8.4 Acid Base Titrations

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Acid–Base Equilibrium 59NEL

8.4 8.4 Acid–Base Titration

In your previous chemistry course, you became familiar with acid–base titrations. A

titration (Figure 1) is a chemical analysis involving the progressive addition of a solu-

tion of known solute concentration, called the titrant, into a solution of unknown con-

centration, called the sample. The purpose is to determine the amount of a specified

chemical in the sample, from which the molar mass and the concentration of the chem-

ical may be determined. This is possible because the titrant and the sample contain sub-

stances that react according to known stoichiometry. In general, the sample is placed in

a receiving flask, and the titrant is dispensed from a buret.

Before the sample can be analyzed, it is important that we know, to a considerable

degree of accuracy, the concentration of the titrant, because this concentration is used

to calculate the concentration of the sample. Measuring the titrant’s concentration is

called “standardizing” the titrant, and is often the first stage of a titration. Common

primary standards are sodium carbonate, Na2CO3(s) (a base used to standardize an acid

titrant), and potassium hydrogen phthalate, KHC7H4O4(s) (an acid used to standardize

a basic titrant). These are appropriate choices because, due to their purity, we can be

confident that their stated concentrations are accurate. The standardization process is itself a titration.

Hydrochloric acid and sodium hydroxide are not used as primary standards because

hydrogen chloride gas vaporizes, especially from concentrated hydrochloric acid solu-

tions, and solid sodium hydroxide is hygroscopic—it gains mass by absorbing water

from the air.

Notice that the primary standards are solids at SATP; they are not hygroscopic like

sodium hydroxide, and they do not vaporize like hydrochloric acid. They are available

in very pure form, and produce colourless aqueous solutions.

titration the precise addition of a

solution in a buret into a measured

volume of a sample solution

titrant the solution in a buret durin

a titration

sample the solution being analyze

in a titration

midway

volume of

titrant used

= V f - V iV f

V i

stopcock

endpoint

(indicator changes colour)

start

sample

titrant

buret

receiving

flask Figure 1

In an acid–base titration, the con-

centration of an acid or base solu-

tion of unknown concentration is

determined by the delivery (from a

buret) of a measured volume of a

solution of known concentration(the titrant). If the sample in the

flask is an acid, the titrant used is

base, and vice versa.

primary standard a chemical,

available in a pure and stable form

for which an accurate concentratio

can be prepared; the solution isthen used to determine precisely, b

means of titrating, the concentratio

of a titrant



596 Chapter 8 NEL

An acid–base titration involves the reaction between an acid and a base. In a typical

titration, a measured volume of standardized titrant is added to a known volume of the

sample. The addition continues until the amount of reactant in the sample is just con-

sumed by the reactant in the titrant. This is called the equivalence point or the stoi-

chiometric point.

Before beginning an acid–base titration, a drop or two of an acid–base indicator is

added to the sample. The acid–base indicator signals the end of the titration by sharply

and permanently changing colour when the equivalence point is reached. At this point,the volume of titrant added is recorded, and the number of moles of titrant used to

reach the equivalence point is calculated. Ideally, an acid–base indicator is chosen such

that the endpoint occurs precisely at the equivalence point (so the colour change occurs

sharply at the point where a complete reaction is attained). However, it is virtually impos-

sible to achieve such precision in the laboratory. Consequently, titrations are subject to

considerable experimental error. Bromothymol blue and phenolphthalein are common

indicators used in acid–base titrations. Bromothymol blue changes from yellow to blue

in the range pH 6.0 to 7.6. Phenolphthalein changes from colourless to pink in the range

pH 8.2 to 10.0.

Titrating a Strong Acid with a Strong BaseNo doubt you are familiar with the titration of a strong acid with a strong base from

previous chemistry courses. Consider the reaction between hydrochloric acid, HCl(aq),

a strong acid, with sodium hydroxide, NaOH(aq), a strong base:

HCl(aq) NaOH(aq) → H2O(l) NaCl(aq) (molecular equation)

H3O

(aq) OH

(aq) → 2 H2O(l) (net ionic equation)

H

(aq) OH

(aq) → H2O(l) (abbreviated net ionic equation)

What chemical changes occur during a reaction such as this?

The stoichiometric analysis that follows a titration usually involves the average of at

least three consistent titration trials. Chemists demand high reproducibility from titra-

tion results. Equivalence points that are more than ±0.2 mL from a set of consistentresults are recorded but not included in the average volume of titrant used.Titrations must

involve reactions that obey the assumptions required of stoichiometric calculations—

the reactions must be stoichiometric (reactants react according to the ratio of coeffi-

cients in the reaction equation), spontaneous,fast, and quantitative (the reaction proceeds

until all reacting entities are consumed).

endpoint the point in a titration at

which a sharp change in a measur-

able and characteristic property

occurs; e.g, a colour change in an

acid–base indicator

In a titration, 20.00 mL of 0.300 mol/L HCl (aq) is titrated with standardized

0.300 mol/L NaOH (aq). What is the amount of unreacted HCl (aq) and the pH of the

solution after the following volumes of NaOH (aq) have been added?

(a) 0 mL(b) 10.0 mL

(c) 20.0 mL

Since HCl(aq) is a strong acid, the major entities in the sample solution are

H

(aq), Cl(aq), and H2O(l).

(a) Before adding titrant (NaOH(aq)), the pH of the solution is equal to the pH of

0.300 mol/L HCl(aq).

Chemical Changes During a Strong Acid/Strong Base Titration SAMPLE problem

equivalence point in a titration,

the measured quantity of titrant

recorded at the point at which

chemically equivalent amounts have

reacted




Section 8

Since HCl(aq) ionizes completely,

[H

(aq)] 0.300 mol/L

pH log(0.300)

pH 0.5

Now, we calculate the amount of unreacted HCl(aq), n HCl.

V HCl 20.00 mL

C HCl 0.300 mol/L

n HCl v HCl C HCl

20.00 mL 0.300 mol/L

n HCl 6.00 mmol

Since HCl is a strong acid, 6.00 mmol HCl(aq) contains 6.00 mmol H(aq) and

6.00 mmol Cl(aq).

(b) Adding 10.00 mL of 0.300 mol/LNaOH(aq) introduces the following amount of

NaOH(aq) to the solution.

