2.Acid-base Equilibria Final Ppt 2013

Acid-Base Equilibria

Chapter 2 – Part A

1

Solutions of a Weak Acid or Base

The simplest acid-base equilibria are those in which a single acid or base solute reacts with water.

2

In this chapter, we will first look at solutions of weak acids and bases.

We must also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water.

Acid-Ionization Equilibria

Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion.

3

Because acetic acid is a weak electrolyte, it ionizes to a small extent in water.

(aq)OHC(aq)OH 2323

When acetic acid is added to water it reacts as follows.

)l(OH)aq(OHHC 2232


For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).

4

Consider the generic monoprotic acid, HA.

)aq(A)aq(OH )l(OH)aq(HA 32



5

The corresponding equilibrium expression is:

]OH][HA[]A][OH[

K2

3c



6

Since the concentration of water remains relatively constant, we rearrange the equation to get:

]HA[]A][OH[

K]OH[K 3c2a



7

Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.

]HA[]A][OH[

K 3a



8

Table 17.1 lists acid-ionization constants for various weak acids.

Experimental Determination

of Ka

The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions.

9

Electrical conductivity or some other colligative property can be measured to determine the degree of ionization.

With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution.

A Problem To Consider

Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25 oC. Calculate the acid-ionization constant for this acid at 25 oC.

10

It is important to realize that the solution was made 0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M.

We will abbreviate the formula for nicotinic acid as HNic.



11

Initial 0.012 0 0

Change -x +x +x

Equilibrium 0.012-x x x

Let x be the moles per liter of product formed.

)aq(Nic)aq(OH )l(OH)aq(HNic 32



12

The equilibrium-constant expression is:

]HNic[]Nic][OH[

K 3a



13

Substituting the expressions for the equilibrium concentrations, we get

)x012.0(x

K2

a



14

We can obtain the value of x from the given pH.

)pHlog(anti]OH[x 3

)39.3log(antix

00041.0101.4x 4



15

Substitute this value of x in our equilibrium expression.

Note first, however, that

012.001159.0)00041.0012.0()x012.0( the concentration of unionized acid remains virtually unchanged.



16

Substitute this value of x in our equilibrium expression.

522

a 104.1)012.0()00041.0(

)x012.0(x

K



17

To obtain the degree of dissociation:

034.00.012

0.00041 ondissociati of Degree

The percent ionization is obtained by multiplying by 100, which gives 3.4%.

Calculations With Ka

Once you know the value of Ka, you can calculate the equilibrium concentrations of species HA, A-, and H3O+ for solutions of different molarities.

18

The general method for doing this was discussed in Chapter 15.


Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid.

19

It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression.

The degree of ionization of a weak acid depends on both the Ka and the concentration of the acid solution (see Figure 17.3).


How do you know when you can use this simplifying assumption?

20

then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%.

It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is,

100KC if

a

a


How do you know when you can use this simplifying assumption?

21

If the simplifying assumption is not valid, you can solve the equilibrium equation exactly by using the quadratic equation.

The next example illustrates this with a solution of aspirin (acetylsalicylic acid), HC9H7O4, a common headache remedy.


What is the pH at 25 oC of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25 oC.

22

The molar mass of HC9H7O4 is 180.2 g.

From this we find that the sample contained 0.00180 mol of the acid.



23

The molar mass of HC9H7O4 is 180.2 g.

Hence, the concentration of the acetylsalicylic acid is 0.00180 mol/0.500 L = 0.0036 M (Retain two significant figures, the same number of significant figures in Ka).



24

Note that

which is less than 100, so we must solve the equilibrium equation exactly.

11103.3

0036.0K

C4

a

a



25

We will abbreviate the formula for acetylsalicylic acid as HAcs and let x be the amount of H3O+ formed per liter.The amount of acetylsalicylate ion is also x mol; the amount of nonionized acetylsalicylic acid is (0.0036-x) mol.



26

Initial 0.0036 0 0

Change -x +x +x


)aq(Acs)aq(OH )l(OH)aq(HAcs 32

These data are summarized below.



