider equilibria further.

1. Thus, at the start, before any base (from the burette) is added to the beaker, the concentration of H⁺ = C_a; so pH = pC_a = 1.
2. As base is added from the burette, the acid already in the beaker reacts with it and keeps the pH from rising. This is called the buffer region. In this region, we can calculate [H⁺] directly by simply subtracting the number of moles of added base from original number of moles of acid and dividing by the new total volume.
   
   
3. When all the acid is used up, the pH rises rapidly with further addition of base. The steepest part of this rapid rising region is occurs at the equivalence point. Since at this point, we have exactly used up the acid (and haven't yet added extra base), we have only a salt solution. The pH of the equivalence point is therefore determined by the salt. In a strong-acid-strong-base reaction, the salt is exactly neutral and so the pH of the equivalence point is 7. This is not the case if weak acids or weak bases are involved.
4. Beyond the equivalence point, the pH asymptotically approaches the value 14–pC_b.

A strong base titrated with a strong acid would give a curve that is similar but upside-down compared to this one.

 

See if you can figure out the relationships for the four regions discussed in the acid/base case above.

If we titrate a weak acid with a strong base, the resulting titration curve depends on the K_a values for the acid being titrated. In this case, we must consider the equilibrium process at all times so the equations to determine the concentrations of H⁺ (and hence, pH) are much more complicated. We've already explored all the equations we need here so there's nothing new to learn.

Let's titrate 25 cm³ of 0.1 M acetic acid with 0.1 M KOH.

1. At the start, the pH is simply that of a 0.1 M solution of acetic acid. Using the shortcut method we learned earlier (pC_a < pK_a–2) we have
   pH = (pC_a + pK_a)/2 = (1 + 4.7)/2 = 2.85.
2. As base is added, the acetic acid is converted into acetate (strong base reacts 100% with any acid). This is the buffer region. The amount of acetic acid remaining (not counting the equilibrium) is simply
       acid = C_aV_a – C_bV_b
   The amount of acetate is equal to the amount of base added =
       base = C_bV_b.
   Thus, we can use the Henderson-Hasselbalch equation to calculate pH in the buffer region
   
   .
   A special point along the buffer region occurs when the amount of base added is exactly half of the amount of acid we had originally. At that point, the concentration of the acetic acid is exactly equal to the concentration of the acetate and the Henderson-Hasselbalch equation reduces to
   pH = pK_a.
   This is called the half-titration point and exactly corresponds to the crossover point observed in the distribution diagram for acetic acid.
3. At the equivalence point all the acetic acid has been converted to acetate. Hence, we simply have a solution of 0.050 M acetate (volume doubled because of added base). Hence, the pH can be determined using standard methods for the solution of a weak base. Since 0.050 is definitely more than 100 times larger than K_b (acetate) we can use the shortcut again.
   pOH = (pC_b + pK_b)/2 = (1.30 + 9.24)/2 = 5.27
   pH = 14 – pOH = 14 – 5.27 = 8.73.
4. As we add more base, just as in the strong acid case, the pH now only depends on the concentration of the base added and asymptotically approaches the value 14 – pC_b.

Now we determine a way to observe the titration experimentally. Several methods can be used to determine the equivalence point of an acid-base reaction.

• Indicator added to solution. An indicator is a weak acid that is coloured and whose colour is different from that of its conjugate base (or vice versa). For example, the indicator phenolphthalein is colourless in acid solution but is pink in basic solution. We can see this if we look at the distribution diagram for phenolphthalein.
  
  We see that the colourless species (light grey) exists primarily at pH values below about 8 and the pink species (red line) exists primarily at pH values above 10. In between, the colours are mixed and the colour change is in process. It may seem that indicators are not very accurate in that the pH has to shift over a pH range of 2 or so before the colour change can be seen. However, at the equivalence points of many acids, that is exactly what happens. Recall that while there is still some acid (and it's conjugate base) in solution, the solution will be somewhat buffered against pH changes. Once the last of the acid (or conjugate base) is used up, there can be rapid pH changes with small additions of titrant from the burette. Hence, a rapid colour change occurs with as little as one drop of titrant past the equivalence point. This colour change is called the end point and if our indicator is chosen well, it will coincide closely with the equivalence point. If the indicator is poorly chosen, the two points will not match and any calculations based on the endpoint will be inaccurate. Thus, the pH range of this indicator is nearly the same as the equivalence point for acetic acid. Hence, phenolphthalein is a good indicator for the titration of acetic acid with a strong base.
• One way to avoid the pitfalls of using indicators is to measure the pH directly and look for the steep change in pH as a function of volume of titrant added. This steep slope will represent the equivalence point. Computer programming can allow us to automatically titrate accurately determine equivalence points with no human intervention and no potential for errors due to missed endpoints or poorly chosen indicators.

Click Here for an Excel spreadsheet that plots titration curves for monoprotic and diprotic acids.

Video: Universal indicator (dialup)(broadband)

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