V NaOH 10.00 mL

C NaOH 0.300 mol/L NaOH(aq)

n NaOH v NaOH C NaOH

10.00 mL 0.300 mol/L

n NaOH 3.00 mmol

Since NaOH is a strong base, 3.00 mmol NaOH(aq) introduces 3.00 mmol Na

(aq) and

3.00 mmol OH(aq) into the solution.

Before reaction, the solution contains the following amounts of the major entities

(in addition to H2O(l)):

From HCl:

H

(aq) Cl

(aq)

6.00 mmol 6.00 mmol

From NaOH:

OH

(aq) Na

(aq)

3.00 mmol 3.00 mmol

Hydrogen ions react with hydroxide ions, according to the neutralization reaction:

H

(aq) OH

(aq) → H2O(l)

Before reaction: 6.00 mmol 3.00 mmol

After reaction: 6.00 mmol 3.00 mmol 3.00 mmol 3.00 mmol

3.00 mmol 0 mmol

(excess)

Sodium ions and chloride ions remain in solution. As you learned in Section 8.3,

Na(aq) does not hydrolyze and thus cannot affect the pH of an aqueous solution.

Chloride ions also do not affect the pH because as the conjugate bases of the strong

acid, HCl, they do not hydrolyze.

Notice that the total volume of the solution after the addition of 10 mLNaOH(aq) is

20.00 mL 10.00 mL 30.00 mL.



598 Chapter 8 NEL

The pH of the sample is calculated using the amount of excess H(aq) and the total

volume of the sample:

[H

(aq)] 3

3

.0

0

0

.0

m0 m

m

L

ol

[H

(aq)] 0.100 mol/L

We can now calculate the pH:

pH log[H

(aq)]

log(0.100 mol/L)

pH 1.000

The pH rises as more and more titrant is added to the receiving flask. Table 1 records

the results of calculations like those above for successive additions of base to the mixture.

(c) From Table 1 you can see that, when 20.00 mLof NaOH(aq) has been added, all of

the HCl(aq) is neutralized and the equivalence point is reached. The resulting solution

contains only H2O, Na(aq), and Cl(aq) ions. Earlier, you learned that neither of these

ions is able to hydrolyze. Therefore, the concentration of hydrogen ions is equal to

that in pure water, 1.0 107 mol/L, and the pH is 7.00.

If we were to add more NaOH(aq) to the receiving flask, we would simply be adding

more Na(aq) and OH

(aq) ions. The pH of the resulting solution is determined by the

amount of OH(aq) added, taking into account the increase in volume of the solution.

Table 1 Titration of 20.00 mL of 0.300 mol/L HCl(aq) with 0.300 mol/L NaOH(aq)

Volume Total Molar

Initial Initial of Amount Amount Volume Concentration

Volume Amount NaOH(aq)

of of Excess of of Ion in

of HCl(aq) of HCl(aq) Added NaOH(aq) Reagent Solution Excess

(mL) (mol) (mL) (mol) (mol) (mL) (mol/L) pH

20.00 6.000 103 0 0 6.000 103 (H) 20.00 0.3000 (H) 0.52

20.00 6.000 103 5.00 1.500 103 4.500 103 (H) 25.00 0.1400 (H) 0.85

20.00 6.000 103 15.00 4.500 103 1.500 103 (H) 35.00 4.286 102 (H) 1.36

20.00 6.000 103 19.00 5.700 103 3.000 104 (H) 39.00 7.692 103 (H) 2.11

20.00 6.000 103 19.90 5.970 103 3.000 105 (H) 39.90 7.519 104 (H) 3.12

20.00 6.000 103 19.99 5.997 103 3.000 106 (H) 39.99 7.502 105 (H) 4.12

20.00 6.000 103 20.00 6.000 103 0 40.00 0 7.00

20.00 6.000 103 20.01 6.003 103 3.000 106 (OH) 40.01 7.498 105 (OH) 9.87

20.00 6.000 10

3 20.10 6.030 10

3 3.000 10

5 (OH

) 40.10 7.481 10

4 (OH

) 10.87

20.00 6.000 103 21.00 6.300 103 3.000 104 (OH) 41.00 7.317 103 (OH) 11.86

20.00 6.000 103 40.00 1.200 102 6.000 103 (OH) 80.00 7.500 102 (OH) 12.88

Consider the acid–base titration in the above Sample Problem. When the pH of the

solution in the receiving flask is plotted against the volume of 0.300 mol/L NaOH(aq)

added, the result is a titration curve (Figure 2). The curve for the titration of

0.300 mol/L HCl(aq) with 0.300 mol/L NaOH(aq) is typical of that for the titration of any

strong acid with any strong base.




Notice the shape of the titration curve. At the beginning, the pH rises gradually as

base is added. In this section, the pH remains relatively constant even though smallamounts of base are being added. This occurs because the first amount of titrant is

immediately consumed, leaving an excess of strong acid, and the pH is changed very

little.Following this flat region there is a very rapid increase in pH for a very small addi-

tional volume of the titrant. The midpoint of the sharp increase in pH occurs at a pH of

7.00. This is the equivalence point of the reaction and corresponds (with an appropriate

indicator) to the endpoint of the titration. Theoretically, the equivalence point represents

the stoichiometric quantity of titrant required by the balanced chemical equation. This

is true of all titrations of a strong monoprotic acid with any strong base. Since the con-

jugates of a strong acid and a strong base cannot hydrolyze (being a weak conjugate

base and weak conjugate acid, respectively), the pH is determined by the auto-ionization

of water only, and is thus 7.00. The curve then bends to reflect a more gradual increase

in pH as excess base is added.

Section 8

pH

0

Volume (mL) of 0.300 mol/L NaOH(aq) Added

321

5 15 2010

4

14

25 30 35 40 45 50

765

89

10111213

15

Titration Curve for Titration of 20 mL of 0.300 mol/L HCl(aq)

with 0.300 mol/L Standardized NaOH(aq)

volume of NaOH(aq) used

to reach equivalence

point is 20 mL

equivalence point,

pH = 7.0

Figure 2

This curve is typical of curves

depicting the titration of a strong

acid with a strong base. Notice tha

the curve sweeps up and to the

right as NaOH(aq) is added, begin-

ning at a pH below 7and ending a

a pH above 7. The equivalence poi

is reached at pH 7.

Practice

Understanding Concepts

1. (a) When 25 mLof 0.10 mol/LHBr (aq) is titrated with 0.10 mol/LNaOH(aq), what is the

pH at the equivalence point?