27

The equilibrium constant expression is

a3 K

]HAcs[]Acs][OH[



28

If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get

42

103.3)x0036.0(

x



29

Rearranging the preceding equation to put it in the form ax2 + bx + c = 0, we get

0)102.1(x)103.3(x 642

You can solve this equation exactly by using the quadratic formula.



30

Now substitute into the quadratic formula.

a2ac4bb

x2



31

Now substitute into the quadratic formula.

2)102.1(4)103.3()103.3(

x6244

The lower sign in ± gives a negative root which we can ignore



32

Taking the upper sign, we get4

3 104.9]OH[x

Now we can calculate the pH.

03.3)104.9log(pH 4

Polyprotic Acids

Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.

33

The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO4

-.

Sulfuric acid, for example, can lose two protons in aqueous solution.

)aq(HSO)aq(OH)l(OH)aq(SOH 43242

)aq(SO)aq(OH )l(OH)aq(HSO 24324

Polyprotic Acids


34

For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered.

)aq(HCO)aq(OH )l(OH)aq(COH 33232

)aq(CO)aq(OH )l(OH)aq(HCO 23323

Polyprotic Acids


35

Each equilibrium has an associated acid-ionization constant.

For the loss of the first proton

7

32

331a 103.4

]COH[]HCO][OH[

K

Polyprotic Acids


36

Each equilibrium has an associated acid-ionization constant.

For the loss of the second proton

11

3

233

2a 108.4]HCO[

]CO][OH[K

Polyprotic Acids


37

In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2.

In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1.

Polyprotic Acids


38

However, reasonable assumptions can be made that simplify these calculations as we show in the next example.

When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.


Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

39

For diprotic acids, Ka2 is so much smaller than Ka1 that the smaller amount of hydronium ion produced in the second reaction can be neglected.



2- ?


40

The pH can be determined by simply solving the equilibrium problem posed by the first ionization.



2- ?


41

If we abbreviate the formula for ascorbic acid as H2Asc, then the first ionization is:

(aq)AscH(aq)OH )l(OH)aq(AscH 322



2- ?


42

(aq)AscH(aq)OH )l(OH)aq(AscH 322

Initial 0.10 0 0

Change -x +x +x




2- ?


43


1a2

3 K]AscH[

]HAsc][OH[



2- ?


44

Substituting into the equilibrium expression

52

109.7)x10.0(

x



2- ?


45

Assuming that x is much smaller than 0.10, you get

0028.0108.2x

)10.0()109.7(x3

52



2- ?


46

The hydronium ion concentration is 0.0028 M, so

55.2)0028.0log(pH



2- ?


47

The ascorbate ion, Asc2-, which we will call y, is produced only in the second ionization of H2Asc.

(aq)Asc(aq)OH )l(OH)aq(HAsc 232



2- ?


48

Assume the starting concentrations for HAsc- and H3O+ to be those from the first equilibrium.




2- ?


49


Initial 0.0028 0.0028 0

Change -y +y +y

Equilibrium 0.0028-x 0.0028+y y



2- ?


50


2a

23 K

]HAsc[

]Asc][OH[



2- ?


51

Substituting into the equilibrium expression

12106.1)y0028.0(

)y)(y0028.0(



2- ?


52

Assuming y is much smaller than 0.0028, the equation simplifies to

12106.1)0028.0(

)y)(0028.0(



2- ?


53

Hence, 122 106.1]Asc[y

The concentration of the ascorbate ion equals Ka2.

Base-Ionization Equilibria

Equilibria involving weak bases are treated similarly to those for weak acids.

54

Ammonia, for example, ionizes in water as follows.

)aq(OH)aq(NH )l(OH)aq(NH 423

The corresponding equilibrium constant is:

]OH][NH[]OH][NH[

K23

4c



55

Ammonia, for example, ionizes in water as follows.


The concentration of water is nearly constant.

]OH][NH[]OH][NH[

K]OH[K23

4c2b



56

In general, a weak base B with the base ionization

)aq(OH)aq(HB )l(OH)aq(B 2

has a base ionization constant equal to

]OH][B[]OH][HB[

K2

b


What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.

