CO(aq) + 3OH – (aq) + I 3 – (aq) t HCO 2 – (aq) + 3I – (aq) + 2 O(l)
H 2 CO(aq) + 3OH – (aq) + I 3 – (aq) t HCO 2 – (aq) + 3I – (aq) + 2H 2 O(l)
is a useful reaction, except that it is too slow for a direct titration. If we add a known amount of I 3 – , such that it is in excess, we can allow the reaction to go to comple-
tion. The I 3 – remaining can then be titrated with thiosulfate, S 2 O 3 2– .
I 3 – (aq) + 2S 2 O 3 2– (aq) t S 4 O 6 2– (aq) + 3I – (aq)
This type of titration is called a back titration.
back titration
Calcium ion plays an important role in many aqueous environmental systems.
A titration in which a reagent is added to
A useful direct analysis takes advantage of its reaction with the ligand ethylenedi- a solution containing the analyte, and aminetetraacetic acid (EDTA), which we will represent as Y 4–
the excess reagent remaining after its
reaction with the analyte is determined
Ca 2+ (aq) + Y 4–
(aq) by a titration.
t (aq)
CaY 2–
Unfortunately, it often happens that there is no suitable indicator for this direct titration. Reacting Ca 2+ with an excess of the Mg 2+ –EDTA complex
Ca 2+ (aq) + MgY 2– (aq) t CaY 2– (aq) + Mg 2+ (aq)
releases an equivalent amount of Mg 2+ . Titrating the released Mg 2+ with EDTA
Mg 2+ (aq) + Y 4– (aq)
t 2– MgY (aq)
gives a suitable end point. The amount of Mg 2+ titrated provides an indirect mea- sure of the amount of Ca 2+ in the original sample. Since the analyte displaces a species that is then titrated, we call this a displacement titration.
displacement titration
When a suitable reaction involving the analyte does not exist it may be possible
A titration in which the analyte displaces
to generate a species that is easily titrated. For example, the sulfur content of coal can
a species, usually from a complex, and
be determined by using a combustion reaction to convert sulfur to sulfur dioxide. the amount of the displaced species is
determined by a titration.
S(s) + O 2 (g) → SO 2 (g) Passing the SO 2 through an aqueous solution of hydrogen peroxide, H 2 O 2 , SO 2 (g) + H 2 O 2 (aq) → H 2 SO 4 (aq)
produces sulfuric acid, which we can titrate with NaOH,
H 2 SO 4 (aq) + 2OH – (aq) t SO 4 2– (aq) + 2H 2 O(l)
providing an indirect determination of sulfur.
9A.3 Titration Curves
To find the end point we monitor some property of the titration reaction that has a well-defined value at the equivalence point. For example, the equivalence point for
a titration of HCl with NaOH occurs at a pH of 7.0. We can find the end point,
276 Modern Analytical Chemistry
Equivalence point
Acid–base titration curve for 25.0 mL of 0.00 10.00 20.00 30.00 40.00 50.00 0.100 M HCl with 0.100 M NaOH.
Volume NaOH (mL)
therefore, by monitoring the pH with a pH electrode or by adding an indicator that changes color at a pH of 7.0.
Suppose that the only available indicator changes color at a pH of 6.8. Is this end point close enough to the equivalence point that the titration error may be safely ignored? To answer this question we need to know how the pH changes dur- ing the titration.
titration curve
A titration curve provides us with a visual picture of how a property, such as
A graph showing the progress of a
pH, changes as we add titrant (Figure 9.1). We can measure this titration curve ex-
titration as a function of the volume of titrant added.
perimentally by suspending a pH electrode in the solution containing the analyte, monitoring the pH as titrant is added. As we will see later, we can also calculate the expected titration curve by considering the reactions responsible for the change in pH. However we arrive at the titration curve, we may use it to evaluate an indica- tor’s likely titration error. For example, the titration curve in Figure 9.1 shows us that an end point pH of 6.8 produces a small titration error. Stopping the titration at an end point pH of 11.6, on the other hand, gives an unacceptably large titration error.
The titration curve in Figure 9.1 is not unique to an acid–base titration. Any titration curve that follows the change in concentration of a species in the titration reaction (plotted logarithmically) as a function of the volume of titrant has the same general sigmoidal shape. Several additional examples are shown in Figure 9.2.
Concentration is not the only property that may be used to construct a titration curve. Other parameters, such as temperature or the absorbance of light, may be used if they show a significant change in value at the equivalence point. Many titra- tion reactions, for example, are exothermic. As the titrant and analyte react, the temperature of the system steadily increases. Once the titration is complete, further additions of titrant do not produce as exothermic a response, and the change in temperature levels off. A typical titration curve of temperature versus volume of titrant is shown in Figure 9.3. The titration curve contains two linear segments, the intersection of which marks the equivalence point.
Chapter 9 Titrimetric Methods of Analysis
Volume of titrant (mL)
(a) 1.800
otential (V) P 0.600 0.400 0.200 0.000
Volume of titrant (mL)
Equivalence point
pCl
2.0 T emperature (
Figure 9.2
Examples of titration curves for (a) a
0.00 10.00 20.00 30.00 40.00 50.00 Volume of titrant (mL)
complexation titration, (b) a redox titration, and (c) a precipitation
Volume of titrant (mL)
Example of a thermometric titration curve.