(b) Select an appropriate indicator for this titration from Appendix C10.

2. In a titration, how many millilitres of 0.23 mol/L NaOH(aq) must be added to 11 mL of

0.18 mol/L HI(aq) to reach the equivalence point?

Applying Inquiry Skills

3. (a) When a titration is being performed, it is common to wash a clinging drop of

titrant into the receiving flask with a stream of distilled water from a wash bottle.

Why is it important that a drop of titrant clinging to the tip of the buret be forced

into the sample solution?

(b) Will the added water affect the results of the titration? If so, why? If not, why not?

Answers

1. (a) 7

2. 8.6 mL



600 Chapter 8 NEL

Titrating a Weak Acid with a Strong BaseNow we will consider the titration of a weak acid with a strong base. In this titration we

place 20.00 mL of 0.300 mol/L HC2H3O2(aq) in the receiving flask and a standardized solu-

tion of 0.300 mol/L NaOH(aq) in the buret (Figure 3). Acetic acid ionizes very little in

aqueous solution, forming the following equilibrium:

HC2H3O2(aq) e H

(aq) C2H3O2

(aq) K a 1.8 105

The low K a indicates that acetic acid exists primarily as HC2H3O2 molecules in solu-

tion. When NaOH(aq) is added drop by drop to the acetic acid solution in the receiving

flask, the OH(aq) ions react with the HC2H3O2(aq) molecules according to the following

neutralization equation.

HC2H3O2(aq) OH

(aq) → C2H3O2

(aq) H2O(l)

As OH(aq) from NaOH are slowly added to the acetic acid solution, more and more

acetic acid molecules are consumed in the neutralization reaction. It is important to

remember that although acetic acid is weak, it reacts quantitatively with the hydroxide

ions until essentially all of the molecules are consumed.

In a titration of 20.00 mLof 0.300 mol/L HC 2 H

3 O

2(aq)with standardized

0.300 mol/L NaOH (aq) , what is the amount of unreacted HC 2 H 3 O 2(aq) and the pH of

the solution:

(a) before titration begins;

(b) during titration but before the equivalence point (10.00 mLof

0.300 mol/L NaOH (aq) added);

(c) at the equivalence point (20.00 mL of 0.300 mol/L NaOH (aq) added); and

(d) beyond the equivalence point.

This reaction consumes HC2H3O2(aq) and continually shifts the equilibrium to the right until

all HC2H3O2(aq) molecules have been consumed and the reaction reaches the equivalence

point. Beginning with 20.00 mL of 0.300 mol/L HC2H3O2(aq) in the flask, we know that an

equivalent amount of OH

(aq) will be contained in 20.00 mL of 0.300 mol/L NaOH(aq).

Calculating the pH of the acetic acid sample before adding sodium hydroxide is staight-

forward and similar to pH calculations you performed earlier in this chapter. However,

when sodium hydroxide is added to the solution, it reacts with acetic acid and causes theacetic acid equilibrium to shift to the right. The extent of this shift determines the pH of

the solution. The calculation of the solution’s pH will be simplified if you deal with the stoi-

chiometry of the acid–base (acetic acid–sodium hydroxide) reaction separately from the

shift in the acetic acid equilibrium. Therefore, we will carry out two separate calculations:

1. A stoichiometry calculation to determine the concentration of weak acid (acetic acid)

remaining after the acid–base reaction, and

2. An equilibrium calculation to determine the new position of the weak acid (acetic acid)

equilibrium and the solution’s pH.

Chemical Changes During a Weak Acid/Strong Base Titration SAMPLE problem

10.00 mL

0.300 mol/L

HC2H3O2(aq)

readings (mL)

0.35

12.10

0.300 mol/L

NaOH(aq)

23.65

35.10

46.55

Figure 3

Even though the base is strong and

the acid is weak, if they are in the

same concentrations, an equivalent

amount of base will be contained in

the same volume as the sample of

acid.




Section 8

Remember to always carry out these two calculations separately.

(a) Before titration, the pH of the solution in the receiving flask is the pH of a

0.300 mol/L HC2H3O2(aq) solution, so we do not need to perform a stoichiometric cal-

culation, only the equilibrium calculation familiar from Section 8.2.

Equilibrium Calculation

The major entities in solution (before ionization occurs) are

HC2H3O2(aq) and H2O(l).

We will begin by constructing an ICE table for the ionization process, letting x represent

the changes in concentration that occur as equilibrium is established.

Now, substitute the equilibrium concentration values into the K a expression for this

equilibrium, and solve for x .

[H

[

(a

H

q

C

)]

2

[C

H

2

3

H

O

3

2

O

(a

2

q

(

)

a

]

q)] K a

(From Appendix C9, K a 1.8 105)

0.30

x

0

2

x 1.8 105

If we assume that 0.300 x 0.300 (use the hundred rule), the equilibrium expres-

sion becomes

0.

x

3

2

00 1.8 105

x 2 5.4 106

x 2.3 103

Validating the assumption ...

2.3

0

.30

1

0

03

100% 0.8%

Since 0.8% 5%, the assumption is valid, and

x 2.3 103 mol/L

[H

(aq)] 2.3 103 mol/L

pH log[H

(aq)] log(2.3 103)

pH 2.62

Before titration begins, the pH of the sample is 2.62.

Table 2 ICETable for the Ionization of HC2H3O2(aq)

HC2H3O2(aq) e H(aq) C2H3O2

(aq)

Initial concentration (mol/L) 0.300 0.00 0.00

Change in concentration (mol/L) x x x

Equilibrium concentration (mol/L) 0.300 x x x



602 Chapter 8 NEL

Calculating the amount of HC2H3O2(aq) in the initial sample is also familiar.

V HC2H

3O

2 20.00 mL

C HC2H

3O

2 0.300 mol/L

n HC2H

3O

2 V HC

2H

3O

2 C HC

2H

3O

2

20.00 mL 0.300 mol/L

n HC2H

3O

2 6.00 mmol

Notice, we are assuming that none of the HC2H3O2(aq) is in ionized form (owing to the

small K a value for acetic acid).

(b) During titration, remember to perform stoichiometry calculations separately from

equilibrium calculations.