57

1. Write the equation and make a table of concentrations.

2. Set up the equilibrium constant expression.

3. Solve for x = [OH-].

As before, we will follow the three steps in solving an equilibrium.



58

Pyridine ionizes by picking up a proton from water (as ammonia does).

)aq(OH)aq(NHHC )l(OH)aq(NHC 55255

Initial 0.20 0 0

Change -x +x +x




59

Note that

which is much greater than 100, so we may use the simplifying assumption that (0.20-x) (0.20).

89

a

a 104.1104.1

20.0K

C



60

The equilibrium expression is

b55

55 K]NHC[

]OH][NHHC[



61

If we substitute the equilibrium concentrations and the Kb into the equilibrium constant expression, we get

92

104.1)x20.0(

x



62

Using our simplifying assumption that the x in the denominator is negligible, we get

92

104.1)20.0(

x



63

Solving for x we get

)104.1()20.0(x 92 59 107.1)104.1()20.0(]OH[x



64

Solving for pOH

8.4)107.1log(]OHlog[pOH 5

2.98.400.14pOH00.14pH

Since pH + pOH = 14.00

Acid-Base Properties of a Salt Solution

One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.

65

A 0.1 M solution has a pH of 11.1 and is therefore fairly basic.

Consider a solution of sodium cyanide, NaCN.

)aq(CN)aq(Na)s(NaCN OH 2



66

Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH-.

)aq(OH)aq(HCN )l(OH)aq(CN 2



67

From the Brønsted-Lowry point of view, the CN- ion acts as a base, because it accepts a proton from H2O.




68

You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic.


The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.


The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

69

The CN- ion hydrolyzes to give the conjugate acid and hydroxide.




70

The hydrolysis reaction for CN- has the form of a base ionization so you writ the Kb expression for it.




71

The NH4+ ion hydrolyzes to the conjugate base

(NH3) and hydronium ion.




72

This equation has the form of an acid ionization so you write the Ka expression for it.


Predicting Whether a Salt is Acidic, Basic, or Neutral

How can you predict whether a particular salt will be acidic, basic, or neutral?

73

The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs.

Consequently, the anions of weak acids (poor proton donors) are good proton acceptors.

Anions of weak acids therefore, are basic.



74

One the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze. For example, the Cl- ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water.

reaction no)l(OH)aq(Cl 2



75

Conversely, the cations of weak bases are acidic.One the other hand, the cations of strong bases have virtually no acidic character, that is, they do not hydrolyze. For example,

reaction no)l(OH)aq(Na 2


To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.

76

Consider potassium acetate, KC2H3O2.

The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

reaction no)l(OH)aq(K 2


To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.

77

Consider potassium acetate, KC2H3O2.

The acetate ion, however, is the anion of a weak acid (HC2H3O2) and is basic.

OHOHCH )l(OH)aq(OHC 2322232

A solution of potassium acetate is predicted to be basic.


These rules apply to normal salts (those in which the anion has no acidic hydrogen)

78

1. A salt of a strong base and a strong acid.

The salt has no hydrolyzable ions and so gives a neutral aqueous solution.

An example is NaCl.



79

The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution.

An example is NaCN.

2. A salt of a strong base and a weak acid.



80

The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an acidic solution.

An example is NH4Cl.

3. A salt of a weak base and a strong acid.



81

Both ions hydrolyze. You must compare the Ka of the cation with the Kb of the anion.

If the Ka of the cation is larger the solution is acidic.

If the Kb of the anion is larger, the solution is basic.

4. A salt of a weak base and a weak acid.

The pH of a Salt Solution

To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion. (see Figure 17.8)

82

The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.

Thus the Kb for CN- is related to the Ka for HCN.


To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

83

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

When these two reactions are added you get the ionization of water.

)aq(OH)aq(OH )l(OH2 32 Kw



84



Ka

Kb

When two reactions are added, their equilibrium constants are multiplied.




85



Ka

Kb

Therefore,


wba KKK


For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases.

86

The only difference is first obtaining the Ka or Kb for the ion that hydrolyzes.

The next example illustrates the reasoning and calculations involved.


What is the pH of a 0.10 M NaCN solution at 25 oC? The Ka for HCN is 4.9 x 10-10.