9A.4 The Buret
The only essential piece of equipment for an acid–base titration is a means for deliv- ering the titrant to the solution containing the analyte. The most common method for delivering the titrant is a buret (Figure 9.4). A buret is a long, narrow tube with
buret
graduated markings, and a stopcock for dispensing the titrant. Using a buret with a
Volumetric glassware used to deliver
small internal diameter provides a better defined meniscus, making it easier to read
variable, but known volumes of solution.
the buret’s volume precisely. Burets are available in a variety of sizes and tolerances
278 Modern Analytical Chemistry
0 Table 9.1 Specifications for Volumetric
Class a (mL)
(mL)
2 5 A 0.01 ±0.01 B 0.01 ±0.02
10 A 0.02 ±0.02 B 0.02 ±0.04 25 A 0.1 ±0.03 B 0.1 ±0.06 50 A 0.1 ±0.05 B 0.1 ±0.10
A 0.2 ±0.10 B 0.2 ±0.20
a Specifications for class A and class B glassware are taken from American Society for Testing Materials (ASTM) E288, E542, and E694 standards.
(Table 9.1), with the choice of buret determined by the demands of the analysis. The accuracy obtainable with a buret can be improved by calibrating it over several intermediate ranges of volumes using the same method described in Chapter 5 for calibrating pipets. In this manner, the volume of titrant delivered can be corrected for any variations in the buret’s internal diameter.
Figure 9.4
Titrations may be automated using a pump to deliver the titrant at a constant
Volumetric buret showing a portion of its
flow rate, and a solenoid valve to control the flow (Figure 9.5). The volume of
graduated scale.
titrant delivered is determined by multiplying the flow rate by the elapsed time. Au- tomated titrations offer the additional advantage of using a microcomputer for data storage and analysis.
9B Titrations Based on Acid–Base Reactions
acid–base titration
The earliest acid–base titrations involved the determination of the acidity or alka-
A titration in which the reaction between
linity of solutions, and the purity of carbonates and alkaline earth oxides. Before
the analyte and titrant is an acid–base reaction.
1800, acid–base titrations were conducted using H 2 SO 4 , HCl, and HNO 3 as acidic titrants, and K 2 CO 3 and Na 2 CO 3 as basic titrants. End points were determined using visual indicators such as litmus, which is red in acidic solutions and blue in basic solutions, or by observing the cessation of CO 2 effervescence when neutraliz- ing CO 3 2– . The accuracy of an acid–base titration was limited by the usefulness of the indicator and by the lack of a strong base titrant for the analysis of weak acids. The utility of acid–base titrimetry improved when NaOH was first introduced as a strong base titrant in 1846. In addition, progress in synthesizing organic dyes led to the development of many new indicators. Phenolphthalein was first synthe- sized by Bayer in 1871 and used as a visual indicator for acid–base titrations in 1877. Other indicators, such as methyl orange, soon followed. Despite the increas- ing availability of indicators, the absence of a theory of acid–base reactivity made se- lecting a proper indicator difficult.
Developments in equilibrium theory in the late nineteenth century led to sig-
Chapter 9 Titrimetric Methods of Analysis
Figure 9.5
Typical instrumentation for performing an automatic titration. Courtesy of Fisher Scientific.
in turn, of acid–base titrimetry. Sørenson’s establishment of the pH scale in 1909 provided a rigorous means for comparing visual indicators. The determination of acid–base dissociation constants made the calculation of theoretical titration curves possible, as outlined by Bjerrum in 1914. For the first time a rational method ex- isted for selecting visual indicators, establishing acid–base titrimetry as a useful al- ternative to gravimetry.
9B.1 Acid–Base Titration Curves
In the overview to this chapter we noted that the experimentally determined end point should coincide with the titration’s equivalence point. For an acid–base titra- tion, the equivalence point is characterized by a pH level that is a function of the acid–base strengths and concentrations of the analyte and titrant. The pH at the end point, however, may or may not correspond to the pH at the equivalence point. To understand the relationship between end points and equivalence points we must know how the pH changes during a titration. In this section we will learn how to
280 Modern Analytical Chemistry
approach will make use of the equilibrium calculations described in Chapter 6. We also will learn how to sketch a good approximation to any titration curve using only
a limited number of simple calculations. Titrating Strong Acids and Strong Bases For our first titration curve let’s consider
the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH. For the reaction of a strong base with a strong acid the only equilibrium reaction of importance is
H 3 O + (aq) + OH – (aq) t 2H 2 O(l)
The first task in constructing the titration curve is to calculate the volume of NaOH needed to reach the equivalence point. At the equivalence point we know from re- action 9.1 that
Moles HCl = moles NaOH or M a V a =M b V b
where the subscript ‘a’ indicates the acid, HCl, and the subscript ‘b’ indicates the base, NaOH. The volume of NaOH needed to reach the equivalence point, there- fore, is
Before the equivalence point, HCl is present in excess and the pH is determined by the concentration of excess HCl. Initially the solution is 0.100 M in HCl, which, since HCl is a strong acid, means that the pH is
pH = –log[H 3 O + ] = –log[HCl] = –log(0.100) = 1.00 The equilibrium constant for reaction 9.1 is (K w ) –1 , or 1.00 × 10 14 . Since this is such
a large value we can treat reaction 9.1 as though it goes to completion. After adding
10.0 mL of NaOH, therefore, the concentration of excess HCl is
moles excess HCl
total volume
(. 0 100 M )( . 50 0 mL ) − ( 0 .200 M )( . 10 0 mL )
= . 0 050 M
. 50 0 mL + . 10 0 mL giving a pH of 1.30.
At the equivalence point the moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissoci- ation of water.