Stoichiometry Calculations

The major entities in the solution (before reaction) are:

HC2H3O2(aq), OH(aq), Na

(aq), and H2O(l)

The OH(aq) ions from the added NaOH(aq) will react with the strongest proton donor

(acid) in solution. Although water may act as an acid, it is a much weaker acid( K w 1.0 1014) than acetic acid ( K a 1.8 105). Therefore, OH

(aq) ions react

with HC2H3O2(aq), according to the following neutralization reaction equation:

HC2H3O2(aq) OH

(aq) → C2H3O2

(aq) H2O(l)

Adding 10.00 mL of 0.300 mol/L NaOH(aq) introduces the following amount of NaOH(aq)

to the solution.

V NaOH 10.00 mL

C NaOH 0.300 mol/L

n NaOH V NaOH C NaOH

10.00 mL 0.300 mol/L

n NaOH 3.00 mmol

Since NaOH is a strong base, 3.00 mmol NaOH(aq) introduces 3.00 mmol Na(aq) and

3.00 mmol OH(aq) into the solution. This amount of NaOH(aq) reacts with an equal amount

(3.00 mmol) of HC2H3O2(aq), according to the neutralization reaction, leaving 3.00 mmol of

acetic acid unreacted. (Remember that Na(aq) does not affect the acid–base characteris-

tics of an aqueous solution, and water does not react with OH(aq) because it is a much

weakeracid than acetic acid.)

The volume of the solution is now

20.00 mL 10.00 mL 30.00 mL

(volume of original solution) (volume ofN aOH(aq) added) (total volume)

Equilibrium CalculationsFirst, we list the majorentities in the solution after the neutralization reaction has taken

place . The major entities are:

HC2H3O2, C2H3O2(aq), Na

(aq), and H2O(l).

Notice that there are no OH(aq) ions in the solution; they were all consumed in the

neutralization.

The acetic acid and acetate ions are components of the following equilibrium:

HC2H3O2(aq) e H(aq) C2H3O2

(aq)




Section 8

We can now construct an ICE table to monitor the changes that occur as the equilib-

rium shifts in response to the neutralization we analyzed above. However, before we con-

struct the ICE table, we must calculate the [HC2H3O2(aq)] and [C2H3O2(aq)] in the new

volume after the neutralization reaction has taken place, but before a shift in equilibrium

occurs . (This is a purely theoretical condition, as these two changes actually occur at the

same time. However, it simplifies the calculations considerably.)

[HC2H3O2(aq)]

3.

3

0

0

0

.0

mmm

L

ol

[HC2H3O2(aq)] 0.100mol/L

and

[C2H3O2

(aq)] 3.

3

0

0

0

.0

mmm

L

ol

[C2H3O2

(aq)] 0.100 mol/L

Let x represent the changes in concentration that occur as the system re-establishes

equilibrium. The ICE table looks like this:

Substituting the equilibrium values into the following K a expression and solving for x ,

we get:

[H

[

(a

H

q

C

)]

2

[C

H

2

3

H

O

3

2

O

(a

2

q

(

)

a

]

q)] K a

x (

0

0

.1

.1

0

0

0

0

x

x ) 1.8 105

Apply the hundred rule to show that a simplifying assumption is warranted:

[HC2H

K

3

a

O2(aq)]

1.8

0

.10

1

0

05

[HC2H

K

3

a

O2(aq)] 5.6 103

Since 5.6 103 100, we can assume that

0.100 x 0.100, and

0.100 x 0.100.

The equilibrium equation simplifies to:

( x )

0

(

.

0

1

.

0

1

0

00) 1.8 105

x 1.8 105

Now we validate the simplifying assumption with the 5% rule:

1.8

0

.10

1

0

05

100% 0.018%

Since 0.018% 5%, the simplifying assumption is justified.

Table 3 ICETable for the Ionization of HC2H3O2(aq)

HC2H3O2(aq) e H

(aq) C2H3O2

(aq)


Change in concentration (mol/L) x x 0.100 x

Equilibrium concentration (mol/L) 0.100 x x 0.100 x



604 Chapter 8 NEL

x 1.8 105

[H

(aq)] 1.8 105

pH log(1.8 105)

pH 4.74

The pH of the solution after adding 10.00 mL of 0.300 mol/L NaOH(aq) is 4.74.

(c) At the equivalence point, 20.00 mL of 0.300 mol/L NaOH(aq) has been added.

The H ions of all HC2H3O2(aq) molecules in the flask have combined with the OH(aq) ions

of all the added NaOH(aq) to form H2O(l). What remains is a solution containing the fol-

lowing major entities:

Na(aq), C2H3O2

(aq), and H2O(l).

Being a Group 1 metal ion, Na(aq) does not hydrolyze, so does not affect the pH of the

solution. However, the C2H3O2(aq) ion is the conjugate base of a weak acid (acetic acid),

and does hydrolyze. The pH of the solution will therefore be determined by the extent of

this hydrolysis reaction. As in part (b), we begin with stiochiometry.

Stoichiometry Calculations

The reaction of acetate ions with water is used to determine the pH of the solution:

C2H3O2

(aq) H2O(l) e HC2H3O2(aq) OH

(aq)

This is a typical weak base equilibrium characterized by the following base ionization

constant equation:

K b

The value of K b forC2H3O2(aq) can be determined from the K a of its conjugate acid,

HC2H3O2(aq), and K w, as follows.

K a 1.8 105

K w 1.0 1014

K aK b K w

K b

K

K

w

a

1

1

.0

.8

1

1

0

0

1

5

4

K b 5.6 1010

Therefore,

5.6 1010

To reach the equivalence point, we added 20.00 mL of NaOH(aq) to the original 20.00 mL

of solution. At the equivalence point, the total volume of the solution is 40.00 mL.Since we began with 6.00 mmol HC2H3O2(aq), we will end up with 6.00 mmol of

C2H3O2(aq) at the equivalence point. This amount of C2H3O2

(aq) is dissolved in 40.00 mL of

solution:

[C2H3O2

(aq)] 6

4

.0

0

0

.0

m0 m

m

L

ol

[C2H3O2

(aq)] 0.150 mol/L

Notice that this is the acetate ion concentration before equilibrium is established.

[HC2H3O2(aq)][OH

(aq)]

[C2H3O2

(aq)]

[HC2H3O2(aq)][OH

(aq)]

[C2H3O2

(aq)]




Section 8

Equilibrium Calculations

We construct an ICE table to monitor the changes in concentrations as equilibrium is

established.