87

Sodium cyanide gives Na+ ions and CN- ions in solution.Only the CN- ion hydrolyzes.




88

The CN- ion is acting as a base, so first, we must calculate the Kb for CN-.

Now we can proceed with the equilibrium calculation.

510

14

a

wb 100.2

109.4

100.1KK

K



89

Let x = [OH-] = [HCN], then substitute into the equilibrium expression.

bK]CN[

]OH][HCN[



90

This gives

52

100.2)x10.0(

x



91

Solving the equation, you find that

3104.1]OH[x

2.11)104.1log(00.14pOH00.14pH 3

Hence,

As expected, the solution has a pH greater than 7.0.

Calculate PH of salt solutions?

92

93

The ions of salts can have an influence on the pH of a solution.

Ions that come from a strong acid or base do not influence the pH. WHY?

Since strong acids and bases are 100% ionized in water, the ions

are unable to reform the molecular acid or the base in

water.

HCl + H2O H3O+ + Cl- “one way arrow”

NaOH + H2O Na+ + OH- “one way arrow”

NaCl is the salt that comes from a strong acid and a strong base.

94

What would the pH of a sodium chloride solution (at 25oC)?

pH = 7

What gives rise to this pH?

Auto hydrolysis of water.

H2O + H2O ↔ H3O+ + OH-

95

Salts that contain ions that come from a weak acid or base.

weak acid: HNO2

A salt containing the anion of the weak acid and the cation from a strong base.

KNO2

Add water: KNO2(s) + H2O K+ + NO2- + H2O

Hydrolysis: NO2- + H2O ↔ HNO2 + OH-

A basic solution.

96

Calculate the pH of a 0.10 M KNO2 solution. Ka(HNO2) = 4.5 10-4.

CHEMISTRY: KNO2(s) K+ + NO2-

More Chemistry: NO2- + H2O ↔ HNO2 + OH-

Equilibrium: ICE

0.10 0 0

-X +X +X

0.10-X +X +X

]NO[

]OH][HNO[K

2

2b

x]-[0.10

[x][x] 10x 22.2 -11

KaKb = 1.0 x 10-14

x =1.49 x 10-6 = [OH-]

pOH = -log[OH-] = 5.83

pH = 14 – 5.83 = 8.17

Try dropping

97

Salts that contain ions that come from a weak acid or base.

Weak Base: (CH3)3N trimethylamine

A salt containing the cation of the weak base and the anion from astrong acid. (CH3)3NHCl trimethylammonium chloride

Add water: (CH3)3NHCl (s) + H2O (CH3)3NH+ + Cl- + H2O

Hydrolysis: (CH3)3NH+ + H2O ↔ (CH3)3NHOH + H+

An acidic solution.

98

Calculate the pH of a 0.10 (CH3)3NHCl solution. Kb((CH3)3NHCl ) = 7.4 10-5.

CHEMISTRY: (CH3)3NHCl(s) + H2O (CH3)3NH+ + Cl- + H2O

More Chemistry: (CH3)3NH+ + H2O ↔ (CH3)3NHOH- + H+

Equilibrium: ICE

0.10 N/A 0 0

-x N/A +x +x

0.10-x N/A +x +x

]NH)CH[(

]H][NHOH)CH[(K

3

3a

KaKb = 1.0 x 10-14

x]-[0.10

[x][x] 10x 35.1 -10

Try dropping

x =3.68 x 10-6 = [H+]

pH = -log[H+] = 5.43

99

What if both ions of a salt come from weak acid and a weak base?

Then the Ka and Kb of the acid or base from which the ions come from must be

compared.

NH4CN(aq) Ka(NH4+) = 5.6 x 10-10 Kb(CN-) = 2.04 x 10-5

note: Ka was calculated from Kb(NH3) and Kb from Ka HCN

Since Kb(CN-) is greater than Ka(NH4+), the solution is basic.

Exercise 1.

1.Calculate the PH of the solution in which 4.9g of NaCN was dissolved in 500 ML of the solution.Ka for HCN=5.0x10-10.