K w = 1.00 × 10 –14 = [H 3 O + ][OH – ] = [H 3 O + ] 2
[H 3 O + ] = 1.00 × 10 –7 M Thus, the pH at the equivalence point is 7.00.
Finally, for volumes of NaOH greater than the equivalence point volume, the pH is determined by the concentration of excess OH – . For example, after adding
30.0 mL of titrant the concentration of OH – is
Chapter 9 Titrimetric Methods of Analysis
Table 9.2 Data for Titration of 50.00 mL of
0.100 M HCl with 0.0500 M NaOH
Volume (mL) of Titrant
moles excess NaOH MV − MV [ OH − ] =
bb aa
total volume
(. 0 200 M )( . 30 0 mL ) − (. 0 100 M )( . 50 0 mL ) =
= . 0 0125 M
. 50 0 mL + . 30 0 mL To find the concentration of H 3 O + , we use the K w expression
giving a pH of 12.10. Table 9.2 and Figure 9.1 show additional results for this titra- tion curve. Calculating the titration curve for the titration of a strong base with a strong acid is handled in the same manner, except that the strong base is in excess before the equivalence point and the strong acid is in excess after the equivalence point.
Titrating a Weak Acid with a Strong Base For this example let’s consider the titra-
tion of 50.0 mL of 0.100 M acetic acid, CH 3 COOH, with 0.100 M NaOH. Again, we
start by calculating the volume of NaOH needed to reach the equivalence point; thus
Moles CH 3 COOH = moles NaOH M a V a =M b V b
V eq = V b =
MV aa (. 0 100 M )( . 50 0 mL )
= . 50 0 mL
M b (. 0 100 M )
282 Modern Analytical Chemistry
Before adding any NaOH the pH is that for a solution of 0.100 M acetic acid. Since acetic acid is a weak acid, we calculate the pH using the method outlined in Chapter 6.
CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO t – (aq) [ HO 3 + ][ CH COO 3 − ]
( )( ) xx
[ CH COOH] 3 . 0 100 − x x = [H 3 O + ] = 1.32 × 10 –3 At the beginning of the titration the pH is 2.88.
Adding NaOH converts a portion of the acetic acid to its conjugate base.
CH 3 COOH(aq) + OH – (aq) t H 2 O(l) + CH 3 COO – (aq)
Any solution containing comparable amounts of a weak acid, HA, and its conjugate weak base, A – , is a buffer. As we learned in Chapter 6, we can calculate the pH of a buffer using the Henderson–Hasselbalch equation.
[ A − ] pH = p K a + log
[ HA]
The equilibrium constant for reaction 9.2 is large (K = K a /K w = 1.75 × 10 9 ), so we can treat the reaction as one that goes to completion. Before the equivalence point, the concentration of unreacted acetic acid is
moles unreacted CH COOH 3 MV aa − MV bb
[CH COOH] 3 =
total volume
and the concentration of acetate is
[CH COO ] 3 =
moles NaOH added
MV bb
total volume
V a + V b For example, after adding 10.0 mL of NaOH the concentrations of CH 3 COOH and
CH 3 COO – are
(. 0 100 M )( . 50 0 mL ) − (. 0 100 M )( . 10 0 mL )
[ CH COOH] 3 =
. 50 0 mL + . 10 0 mL giving a pH of
A similar calculation shows that the pH after adding 20.0 mL of NaOH is 4.58. At the equivalence point, the moles of acetic acid initially present and the moles
of NaOH added are identical. Since their reaction effectively proceeds to comple- tion, the predominate ion in solution is CH 3 COO – , which is a weak base. To calcu- late the pH we first determine the concentration of CH 3 COO – .
[ CH COO 3 ] =
moles CH COOH 3 (. 0 100 M )( . 50 0 mL )
= . 0 0500 M
total volume
. 50 0 mL + . 50 0 mL
Chapter 9 Titrimetric Methods of Analysis
CH 3 COO – (aq) + H 2 O(l) t OH – (aq) + CH 3 COOH(aq)
[ OH − ][ CH COOH] 3 ( )( ) xx
x = [OH – ] = 5.34 × 10 –6 M The concentration of H 3 O + , therefore, is 1.87 × 10 –9 , or a pH of 8.73.
After the equivalence point NaOH is present in excess, and the pH is deter- mined in the same manner as in the titration of a strong acid with a strong base. For example, after adding 60.0 mL of NaOH, the concentration of OH – is
giving a pH of 11.96. Table 9.3 and Figure 9.6 show additional results for this titra- tion. The calculations for the titration of a weak base with a strong acid are handled in a similar manner except that the initial pH is determined by the weak base, the pH at the equivalence point by its conjugate weak acid, and the pH after the equiva- lence point by the concentration of excess strong acid.
Table 9.3 14.00
Data for Titration of 50.0 mL of 0.100 M Acetic Acid with 0.100 M
Volume of NaOH
48.00 6.14 Volume NaOH (mL)
50.00 8.73 Figure 9.6
52.00 11.29 Titration curve for 50.0 mL of 0.100 M acetic acid (pK a = 4.76) with
55.00 11.68 0.100 M NaOH. 60.00 11.96 65.00 12.12 70.00 12.22 75.00 12.30 80.00 12.36 85.00 12.41 90.00 12.46 95.00 12.49
284 Modern Analytical Chemistry
The approach that we have worked out for the titration of a monoprotic weak acid with a strong base can be extended to reactions involving multiprotic acids or bases and mixtures of acids or bases. As the complexity of the titration increases, however, the necessary calculations become more time-consuming. Not surpris-
ingly, a variety of algebraic 1 and computer spreadsheet 2 approaches have been de-
scribed to aid in constructing titration curves. Sketching an Acid–Base Titration Curve To evaluate the relationship between an
equivalence point and an end point, we only need to construct a reasonable approx- imation to the titration curve. In this section we demonstrate a simple method for sketching any acid–base titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible.