C2H3O2


(aq) K b 5.6 1010 (from an earlier calculation)

K b

Substituting the equilibrium values into the ionization constant equation, we get:

(0.15

x

0

2

x ) 5.6 1010

If we assume that 0.150 x 0.150 (after using the hundred rule)

0.

x

1

2

50 5.6 1010

x 2 8.4 1011

x 9.2 106

( The 5% rule justifies the assumption that 0.150 x 0.150.)

[OH

(aq)] 9.2 106 mol/L

pOH log(9.2 106)

pOH 5.03

Since

pH pOH 14.00pH 14.00 pOH

14.00 5.03

pH 8.97

The pH of the solution at the equivalence point is 8.97, and there are no HC2H3O2(aq)

molecules left in solution.

It is important to realize that the equivalence point does not necessarily mean the point

at which the sample is neutral (with a pH of 7). While this is true of all strong acid–strong

base titrations, it may not be true of other titrations. Rather, the equivalence point is the

point at which equivalent amounts of reactants have reacted, according to the balanced

chemical equation. At the equivalence point, there may still be ions in solution that affect

the pH.Not surprisingly, the titration of any weak acid with a strong base generally produces a

basic solution with a pH greater than 7.

Since the reactants and products of this titration are all clear, colourless solutions, the

analyst would select an indicator that changes colour at or neara pH of 8.97, to show that

the equivalence point has been reached. Phenolphthalein is an ideal choice since it

changes from colourless to pink between pH 8.2 and 10.0. In order to select an appro-

priate indicator, the pH at the equivalence point must be calculated before the titration.

[HC2H3O2(aq)][OH

(aq)]

[C2H3O2

(aq)]

Table 4 ICETable for the Hydrolysis Reaction of C2H3O2(aq)

C2H3O2(aq) H2O(l) e HC2H3O2(aq) OH(aq)


Change in concentration (mol/L) x x x

Equilibrium concentration (mol/L) 0.150 x x x



606 Chapter 8 NEL

(d) Beyond the equivalence point, adding additional NaOH(aq) increases the [OH(aq)]

in the same way it did in the case of the strong acid–strong base titration. Since

sodium hydroxide is a strong base, the [OH(aq)] will be equal to the number of moles

of NaOH(aq) added, divided by the new volume of the solution. The pOH is determined

first, followed by the pH, as we did at the end of part (c).

Notice, however, that the equilibrium,

C2H3O2


(aq)

continues to exist in the solution, even as more NaOH(aq) is added from the buret.

However, the K b value of this equilibrium is so low ( K b 5.6 1010) it contributes

an insignificant amount of OH(aq) ions to the solution. Also, the relatively large

amounts of OH(aq) entering the solution with every drop from the buret shift the equi-

librium to the left, reducing even farther the tiny amount of OH(aq) ions it produces.

We assume that the amount of OH(aq) produced by the equilibrium reaction is negli-

gible (when compared to the amount of OH(aq) ions added with every drop of titrant).

This is why the pH of the solution beyond the equivalence point can be ascertained

using a [OH(aq)] calculated by simply dividing the amount of NaOH(aq) added (from

the buret) by the new volume of the solution

Table 5 records more of the results of the titration in the Sample Problem above, andFigure 4 illustrates the titration curve. Notice the similarities and the differences between

this titration curve and that of the strong acid–strong base titration (Figure 2). At the

beginning, the pH rises gradually as base is added. This relatively flat region of the

curve is where a buffering action occurs, which means that the pH remains relatively

constant even though small amounts of strong base are being added. This occurs

because the first amount of titrant is immediately consumed, leaving an excess of acid,

and the pH is changed very little. Following this buffering region is a very rapid increase

in pH for a very small additional volume of the titrant. You will learn more about

buffering action in Section 8.5.

Table 5 Titration of 20.00 mL of 0.3000 mol/LHC2H3O2(aq) with 0.3000 mol/LNaOH(aq)

Volume of base Molar concentration of entityadded (mL) in parentheses (mol/L) pH

None 2.3 103 (H) 2.6

5.00 5.5 105 (H) 4.26

19.90 9.1 108 (H) 7.04

19.99 9.1 109 (H) 8.04

20.00 9.3 106 (OH) 8.97

20.01 7.6 105 (OH) 9.88

20.10 7.4 104 (OH) 10.87

21.00 7.3 103 (OH) 11.86

30.00 6.0 102 (OH) 12.78




Section 8

Practice


4. If 25.00 mL of 0.20 mol/L HCO2H(aq) is titrated with 0.20 mol/L NaOH(aq) (the titrant),

determine the pH

(a) before titration begins;

(b) after 10.00 mL of NaOH(aq) has been added;

(c) at the equivalence point.

5. If 10.0 mL of 0.250 mol/L NaOH(aq) is added to 30.0 mL of 0.17 mol/LHOCN(aq), what is

the pH of the resulting solution?

Answers

4. (a) 2.22

(b) 3.57

(c) 8.37

5. 3.74

Titrating a Weak Base witha Strong Acid

The titration of a weak base with a strong acid can be analyzed the same way we analyzedthe titration of a weak acid and a strong base. Consider the titration of 20.0 mL of

0.100 mol/L NH3(aq) with 0.100 mol/L HCl(aq). The relevant chemical equations are:

NH3(aq) HCl(aq) → NH4Cl(aq) (molecular)

NH3(aq) H

(aq) → NH4

(aq) (net ionic)

Like the other titrations we have studied in this section, the titration of a weak base

with a strong acid can be analyzed at the following four points in the titration:

(a) Before titration begins, when the receiving flask contains a dilute solution of

NH3(aq). (We can find the pH by using the K b of NH3(aq).)

(b) During titration, but before the equivalence point, when the solution contains

significant amounts of unreacted NH3(aq) and NH4(aq) ions. At this stage, we canfind the pH by using the K b for NH3(aq) (or the K a of NH4

(aq)) and by per-

forming separate stoichiometry and equilibrium calculations. Remember to use

the total volume of the solution in the flask.

(c) At the equivalence point, where the solution contains H2O(l), NH4(aq), and Cl(aq)

ions only. Since Cl(aq) ions do not hydrolyze, we can calculate the pH of the

solution by using the K a of NH4(aq). Again, keep stoichiometric and equilibrium

calculations separate and remember to use the total volume of the solution in

the receiving flask when calculating final concentrations.

pH

0

Volume of NaOH(aq) (mL)

321

5 15 2010

4

14

25 30 35 40 45 50

765

8

910111213

15

Titration Curve for Titrating 0.300 mol/L HC2H3O2(aq) with 0.300 mol/L NaOH(aq)

half-way toequivalence

point

equivalencepoint

Figure 4


depicting the titration of a weak ac

with a strong base. Notice that the

curve sweeps up and to the right a

NaOH(aq) is added, beginning at a

pH below 7 and ending at a pH

above 7. The equivalence point is

reached at a pH greater than 7.