2.Calculate the PH and precent hydrolysis of a 0.1 mol/L solution of pyridinium chloride,C5H5NHCl.For pyridine C5H5N,Kb=1.7x10-9

3. Derive the formula to calculate the PH of slat of weak acid and bases.

PH=1/2(Pka +14-logkb)

100

The Common Ion Effect

The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

101

Consider a solution of acetic acid (HC2H3O2), in which you have the following equilibrium.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232



102

If we were to add NaC2H3O2 to this solution, it would provide C2H3O2

- ions which are present on the right side of the equilibrium.




103

The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased.




104

This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect.



An aqueous solution is 0.025 M in formic acid, HCH2O and 0.018 M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.

105

Consider the equilibrium below.

(aq)OCH(aq)OH 23 )l(OH)aq(OHCH 22

Initial 0.025 0 0.018

Change -x +x +x

Equilibrium 0.025-x x 0.018+x



106

The equilibrium constant expression is:

a2

23 K]OHCH[

]OCH][OH[



107

Substituting into this equation gives:

4107.1)x025.0()x018.0(x



108

Assume that x is small compared with 0.018 and 0.025. Then

025.0)x025.0(

018.0)x018.0(



109

The equilibrium equation becomes

4107.1)025.0()018.0(x



110

Hence,

44 104.2018.0025.0

)107.1(x



111

Note that x was much smaller than 0.018 or 0.025.

63.3)104.2log(pH 4

For comparison, the pH of 0.025 M formic acid is 2.69.

Exercise 1.

1. A solution was prepared by dissolving 0.01 mole of acetic acid in a litre of a solution.

Calculate

A) the H+ concentration and the precent dissociation of the acid

B) the H+ concentration and the precent dissociation of the same acid if0.01 mole of solid acetate NaC2H3O2 is added to the solution

Ka=1.76x10-5

112

Buffers

A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

113

Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid.

Thus, a buffer contains both an acid species and a base species in equilibrium.

Buffers


114

Consider a buffer with equal molar amounts of HA and its conjugate base A-.

When H3O+ is added to the buffer it reacts with the base A-.

)l(OH)aq(HA)aq(A)aq(OH 23

Buffers


115

Consider a buffer with equal molar amounts of HA and its conjugate base A-.

When OH- is added to the buffer it reacts with the acid HA.

)aq(A)l(OH)aq(HA)aq(OH 2

Buffers


116

Two important characteristics of a buffer are its buffer capacity and its pH.Buffer capacity depends on the amount of acid and conjugate base present in the solution.

The next example illustrates how to calculate the pH of a buffer.

The Henderson-Hasselbalch Equation

How do you prepare a buffer of given pH?

117

A buffer must be prepared from a conjugate acid-base pair in which the Ka of the acid is approximately equal to the desired H3O+ concentration.To illustrate, consider a buffer of a weak acid HA and its conjugate base A-.

The acid ionization equilibrium is:

)aq(A)aq(OH )l(OH)aq(HA 32



118

The acid ionization constant is:

By rearranging, you get an equation for the H3O+ concentration.

]A[

]HA[K]O[H a3

]HA[]A][OH[

K 3a



119

Taking the negative logarithm of both sides of the equation we obtain:

]A[

]HA[log)Klog(]Olog[H- a3

The previous equation can be rewritten

]HA[]A[

logKppH a



120

More generally, you can write

This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation.

]acid[]base[

logKppH a



121

So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pKa close to the desired pH.

The Ka for acetic acid is 1.7 x 10-5, and its pKa is 4.77.

You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid].

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

16.2

Mixture of weak acid and conjugate base!

KKaa for HCOOH = 1.8 x 10 for HCOOH = 1.8 x 10 -4-4

[H+] [HCOO-]Ka = [HCOOH]

x = 1.038 X 10 -4

pH = 3.98

122

OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)Ka =

[H+][A-][HA]

[H+] =Ka [HA]

[A-]

-log [H+] = -log Ka - log[HA]

[A-]

-log [H+] = -log Ka + log [A-][HA]

pH = pKa + log [A-][HA]

pKa = -log Ka

Henderson-Hasselbach equation

16.2

pH = pKa + log[conjugate base]

[acid]

123

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect

0.30 – x 0.30

0.52 + x 0.52

pH = pKa + log [HCOO-][HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log[0.52]

[0.30]= 4.01

16.2

Mixture of weak acid and conjugate base!