To quickly sketch a titration curve we take advantage of the following observa- tion. Except for the initial pH and the pH at the equivalence point, the pH at any point of a titration curve is determined by either an excess of strong acid or strong base, or by a buffer consisting of a weak acid and its conjugate weak base. As we have seen in the preceding sections, calculating the pH of a solution containing ex- cess strong acid or strong base is straightforward.
We can easily calculate the pH of a buffer using the Henderson–Hasselbalch equa- tion. We can avoid this calculation, however, if we make the following assumption. You may recall that in Chapter 6 we stated that a buffer operates over a pH range ex-
tending approximately ± 1 pH units on either side of the buffer’s pK a . The pH is at the lower end of this range, pH = pK a – 1, when the weak acid’s concentration is approxi- mately ten times greater than that of its conjugate weak base. Conversely, the buffer’s pH is at its upper limit, pH = pK a + 1, when the concentration of weak acid is ten times less than that of its conjugate weak base. When titrating a weak acid or weak base, therefore, the buffer region spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume.*
Our strategy for quickly sketching a titration curve is simple. We begin by draw- ing our axes, placing pH on the y-axis and volume of titrant on the x-axis. After calcu- lating the volume of titrant needed to reach the equivalence point, we draw a vertical line that intersects the x-axis at this volume. Next, we determine the pH for two vol- umes before the equivalence point and for two volumes after the equivalence point. To save time we only calculate pH values when the pH is determined by excess strong acid or strong base. For weak acids or bases we use the limits of their buffer region to esti- mate the two points. Straight lines are drawn through each pair of points, with each line intersecting the vertical line representing the equivalence point volume. Finally, a smooth curve is drawn connecting the three straight-line segments. Example 9.1 illus- trates this approach for the titration of a weak acid with a strong base.
EXAMPLE 9.1
Sketch the titration curve for the titration of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. This is the same titration for which we previously calculated the titration curve (Table 9.3 and Figure 9.6).
SOLUTION
We begin by drawing the axes for the titration curve (Figure 9.7a). We have already shown that the volume of NaOH needed to reach the equivalence point is 50 mL, so we draw a vertical line intersecting the x-axis at this volume (Figure 9.7b).
Chapter 9 Titrimetric Methods of Analysis
Percent titrated Percent titrated
Volume of titrant Volume of titrant
(a)
(b)
Percent titrated Percent titrated
Volume of titrant Volume of titrant
(c)
(d)
Percent titrated Percent titrated
Volume of titrant Volume of titrant
How to sketch an acid–base titration curve; see text for explanation.
286 Modern Analytical Chemistry
Before the equivalence point the titrant is the limiting reagent, and the pH is controlled by a buffer consisting of unreacted acetic acid and its conjugate weak base, acetate. The pH limits for the buffer region are plotted by superimposing the ladder diagram for acetic acid on the y-axis (Figure 9.7c) and adding the appropriate points at 10% (5.0 mL) and 90% (45.0 mL) of the equivalence point volume.
After the equivalence point the pH is controlled by the concentration of excess NaOH. Again, we have already done this calculation. Using values from Table 9.3, we plot two additional points.
An approximate sketch of the titration curve is completed by drawing separate straight lines through the two points in the buffer region and the two points in the excess titrant region (Figure 9.7e). Finally, a smooth curve is drawn connecting the three straight-line segments (Figure 9.7f).
This approach can be used to sketch titration curves for other acid–base titra- tions including those involving polyprotic weak acids and bases or mixtures of weak acids and bases (Figure 9.8). Figure 9.8a, for example, shows the titration curve
when titrating a diprotic weak acid, H 2 A, with a strong base. Since the analyte is
Percent titrated
Volume of titrant
(a)
Percent titrated
Sketches of titration curves for (a) 50.00 mL
of 0.0500 M diprotic weak acid (pK a1 = 3,
pK a2 = 7) with 0.100 M strong base; and
(b) 50.00 mL of a mixture of weak acids consisting of 0.075 M HA (pK a,HA = 3) and
0.025 M HB (pK a,HB = 7) with 0.100 M
strong base. The points used to sketch the titration curves are indicated by the dots (•).
Equivalence points are indicated by the
Volume of titrant
Chapter 9 Titrimetric Methods of Analysis
diprotic there are two equivalence points, each requiring the same volume of titrant. Before the first equivalence point the pH is controlled by a buffer consisting
of H 2 A and HA – , and the HA – /A 2– buffer determines the pH between the two equiv- alence points. After the second equivalence point, the pH reflects the concentration of the excess strong base titrant.
Figure 9.8b shows a titration curve for a mixture consisting of two weak acids: HA and HB. Again, there are two equivalence points. In this case, however, the equivalence points do not require the same volume of titrant because the concen- tration of HA is greater than that for HB. Since HA is the stronger of the two weak acids, it reacts first; thus, the pH before the first equivalence point is controlled by the HA/A – buffer. Between the two equivalence points the pH reflects the titration of HB and is determined by the HB/B – buffer. Finally, after the second equivalence point, the excess strong base titrant is responsible for the pH.