608 Chapter 8 NEL

(d) Beyond the equivalence point, where the pH decreases quantitatively. We find

the pH by calculating the [H(aq)] produced by the ionization of the excess

HCl(aq) added.

In general, the pH at the equivalence point, for a titration of a weak base with a strong

acid, will be lower than 7 (Figure 5).

Titration Characteristics SUMMARY

Table 6

Type of titration pH at Entity determining pH

equivalence point at equivalence point

strong acid and weak base < 7 conjugate acid of weak basestrong base and strong acid 7 autoionization of water

strong base and weak acid > 7 conjugate base of weak acid

Practice


6. For the titration of 20.0 mL of 0.1500 mol/LNH3(aq) with 0.1500 mol/L HI(aq) (the

titrant), calculate

(a) the pH before any HI(aq) is added.

(b) the pH at the equivalence point.

Answer

6. (a) 11.21

(b) 5.18

pH

0

Volume (mL) of 0.100 mol/L HCl (aq) added

321

5 15 2010

4

14

25 30 35 40 45 50

765

89

10111213

15

Titration Curve for Titrating 0.100 mol/L NH3(aq) with 0.100 mol/L HCl(aq)

equivalence point,

pH = 5.27

volume of HCl(aq) used


point is 20 mL

Figure 5


depicting the titration of a weak

base with a strong acid. Notice that

the curve sweeps down and to the

right as HCl(aq) is added, beginning

at a pH higher than 7 and ending ata pH below 7. The equivalence point

is reached at a pH lower than 7.

pH curves provide a wealth of

information:

• equivalence points

• initial pH of solution

• numberof quantitative

reactions

• pH endpoints

• transition points for selectingindicators

LEARNING TIP

Acid–Base IndicatorsThe behaviour of acid–base indicators depends, in part, on both the Brønsted-Lowry con-

cept and the equilibrium concept. An indicator is a conjugate weak acid–base pair formed

when an indicator dye dissolves in water.If we use HIn(aq) to represent the acid form and

In–(aq) to represent the base form of any indicator, the following equilibrium can be

written. (The colours of litmus are given below the equation as an example.)




According to Le Châtelier’s principle, an increase in the hydronium ion concentrationshifts the above equilibrium to the left. In acidic solutions, the primary form of the indi-

cator is its un-ionized (acid) form. This happens, for example, when litmus is added to

an acidic solution. Similarly, in basic solutions the

hydroxide ions remove hydronium ions with the

result that the equilibrium shifts to the right. Then

the base colour of the indicator (In–) predominates.

Since different indicators have different acid

strengths, the acidity or pH of the solution at which

an indicator changes colour varies (Figure 6). These

pH values have been measured and are reported in

Table 7 and in Appendix C9.

We can use the indicator equilibrium equationabove to derive the following acid (indicator) ion-

ization constant equation:

K In

[In

(a

[

q

H

)]

I

[

n

H

(a

3

q

O

)]

(aq)]

Section 8

Table 7 Acid–Base Indicators

Common name Suggested Colour of Approximate Colour of pK Inof indicator symbol Hln(aq) pH range In(aq)

methyl violet HMv yellow 0.0–1.6 blue 0.8

thymol blue* H2Tb red 1.2–2.8 yellow 1.6

methyl yellow HMy red 2.9–4.0 yellow 3.3

congo red HCr blue 3.0–5.0 red 4.0

methyl orange HMo red 3.2–4.4 yellow 4.2

bromocresol green HBg yellow 3.8–5.4 blue 4.7

methyl red HMr red 4.8–6.0 yellow 5.0

chlorophenol red HCh yellow 5.2–6.8 red 6.0

bromothymol blue HBb yellow 6.0–7.6 blue 7.1

litmus HLt red 6.0–8.0 blue 7.2

phenol red HPr yellow 6.6–8.0 red 7.4

metacresol purple HMp yellow 7.4–9.0 purple 8.3

thymol blue* HTb yellow 8.0–9.6 blue 8.9

phenolphthalein HPh colourless 8.2–10.0 red 9.4

thymolphthalein HTh colourless 9.4–10.6 blue 9.9

alizarin yellow r HAy yellow 10.1–12.0 red 11.0

indigo carmine Hlc blue 11.4–13.0 yellow 12.2

Clayton yellow HCy yellow 12.0–13.2 amber 12.7

* Thymol blue is a diprotic indicator that changes colour twice

Figure 6

Colour changes of common

acid–base indicators

(a) bromothymol blue (b) phenolphthalein

HIn(aq) + H2O(l) In

(aq) + H3O+(aq)

acid

red

base

blue(litmus colour)

conjugate pair



610 Chapter 8 NEL

In an acid–base titration, the pH changes sharply near the equivalence point. This

large change in pH shifts the indicator’s equilibrium from one colour state to another.

The change in colour actually occurs over a small range in pH but, in a titration, the

change in pH occurs so quickly that we see it as a sudden colour change occurring at the

indicator’s transition point. Note that, because a small amount of indicator is used in a

titration (a few drops at most), it does not contribute to (or affect) the pH of the solu-

tion. Instead,the pH of the solution determines the position of the indicator equilibrium.

The indicator “responds” to the pH conditions of the solution.When selecting an indicator for a particular acid–base titration, the pH at the equiva-

lence point must be known. Ideally, the pH of the titration’s equivalence point should be

reached at the point where half of all indicator molecules have changed colour so that at

the equivalence point there will be equal concentrations of both forms of the indicator.