124

pH= 9.18

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Initial

End

0.30 0.36 0

0.30 - x 0.36 + x x

16.3

[NH4+] [OH-]

[NH3]Kb =

Change - x + x + x

= 1.8 X 10-5

(.36 + x)(x)

(.30 – x) 1.8 X 10-5 =

1.8 X 10-5 0.36x

0.30

x = 1.5 X 10-5 pOH = 4.82

125

pH = 9.20


NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (M)

end (M)

0.29 0.01 0.24

0.28 0.0 0.25

final volume = 80.0 mL + 20.0 mL = 100 mL

NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M

OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M

NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M

Ka= = 5.6 X 10-10[H+] [NH3]

[NH4+]

= 5.6 X 10-10[H+] 0.25

0.28

[H+] = 6.27 X 10 -10

126

= 9.20


NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log[NH3]

[NH4+]

pKa = 9.25 pH = 9.25 + log[0.30]

[0.36]= 9.17

NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (M)

end (M)

0.29 0.01 0.24

0.28 0.0 0.25

pH = 9.25 + log[0.25]

[0.28]

final volume = 80.0 mL + 20.0 mL = 100 mL

127

Exercise 1

1.Calculate the change in PH when 2 mL of 10 M HCl is added to 1.0 L of a buffered solution that is intially 1M in HC2H3O2 and 1M in NaC2H3O2

Ka=1.8 x10-5

128

2.2 Solubility EquilibriaSolubility Equilibria

Solubility Product ConstantSolubility Product Constant

KKspsp

for saturated solutions at equilibriumfor saturated solutions at equilibrium

129

130

Comparing values.Comparing values.

131

Solubility Product (Ksp) = [products]x/[reactants]y but.....

reactants are in solid form, so Ksp=[products]x

i.e. A2B3(s) 2A3+ + 3B2– Ksp=[A3+]2 [B2–]3

Given: AgBr(s) Ag+ + Br–

In a saturated solution at 25oC, the [Ag+] = [Br– ]= 5.7 x 10–7 M. Determine the Ksp value.

13-27- 10 x 3.3 10 x 7.5 BrAgK sp

132

Problem: A saturated solution of silver chromate was to found contain 0.022 g/L of Ag2CrO4. Find Ksp

Eq. Expression: Ag2CrO4 (s) 2Ag+ + CrO42–

Ksp = [Ag+]2[CrO42–]

So we must find theconcentrations of each

ion and then solvefor Ksp.

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Problem: A saturated solution of silver chromate was to contain 0.022 g/L of Ag2CrO4. Find Ksp

Eq. Expression: Ag2CrO4 (s) 2Ag+ + CrO42–

Ksp = [Ag+]2[CrO42–]

Ag+: L

Agmol

0.022 g Ag2CrO4

L g332

mol

42CrO Ag1

Ag2 1.33 x 10–4

CrO4–2: 0.022g Ag2CrO4

L L

CrO mol

-24

g 332

mol4

-24

AgCrO1

CrO 1 6.63 x 10–5

-12-52-4sp 10x 1.16 10x 63.610x 33.1K

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Problems working from Ksp values.

Given: Ksp for MgF2 is 6.4 x 10–9 @ 25 oCFind: solubility in mol/L and in g/L

MgF2(s) Mg2+ + 2F– Ksp = [Mg2+][F–]2

I.C.E.

N/A 0 0

N/A +x +2x

N/A +x +2x

Ksp= [x][2x]2 = 4x3

6.4 x 10–9 = 4x3

223-3

-9

MgF Mg 10x 1.2 4

10x 4.6 x

now for g/L:

L

MgF mol10x 2.1 2-3

L

MgF g 2

mol

g 62.3 7.3 x 10–2

Factors that affect solubility

i. Temperture(solubility and Ksp)

ii. Nature of solvents

iii. Common –ion effects

iv. PH of the solution

v. Diverse-ion(Diverse-salt)effect

vi. Complexing agents

135

136

The common ion effect “Le Chatelier”

What is the effect of adding NaF?