9B.2 Selecting and Evaluating the End Point
Earlier we made an important distinction between an end point and an equivalence point. The difference between these two terms is important and deserves repeating. The equivalence point occurs when stoichiometrically equal amounts of analyte and titrant react. For example, if the analyte is a triprotic weak acid, a titration with NaOH will have three equivalence points corresponding to the addition of one, two, and three moles of OH – for each mole of the weak acid. An equivalence point, therefore, is a theoretical not an experimental value.
An end point for a titration is determined experimentally and represents the analyst’s best estimate of the corresponding equivalence point. Any difference be- tween an equivalence point and its end point is a source of determinate error. As we shall see, it is even possible that an equivalence point will not have an associated end point.
Where Is the Equivalence Point? We have already learned how to calculate the equivalence point for the titration of a strong acid with a strong base, and for the titration of a weak acid with a strong base. We also have learned to sketch a titra- tion curve with a minimum of calculations. Can we also locate the equivalence point without performing any calculations? The answer, as you may have guessed, is often yes!
It has been shown 3 that for most acid–base titrations the inflection point,
which corresponds to the greatest slope in the titration curve, very nearly coincides with the equivalence point. The inflection point actually precedes the equivalence point, with the error approaching 0.1% for weak acids or weak bases with dissocia-
tion constants smaller than 10 , or for very dilute solutions. Equivalence points de-
12.0 (f)
termined in this fashion are indicated on the titration curves in Figure 9.8.
10.0 (e)
The principal limitation to using a titration curve to locate the equivalence
8.0 (d)
point is that an inflection point must be present. Sometimes, however, an inflection
pH 6.0 (c)
point may be missing or difficult to detect. Figure 9.9, for example, demonstrates
4.0 (b) (a)
the influence of the acid dissociation constant, K a , on the titration curve for a weak
acid with a strong base titrant. The inflection point is visible, even if barely so, for
acid dissociation constants larger than 10 –9 , but is missing when K a is 10 –11
Volume of titrant
Another situation in which an inflection point may be missing or difficult to detect occurs when the analyte is a multiprotic weak acid or base whose successive
Figure 9.9
dissociation constants are similar in magnitude. To see why this is true let’s con- Titration curves for 50.00 mL of 0.100 M
weak acid with 0.100 M strong base. The
sider the titration of a diprotic weak acid, H 2 A, with NaOH. During the titration the
pK a s of the weak acids are (a) 1, (b) 3, (c) 5,
288 Modern Analytical Chemistry
H 2 A(aq) + OH – (aq) → HA – (aq) + H 2 O(l)
HA – (aq) + OH – (aq) → A 2– (aq) + H 2 O(l)
12.0 Maleic acid
14.0 Two distinct inflection points are seen if reaction 9.3 is essentially complete
10.0 before reaction 9.4 begins.
8.0 Figure 9.10 shows titration curves for three diprotic weak acids. The titra-
pH 6.0
4.0 tion curve for maleic acid, for which K a1 is approximately 20,000 times larger
2.0 than K a2 , shows two very distinct inflection points. Malonic acid, on the other
0.00 20.00 40.00 60.00 80.00 hand, has acid dissociation constants that differ by a factor of approximately
690. Although malonic acid’s titration curve shows two inflection points, the
Volume of titrant
(a)
first is not as distinct as that for maleic acid. Finally, the titration curve for suc- cinic acid, for which the two K a values differ by a factor of only 27, has only a single inflection point corresponding to the neutralization of HC 4 H 4 O 4 14.0 – to
12.0 Malonic acid
C 4 H 4 O 4 2– . In general, separate inflection points are seen when successive acid
8.0 dissociation constants differ by a factor of at least 500 (a ∆ pK a of at least 2.7).
pH 6.0 4.0
2.0 Finding the End Point with a Visual Indicator One interesting group of
0.0 weak acids and bases are derivatives of organic dyes. Because such com-
0.00 20.00 40.00 60.00 80.00 pounds have at least one conjugate acid–base species that is highly colored,
Volume of titrant
their titration results in a change in both pH and color. This change in color
(b)
can serve as a useful means for determining the end point of a titration, pro- vided that it occurs at the titration’s equivalence point.
14.0 12.0 Succinic acid
The pH at which an acid–base indicator changes color is determined by
10.0 its acid dissociation constant. For an indicator that is a monoprotic weak
pH 6.0 acid, HIn, the following dissociation reaction occurs
HIn(aq) + H 2 O(l) H 3 O t + (aq) + In – (aq)
0.00 20.00 40.00 60.00 80.00 for which the equilibrium constant is
Volume of titrant
Titration curves for (a) maleic acid,
Taking the negative log of each side of equation 9.5, and rearranging to solve for pH
pK a1 = 1.91, pK a2 = 6.33; (b) malonic acid,
gives a familiar equation.
pK a1 = 2.85, pK a2 = 5.70; (c) succinic acid, pK a1 = 4.21, pK a2 = 5.64. Titration curves are
for 50.00 mL of 0.0500 M acid with 0.100 M
[ HIn ]
strong base. Equivalence points for all three
titrations occur at 25.00 and 50.00 mL of titrant.