The equilibrium constant equation for the indicator equilibrium is

K In

[In

(a

[

q

H

)]

I

[

n

H

(a

3

q

O

)]

(aq)]

If, at the equivalence point,

[In

(aq)] [HIn(aq)]

then, K In

[In

(a[qH)

]

I

[

n

H

(a3q

O

)]

(aq)

]

and

K In [H3O

(aq)]

Since K In and [H3O(aq)] (or [H

(aq)]) values are very small, we can conveniently con-

vert them into pK In and pH by taking the negative logarithm of each value as follows:

pK In logK In and, of course,

pH log[H

(aq)]

Thus, for an ideal indicator,

pK In

pH (at the equivalence point)

This means that an indicator will be ideally suited to mark the endpoint of a titration

if its pK In equals the pH at the equivalence point of that particular titration (Figure 7).

pH

0

Volume of 0.300 mol/L NaOH(aq) added (mL)

321

5 15 2010

4

14

25 30 35 40 45 50

765

89

10111213

15

Titration Curve for Titrating 20.00 mL of 0.300 mol/L HCl(aq) with 0.300 mol/L NaOH(aq)

volume of NaOH used


point is 20.00 mL

equivalence point,

pH = pK In

alizarin yellow

bromothymol blue

thymol blue

Figure 7

Thymol blue is an unsuitable indi-

cator for this titration because it

changes colourbefore the equiva-

lence point (pH 7). Alizarin yellow isalso unsuitable because it changes

colour after the equivalence point.

Bromothymol blue is suitable

because its endpoint of pH 6.8

(assume the middle of its pH range)

closely matches the equivalence

point of pH 7, and the colour change

is completely on the vertical portion

of the pH curve, within the range

where there is rapid change in pH.

Quantitative Titration (p. 627)

In this activity you will be given an

opportunity to standardize a sodiumhydroxide solution and then use it to

determine the concentration of an

acid solution of unknown concen-

tration.

INVESTIGATION 8.4.1

transition point the pH at which an

indicator changes colour




Polyprotic Acid TitrationsThe pH curve for the titration of hydrochloric acid with sodium hydroxide has only

one observable endpoint (Figure 7), but the pH curve for the addition of HCl(aq)

(titrant) to Na2CO3(aq) (Figure 8) displays two equivalence points—two rapid changes

in pH. pH curves such as this are typical of the titration of polyprotic acids or bases.

Here, for example, two successive reactions occur. The two endpoints in Figure 9 can

be explained by two different proton transfer equations.

Remember that sodium carbonate is a strong electrolyte and so fully dissociatesinto Na

(aq) and CO32(aq) ions.

Na2CO3(aq) → 2 Na

(aq) CO32

(aq)

Therefore, the major entities in the receiving flask are Na(aq), CO3

2(aq), and H2O(l).

At the beginning of the titration, H(aq) ions from HCl(aq) react with CO3

2(aq) ions,

since carbonate ions are the strongest base present in the initial mixture.

H

(aq) CO32

(aq) → HCO3

(aq)

from HCl(aq) from Na2CO3(aq)

Then, in a second reaction, protons from HCl(aq)

react with the hydrogen carbonate

ions formed in the first reaction.

H

(aq) HCO3

(aq) → H2CO3(aq)

from HCl(aq) from first reaction

Section 8

pH

0

Volume of HCl(aq) added (mL)

2

5 15 2010

4

25 30 35 40 45 50

6

8

10

12

25.0 mL of 0.50 mol/L Na2CO3(aq) Titrated with 0.50 mol/L HCl(aq)

55 60 65 70 75

first endpoint pH

second endpoint pH

methyl

orange

equivalence

points Figure 8

A pH curve for the addition of

0.50 mol/L HCl(aq) to a 25.0 mL

sample of 0.50 mol/LNa2CO3(aq).

Practice


For the following questions, use Appendix C10.

7. Explain why bromocresol green is a better indicator than alizarin yellow in the titration

of dilute ammonia with dilute hydrochloric acid.

8. Why must a very small amount of indicator be used in a titration?

9. If methyl red is used in the titration of dilute benzoic acid, HC7H5O2(aq), with dilutesodium hydroxide,NaOH(aq), will the endpoint of the titration correspond to the

equivalence point? Explain.



612 Chapter 8 NEL

Notice from observing the pH curve that each reaction requires about 25 mL of

hydrochloric acid to reach the equivalence point and that the methyl orange colour

change marks the second endpoint of the titration.

As you know, acids that can donate more than one proton are called polyprotic acids.

This term applies to bases as well. The carbonate ion is a polyprotic base, called diprotic

because it can accept two protons. Other polyprotic bases include sulfide ions and phos-

phate ions. Sulfide ions are diprotic; phosphate ions are triprotic. Their reactions with

a strong acid are shown below.

S2

(aq) H

(aq) → HS

(aq)

HS

(aq) H

(aq) → H2S(aq)

PO43

(aq) H

(aq) → HPO42

(aq)

HPO42

(aq) H

(aq) → H2PO4

(aq)

H2PO4

(aq) H

(aq) → H3PO4(aq)

Polyprotic acids, such as oxalic acid and phosphoric acid, can donate more than one

proton. Oxalic acid is a diprotic acid; phosphoric acid is a triprotic acid. Their reactions

with strong bases are shown below.

H2C2O4(aq) OH

(aq) → HC2O4

(aq)

HC2O4

(aq) OH

(aq) → C2O42

(aq)

H3PO4(aq) OH

(aq) → H2PO4

(aq)

H2PO4

(aq) OH

(aq) → HPO42

(aq)

HPO42

(aq) OH

(aq) → PO43

(aq)

Evidence from pH measurements indicates that polyprotic substances become weaker

acids or bases with every proton donated or accepted. This occurs because it is easier to

remove a H ion (a proton) from neutral H3PO4(aq) than from the negatively charged

H2PO4(aq) ion or the even more negatively charged HPO4

2(aq) ion. According to Le

Châtelier’s principle, with each successive proton removal, there are more H

(aq) ions insolution pushing the reaction back toward the reactants.

Figure 9 shows the pH curve for phosphoric acid titrated with sodium hydroxide.Only

two endpoints are present,corresponding to equivalence points of 25 mL and 50 mL of

NaOH(aq) titrant added.At the first equivalence point, equal amounts of H3PO4(aq) and

OH(aq) have been added.

H3PO4(aq) OH

(aq) → H2PO4

(aq)

Since all the H3PO4(aq) has reacted, the second plateau (30 mL to 50 mL NaOH(aq)

added) must represent the reaction of OH

(aq) with H2PO4

(aq). The second equivalence

point (50 mL NaOH(aq) added) corresponds to the completion of the reaction of H2PO4

(aq)

with an additional 25 mL of NaOH(aq) solution added. No H2PO4

(aq)remains.