CaF2(s) Ca2+ + 2F-

137

Solubility and pH

CaF2(s) Ca2+ + 2F–

Add H+ (i.e. HCl)

2F– + H+ HF

138

Solubility and pH

Mg(OH)2(s) Mg2+ + 2OH–

Adding NaOH?

Adding HCl?

139

Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M. What is its solubility in seawater where the [Cl–] = 0.55 M? (Ksp of AgCl = 1.8 x 10–10)

AgCl(s) Ag+ + Cl–

I.C.E.

N/A 0 0.55

N/A +x +x

N/A +x 0.55 + x

Ksp= [Ag+][Cl–]

Ksp= [x][0.55 + x]try dropping this x

Ksp = 0.55x

1.8 x 10–10 = 0.55x

x = 3.3 x 10–10 = [Ag+]=[AgCl]

“AgCl is much less soluble in seawater”

140

Complex ion formation:

Ag+(aq) + NH3(aq) Ag(NH3)+(aq)

AgCl(s) Ag+ + Cl–

Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2+(aq)

Ag+(aq) + 2NH3(aq) {Ag(NH3)2}+(aq)

formation or stability constant:

72

3

23f 10x 1.7

NH Ag

)NH(AgK

Ksp= 1.8 x 10–10

For Cu2+: Cu2+ + 4NH3 [Cu(NH3)4]2+(aq)

K1 x K2 x K3 x K4 = Kf = 6.8 x 1012

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Solubility and complex ions: Problem: How many moles of NH3 must be added to dissolve 0.050 mol of AgCl in 1.0 L of H2O? (KspAgCl = 1.8 x 10–10 ; Kf[Ag(NH3)2]+ = 1.6 x 107)

AgCl(s) Ag+ + Cl–

Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)

sum of RXNS:AgCl(s) + 2NH3 Ag(NH3)2

+(aq) + Cl–

fsp2

3

23overall K x K

NH

Cl )NH(Ag K

= 2.9 x 10–3

Now use Koverall to solve the problem:

Koverall= 2.9 x 10–3 =

NH

Cl )NH(Ag 2

3

23

2 3NH

0.050 0.050

[NH3]eq = 0.93 but ..... How much NH3 must we add?

[NH3]total= 0.93 + (2 x 0.050) = 1.03 M

2 ammonia’s for each Ag+

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Reaction Quotient (Q): will a ppt. occur?

Use Q (also called ion product) and compare to Ksp

Q < Ksp

reaction goes Q = Ksp

Equilibrium

Q > Kspreaction goes

3.Problems dealing with precipitation(predicting the occurrence of precipitation)

143

Problem: A solution is 1.5 x 10–6 M in Ni2+. Na2CO3 is added to make the solution 6.0 x 10–4 M in CO3

2–. Ksp(NiCO3) = 6.6 x 10–9. Will NiCO3 ppt?

NiCO3 Ni2+ + CO32–

Ksp = [Ni2+][CO32–]

Q = [Ni2+][CO32–]

Q = [1.5 x 10–6][6.0 x 10–4] = 9.0 x 10–10

Q < Ksp no ppt.

We must compare Q to Ksp.

144

Problem: 0.50 L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L of 1.0 x 10–3 M K2CrO4.

a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14

Pb(OAc)2(aq) + K2CrO4(aq) PbCrO4(s) + 2KOAc(aq)

then: PbCrO4(s) Pb2+ + CrO42– Ksp= [Pb2+][CrO4

2–]

[Pb2+]:

L

Pb mol

2

0.50 L

L

Pb(OAc) mol10x 0.1 2-5

2

2

Pb(OAc) 1

Pb 1

1 L5.0 x 10–6

[CrO42-]:

L

CrO mol

-24

0.50 L

L

CrOK mol10x 1.0 42-3

42

-24

CrOK 1

CrO 15.0 x 10-4

Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9

compare to Ksp: Q > Ksp so a ppt. will occur

1 L

Exercise 1

1. The solubility of lead iodate,Pb(IO3)2,is 4.0 x10-5 mol/L at 25 0c.What is the value of Ksp for this salt?

2.Calculate the concentration of Ag+ and CrO42- in saturated solution of Ag2CrO4 at 25 Ksp=1.9x 10-12