The two forms of the indicator, HIn and In – , have different colors. The color of
a solution containing an indicator, therefore, continuously changes as the concen- tration of HIn decreases and the concentration of In – increases. If we assume that both HIn and In – can be detected with equal ease, then the transition between the two colors reaches its midpoint when their concentrations are identical or when the
pH is equal to the indicator’s pK a . The equivalence point and the end point coin- cide, therefore, if an indicator is selected whose pK a is equal to the pH at the equiva- lence point, and the titration is continued until the indicator’s color is exactly halfway between that for HIn and In – . Unfortunately, the exact pH at the equiva- lence point is rarely known. In addition, detecting the point where the concentra- tions of HIn and In – are equal may be difficult if the change in color is subtle.
We can establish a range of pHs over which the average analyst will observe a change in color if we assume that a solution of the indicator is the color of HIn whenever its concentration is ten times more than that of In – , and the color of In –
Chapter 9 Titrimetric Methods of Analysis
whenever the concentration of HIn is ten times less than that of In – . Substi-
pH
tuting these inequalities into equation 9.6
In –
pH = p K a + log
= p K a − 1 pH = p K a,HIn +1
Indicator’s
pH = p K a,HIn
color transition
pH = p K a,HIn –1
shows that an indicator changes color over a pH range of ± 1 units on either
HIn
side of its pK a (Figure 9.11). Thus, the indicator will be the color of HIn when
the pH is less than pK a – 1, and the color of In – for pHs greater than pK a + 1.
Figure 9.11
The pH range of an indicator does not have to be equally distributed on either
Ladder diagram showing the range of pH
side of the indicator’s pK a . For some indicators only the weak acid or weak base is col-
levels over which a typical acid–base
ored. For other indicators both the weak acid and weak base are colored, but one indicator changes color. form may be easier to see. In either case, the pH range is skewed toward those pH lev-
els for which the less colored form of the indicator is present in higher concentration.
A list of several common acid–base indicators, along with their pK a s, color
changes, and pH ranges, is provided in the top portion of Table 9.4. In some cases,
Table 9.4 Properties of Selected Indicators, Mixed Indicators,
and Screened Indicators for Acid–Base Titrations
Indicator
Acid Color
Base Color
pH Range
pK a
cresol red
thymol blue
bromophenol blue
methyl orange
Congo red
bromocresol green
methyl red
bromocresol purple
bromothymol blue
phenol red
cresol red
thymol blue
alizarin yellow R
yellow
orange/red
Mixed Indicator
Acid Color
Base Color
pH Range
bromocresol green and methyl orange
bromocresol green and chlorophenol red
bromothymol blue and phenol red
Screened Indicator
Acid Color
Base Color
pH Range
dimethyl yellow and methylene blue
methyl red and methylene blue
neutral red and methylene blue
violet-blue
green
290 Modern Analytical Chemistry
Bromothymol blue
Titration curve for 50.00 mL of 0.100 M CH 3 COOH with 0.100 M NaOH showing the
range of pHs and volumes of titrant over which the indicators bromothymol blue and
phenolphthalein are expected to change 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 color.
Volume of titrant
mixed indicators, which are a mixture of two or more acid–base indicators, provide
a narrower range of pHs over which the color change occurs. A few examples of such mixed indicators are included in the middle portion of Table 9.4. Adding a neutral screening dye, such as methylene blue, also has been found to narrow the pH range over which an indicator changes color (lower portion of Table 9.4). In this case, the neutral dye provides a gray color at the midpoint of the indicator’s color transition.
The relatively broad range of pHs over which any indicator changes color places additional limitations on the feasibility of a titration. To minimize a determi- nate titration error, an indicator’s entire color transition must lie within the sharp transition in pH occurring near the equivalence point. Thus, in Figure 9.12 we see that phenolphthalein is an appropriate indicator for the titration of 0.1 M acetic acid with 0.1 M NaOH. Bromothymol blue, on the other hand, is an inappropriate indicator since its change in color begins before the initial sharp rise in pH and, as a result, spans a relatively large range of volumes. The early change in color increases the probability of obtaining inaccurate results, and the range of possible end point volumes increases the probability of obtaining imprecise results.
The need for the indicator’s color transition to occur in the sharply rising por- tion of the titration curve justifies our earlier statement that not every equivalence point has an end point. For example, trying to use a visual indicator to find the first equivalence point in the titration of succinic acid (see Figure 9.10c) is pointless since any difference between the equivalence point and the end point leads to a large titration error.
Finding the End Point by Monitoring pH An alternative approach to finding a titration’s end point is to monitor the titration reaction with a suitable sensor whose signal changes as a function of the analyte’s concentration. Plotting the data gives us the resulting titration curve. The end point may then be determined from the titration curve with only a minimal error.
The most obvious sensor for an acid–base titration is a pH electrode.* For ex- ample, Table 9.5 lists values for the pH and volume of titrant obtained during the titration of a weak acid with NaOH. The resulting titration curve, which is called a potentiometric titration curve, is shown in Figure 9.13a. The simplest method for finding the end point is to visually locate the inflection point of the titration curve. This is also the least accurate method, particularly if the titration curve’s slope at the equivalence point is small.
Chapter 9 Titrimetric Methods of Analysis
Table 9.5 Data for the Titration of a Weak Acid with
0.100 M NaOH Normal Titration
First Derivative
Second Derivative
(mL) pH
(mL)
∆ V (mL)
Another method for finding the end point is to plot the first or second deriva- tive of the titration curve. The slope of a titration curve reaches its maximum value at the inflection point. The first derivative of a titration curve, therefore, shows a separate peak for each end point. The first derivative is approximated as ∆ pH/ ∆ V, where ∆ pH is the change in pH between successive additions of titrant. For exam- ple, the initial point in the first derivative titration curve for the data in Table 9.5 is
and is plotted at the average of the two volumes (1.00 mL). The remaining data for
292 Modern Analytical Chemistry
Volume of titrant (mL)
Volume of titrant (mL)
Titration curves for a weak acid with
10 11 12 13 14 15 16 curve; (b) first derivative titration curve;
0.100 M NaOH—(a) normal titration
Volume of titrant (mL)
(c) second derivative titration curve;
Volume of titrant (mL)
(d) Gran plot.