H2PO4

(aq) OH

(aq) → HPO42

(aq) H2O(l)

Notice that there is no apparent endpoint at 75 mL for the possible reaction of HPO42(aq)

with OH

(aq). A clue to this missing third endpoint can be obtained from Appendix C9.

The hydrogen phosphate ion is an extremely weak acid (K a 4.2 1013) and appar-

ently does not quantitatively lose its proton to OH

(aq). As a general rule, only

quantitative reactions produce detectable endpoints in an acid–base titration .

Some chemists and textbooks

use the term polyprotic only in

conjunction with acids that may

donate more than one proton.

Bases that may accept more

than one proton are called

polyfunctional bases. We will usethe term polyprotic for acids and

bases.

LEARNING TIP




Section 8

pH

1

Volume of NaOH(aq) added (mL)

2

5 15 2010

4

25 30 35 40 45 50

6

8

1011

25.0 mL of 0.50 mol/L H3PO4(aq) Titrated with 0.50 mol/L NaOH(aq)

55 60 65 70 75

3

5

7

9

80

Figure 9

A pH curve for the addition of

0.50 mol/L NaOH(aq) to a 25.0-mLsample of 0.50 mol/LH3PO4(aq) dis

plays only two rapid changes in pH

This means that there are only two

quantitative reactions for phospho

acid with sodium hydroxide.

Practice


10. In an acid–base titration, 25.0

mLof 0.50 mol/L Na3PO4(aq)

was titrated with

0.50 mol/L HCl(aq)

( Figure 10).

(a) Write three Brønsted-

Lowry equations that

describe the reactions

that may take place

during the titration.

(b) At what volumes of HCl(aq)

added do the equivalence

points occur?

(c) Why do only two equiva-

lence points show in

Figure 10?

Answer

10 (b) 25 mL, 50 mL

Section 8.4 Questions


1. An acetic acid sample is titrated with sodium hydroxide.

(a) Based on Figure 11, estimate the endpoint and the

equivalence point.

(b) Choose an appropriate indicator for this titration.

(c) Write a Brønsted-Lowry equation for this reaction.

2. Predict whether the pH endpoint is 7, > 7, or < 7 for each

of the following acid–base titrations. Justify your predictions.

(a) hydroiodic acid with sodium hydroxide

(b) boric acid with sodium hydroxide

(c) hydrochloric acid with magnesium hydroxide

(d) hydrochloric acid with aqueous ammonia

3. Predict the pH of the following solutions. Justify your

predictions.

(a) NH4Cl(aq) (b) Na2S(aq) (c) KNO3(aq)

4. How is a pH curve used to choose an indicator fora titration?

5. According to the table of acid–base indicators in Appendix

C10, what is the colour of each of the following indicators inthe solutions of given pH?

(a) phenolphthalein in a solution with a pH of 11.7

(b) bromothymol blue in a solution with a pH of 2.8

(c) litmus in a solution with a pH of 8.2

(d) methyl orange in a solution with a pH of 3.9

pH

Volume of NaOH(aq) added (mL)

4

25 15 2010

6

25 30 35 40 45 50

810

12

14

25.0 mL of 0.50 mol/L HC2H3O2(aq)

Titrated with 0.50 mol/L NaOH(aq)

Figure 11

pH

Volume of HCl(aq) added (mL)

2

10 30 4020

4

50 60 70 80 90 100

6

8

10

12

25.0 mL of 0.50 mol/L Na3PO4(aq)

Titrated with 0.50 mol/L HCl(aq)

0

14

Figure 10



6. For the titration of 25.00 mL of 0.100 mol/L benzoic acid,

HC7H5O2(aq), with 0.100 mol/L of sodium hydroxide,

NaOH(aq), calculate the pH

(a) before adding any NaOH(aq);

(b) after 10 mLof NaOH(aq) has been added;

(c) at the equivalence point.

7. Select an appropriate indicator for the titration in question 7.

8. (a) What is the pH at the equivalence point for each of thetitrations in Table 8?

(b) Select appropriate indicators for the titrations in Table 8.

9. If 25 mL of 0.23 mol/L NaOH(aq) were added to 45 mLof

0.10 mol/L HC2H3O2(aq), what would be the pH of the

resulting solution?

10. In an investigation, separate samples of an unknown solu-

tion turned both methyl orange and bromothymol blue to

yellow, and turned bromocresol green to blue.

(a) Estimate the pH of the unknown solution.

(b) Calculate the approximate hydronium ion concentration.

Applying Inquiry Skills

11. Oxalic acid reacts quantitatively in a two-step reaction with

a sodium hydroxide solution. Assuming that an excess of

sodium hydroxide is added drop by drop, sketch a pH curve

(without any numbers) for the titration.

12. Given the following experimental design and evidence,

determine the approximate pH of three unknown solutions.

Experimental Design

The unknown solutions were labelled A, B, and C. Each

solution was tested by adding each of several indicators to

samples.

Evidence

13. Design an experiment that uses indicators to identify which

of three unknown solutions labelled X, Y, and Z have pH

values of 3.5, 5.8, and 7.8. There are several acceptable

designs!

14. Given the following experimental evidence, determine the

relative strengths of the acids in Table 10. All acid solu-

tions were of equal molar concentration

Table 9

Solution Indicator Indicator colour A methyl violet blue

methyl orange yellow

methyl red red

phenolphthalein colourless

B indigo carmine blue

phenol red yellow

bromocresol green blue

methyl red yellow

C phenolphthalein colourless

thymol blue yellowbromocresol green yellow

methyl orange red

Table 10 Indicator Colours with Various Acids

Acid Methyl violet Thymol blue Benzopurpurine-48 Congo red Chlorophenol red

hydrofluoric, HF(aq) blue orange violet blue yellow

acetic, HC2H3O2(aq) blue yellow purple blue yellow

nitric, HNO3(aq) green red violet blue yellow

hydrocyanic, HCN(aq) blue yellow red red yellow

methanoic, HCHO2(aq) blue orange purple blue yellow

hydrochloric HCl(aq) green red violet blue yellow

Table 8

Titrant Sample

(i) 0.200 mol/LHCl(aq) 20.0 mL of 0.100 mol/LNH3(aq)

(ii) 0.150 mol/L NaOH(aq) 10.0 mLof 0.350 mol/LHC2H3O2(aq)

(iii) 0.250 mol/LHBr (aq) 15.0 mL of 0.150 mol/L N2H4(aq)

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8.4 Acid Base Titrations

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