(c)
(d)
The second derivative of a titration curve may be more useful than the first de- rivative, since the end point is indicated by its intersection with the volume axis. The second derivative is approximated as ∆ ( ∆ pH/ ∆ V)/ ∆
∆ V, or 2 pH/ ∆ V 2 . For the titration data in Table 9.5, the initial point in the second derivative titration curve is
and is plotted as the average of the two volumes (2.00 mL). The remainder of the data for the second derivative titration curve are shown in Table 9.5 and plotted in Figure 9.13c.
Derivative methods are particularly well suited for locating end points in multi- protic and multicomponent systems, in which the use of separate visual indicators for each end point is impractical. The precision with which the end point may be located also makes derivative methods attractive for the analysis of samples with poorly defined normal titration curves.
Derivative methods work well only when sufficient data are recorded during the sharp rise in pH occurring near the equivalence point. This is usually not a problem when the titration is conducted with an automatic titrator, particularly when operated under computer control. Manual titrations, however, often contain only a few data points in the equivalence point region, due to the limited range of volumes over which the transition in pH occurs. Manual titrations are, however, information-rich during the more gently rising portions of the titration curve be- fore and after the equivalence point.
Consider again the titration of a monoprotic weak acid, HA, with a strong base. At any point during the titration the weak acid is in equilibrium with H 3 O + and A –
HA(aq) + H 2 O(l) H 3 O + (aq) + A t –
Chapter 9 Titrimetric Methods of Analysis
for which
[ HO 3 + ][ A − ] K a =
[ HA]
Before the equivalence point, and for volumes of titrant in the titration curve’s buffer region, the concentrations of HA and A – are given by the following equations.
moles HA − moles OH added −
MV aa − MV bb
[HA] =
total volume
moles OH added −
MV bb
total volume
Substituting these equations into the K a expression for HA, and rearranging gives [ HO 3 + ]( MV bb )/( V a + V b )
Finally, recognizing that the equivalence point volume is
V eq =
MV aa
leaves us with the following equation.
V b × [H 3 O + ]=K a × V eq –K a × V b For volumes of titrant before the equivalence point, a plot of V b × [H 3 O + ] versus V b
is a straight line with an x-intercept equal to the volume of titrant at the end point and a slope equal to –K a .* Results for the data in Table 9.5 are shown in Table 9.6 and plotted in Figure 9.13d. Plots such as this, which convert a portion of a titration curve into a straight line, are called Gran plots.
Finding the End Point by Monitoring Temperature The reaction between an acid
Gran plot
and a base is exothermic. Heat generated by the reaction increases the temperature
A linearized form of a titration curve.
of the titration mixture. The progress of the titration, therefore, can be followed by monitoring the change in temperature.
An idealized thermometric titration curve (Figure 9.14a) consists of three distinct linear regions. Before adding titrant, any change in temperature is due to the cooling or warming of the solution containing the analyte. Adding titrant ini- tiates the exothermic acid–base reaction, resulting in an increase in temperature. This part of a thermometric titration curve is called the titration branch. The
294 Modern Analytical Chemistry
Table 9.6 Gran Plot Treatment of the Data
in Table 9.5
Titration branch
Excess titration
T emperature
V titr
temperature continues to rise with each addition of titrant until the equivalence
point is reached. After the equivalence point, any change in temperature is due to
emperature
the difference between the temperatures of the analytical solution and the titrant, and the enthalpy of dilution for the excess titrant. Actual thermometric titration curves (Figure 9.14b) frequently show curvature at the intersection of the titration
0 branch and the excess titrant branch due to the incompleteness of the neutraliza-
V titr tion reaction, or excessive dilution of the analyte during the titration. The latter
problem is minimized by using a titrant that is 10–100 times more concentrated
(b)
than the analyte, although this results in a very small end point volume and a larger relative error.
Figure 9.14
The end point is indicated by the intersection of the titration branch and
Thermometric titration curves—(a) ideal;
(b) showing curvature at the intersection of
the excess titrant branch. In the idealized thermometric titration curve (see
the titration and excess titrant branches.
Figure 9.14a) the end point is easily located. When the intersection between
Equivalence points are indicated by the dots (•).
the two branches is curved, the end point can be found by extrapolation (Figure 9.14b).
Although not commonly used, thermometric titrations have one distinct ad- vantage over methods based on the direct or indirect monitoring of pH. As dis- cussed earlier, visual indicators and potentiometric titration curves are limited by the magnitude of the relevant equilibrium constants. For example, the titra-
tion of boric acid, H 3 BO 3 , for which K a is 5.8 × 10 –10 , yields a poorly defined equivalence point (Figure 9.15a). The enthalpy of neutralization for boric acid with NaOH, however, is only 23% less than that for a strong acid (–42.7 kJ/mol
2.0 T emperature 24.980
Figure 9.15
0.00 1.00 2.00 3.00 4.00 5.00 6.00 Volume of titrant
Titration curves for 50.00 mL of 0.0100 M
Volume of titrant
H 3 BO 3 with 0.100 M NaOH determined by
Chapter 9 Titrimetric Methods of Analysis
for H 3 BO 3 versus –55.6 kJ/mol for HCl), resulting in a favorable thermometric titration curve (Figure 9.15b).
9B.3 Titrations in Nonaqueous Solvents
Thus far we have assumed that the acid and base are in an aqueous solution. In- deed, water is the most common solvent in acid–base titrimetry. When considering the utility of a titration, however, the solvent’s influence cannot be ignored.
The dissociation, or autoprotolysis constant for a solvent, SH, relates the con-
centration of the protonated solvent, SH 2 + , to that of the deprotonated solvent, S – .
For amphoteric solvents, which can act as both proton donors and proton accep- tors, the autoprotolysis reaction is
2SH
t – SH 2 +S
with an equilibrium constant of
K s = [SH 2 + ][S – ]
You should recognize that K w is just the specific form of K s for water. The pH of a solution is now seen to be a general statement about the relative abundance of pro- tonated solvent
pH = –log[SH 2 + ]
where the pH of a neutral solvent is given as
1 pH neut = p K s
Perhaps the most obvious limitation imposed by K s is the change in pH during
a titration. To see why this is so, let’s consider the titration of a 50 mL solution of
10 –4 M strong acid with equimolar strong base. Before the equivalence point, the pH is determined by the untitrated strong acid, whereas after the equivalence point the concentration of excess strong base determines the pH. In an aqueous solution
the concentration of H 3 O + when the titration is 90% complete is
MV MV
HO ] = aa bb
( 1 × 10 − 4 M)(50 mL) −× ( 1 10 − 4 M)(45 mL)
= 53 . × 10 − 6 M
corresponding to a pH of 5.3. When the titration is 110% complete, the concentra- tion of OH – is
MV bb − MV aa [ OH ] =
( 1 × 10 − 4 M)(55 mL) −× ( 1 10 − 4 M)(50 mL)
= 48 . × 10 − 6 M
or a pOH of 5.3. The pH, therefore, is pH = pK w – pOH = 14.0 – 5.3 = 8.7 The change in pH when the titration passes from 90% to 110% completion is
296 Modern Analytical Chemistry
20.0 If the same titration is carried out in a nonaqueous solvent with a K s of 1.0 × 10 15.0 –20 (b) , the pH when the titration is 90% complete is still 5.3. However, the pH when the titration is 110% complete is now
pH 10.0 (a)
pH = pK s – pOH = 20.0 – 5.3 = 14.7
0.0 In this case the change in pH of
Volume of titrant
Figure 9.16
is significantly greater than that obtained when the titration is carried
Titration curves for 50.00 mL of 10 –4 M HCl
out in water. Figure 9.16 shows the titration curves in both the aque-
with 10 –4 M NaOH in (a) water,
ous and nonaqueous solvents. Nonaqueous solvents also may be used
K w =l × 10 –14 , and (b) nonaqueous solvent,
to increase the change in pH when titrating weak acids or bases (Fig-
K s =1 × 10 –20 .
ure 9.17).
Another parameter affecting the feasibility of a titration is the dis- sociation constant of the acid or base being titrated. Again, the solvent
(b)
plays an important role. In the Brønsted–Lowry view of acid–base be-
15.0 havior, the strength of an acid or base is a relative measure of the ease with which a proton is transferred from the acid to the solvent, or
pH 10.0 (a)
from the solvent to the base. For example, the strongest acid that can
5.0 exist in water is H
3 O + . The acids HCl and HNO 3 are considered
0.0 strong because they are better proton donors than H 3 O + . Strong acids
essentially donate all their protons to H 2 O, “leveling” their acid
Volume of titrant
strength to that of H 3 O + . In a different solvent HCl and HNO 3 may
Figure 9.17
not behave as strong acids.
Titration curves for 50.00 mL of 0.100 M
When acetic acid, which is a weak acid, is placed in water, the dis-
weak acid (pK a = 11) with 0.100 M NaOH in (a) water, K w =1 × 10 –14 ; and
sociation reaction
(b) nonaqueous solvent, K s =1 × 10 –20 . The
CH 3 COOH(aq) + H 2 O(l) t H 3 O + (aq) + CH 3 COO – (aq)
titration curve in (b) assumes that the
change in solvent has no effect on the acid dissociation constant of the weak acid.
does not proceed to a significant extent because acetate is a stronger base than water and the hydronium ion is a stronger acid than acetic acid. If acetic acid is placed in a solvent that is a stronger base than water, such as ammonia, then the reaction
leveling
CH 3 COOH + NH 3 NH 4 t + + CH 3 COO –
Acids that are better proton donors than the solvent are leveled to the acid
proceeds to a greater extent. In fact, HCl and CH 3 COOH are both strong acids in
strength of the protonated solvent; bases
ammonia.
that are better proton acceptors than the
All other things being equal, the strength of a weak acid increases if it is placed
solvent are leveled to the base strength of
in a solvent that is more basic than water, whereas the strength of a weak base in-
the deprotonated solvent.
creases if it is placed in a solvent that is more acidic than water. In some cases, how- ever, the opposite effect is observed. For example, the pK b for ammonia is 4.76 in water and 6.40 in the more acidic glacial acetic acid. In contradiction to our expec- tations, ammonia is a weaker base in the more acidic solvent. A full description of
the solvent’s effect on a weak acid’s pK a or on the pK b of a weak base is beyond the scope of this text. You should be aware, however, that titrations that are not feasible in water may be feasible in a different solvent.