3.What is saturated solution? 4.Will AgI ppt form if 25 mLof 1.4x 10-9 M of NaI and 35.0

mL of 7.9x10-7 MAgNO3 are mixed? Ksp=8.5x10-17 5.What is the Molar solubility of AgCl in a litre of a

1 M,NH3 at 25oc? Ksp=1.7x10-10 and Kf=1.7 x10+7

6.What is diverse-ion(diverse-salt)effect

145

Complexation Equilibria. In the broadest sense, complexation is the association

of two or more chemical species that are capable of independent existence by sharing one or more pairs of electrons.

Although this type of chemical reaction can be classified as a Lewis acid-base reaction, it is more commonly known as a complexation reaction. As applied to chemical analysis, this definition is generally taken to mean the bonding of a central metal ion, particularly a transition-metal atom, capable of accepting an unshared pair of electrons from a ligand that can donate a pair of unshared electrons.

146

Complex compounds are formed by the combination of simple compounds (ions, molecules or atoms) that are other wise capable of independent existence. These compounds on combination lose their separate identity and form an entirely new compound with different physical and chemical properties.

Complex compounds are usually composed of a central ion (or metal) surrounded by a number of ions or molecules called ligands

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Consider the addition of anhydrous copper (II) perchlorate to water. The salt dissolves readily according to the reaction,

Cu(ClO4)2(s) + 4 H:ö:H Cu(H2O)4

2+(aq) + 2ClO4

- in which a pair of electrons on the oxygen atom of each H2O formsa

coordinate covalent bond to Cu2+ ion. In this reaction, Cu2+ acts as a Lewis acid and H2O as a Lewis base. Such binding of solvent molecules to a metal ion is called solvation or, in the special case of solvent water, hydration. The Cu(H2O)4

2- ion is called an aquo-complex.

In a complexation reaction, the product of the reaction is termed a complex. The species which donates the electron pairs by acting as a Lewis base is known as a complexing agent or ligand, and the ion which accepts the donated electrons, the Lewis acid, is called the central ion or central atom. Central ions are generally metallic cations. The ligand can be either a neutral molecule such as water or ammonia or an ion such as chloride, cyanide, or hydroxide. The complex can have either a positive or a negative charge, or it can be neutral.

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For most analytical applications, complexation

occurs between a dissolved metal ion and a dissolved ligand capable of displacing water from the metal ion. This is illustrated for the reaction between a hydrated copper (II) ion and dissolved NH3 ligand below.

Cu(H2O)42+ + NH3 CuNH⇌ 3(H2O)3

2+ + H2O

Normally for reactions that occur in water, H2O is

omitted and the complexation reaction is written simply as

Cu2+ + NH3 ⇌ CuNH32+

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A central metal ion can form a single bond with a ligand which is able to donate a pair of electrons from one of its atoms only, as in the examples given above for the formation of Zn(NH3)4

2+, Cr(NH3)63+ etc. The number of bonds formed by

ligands to the central ion is called the coordination number of the ion.

A central metal ion, however, with multidentate ligands can

form a bond in more than one location to form a ring structure. Generally, ring formation results in increased stability of the complex. A species that is simultaneously bonded to two or more sites on a ligand is called a metal chelate, or simply chelate and the generalization given about the stability of such species is called the chelate effect. Because of the stability of chelates, polydentate ligands (also called chelating agents) are often used to remove metal ions from a chemical system.

150

The formation of complex compounds depends to a large extent on the nature of the metal ion. The most stable complexes are formed with cations of smaller size and high

nuclear charge. Metal atoms or ions with these properties will show the greatest attraction for the electrons of the other donor atoms, ions or molecules. For example Li+ is hydrated to a very large extent

The outer electronic configuration plays an important role in complex formation. The availability of unfilled d-orbitals, into which the electrons donated by the ligand can enter, is a very important factor.

The formation of complex compound requires:i. Atoms, molecules or ions that contain a pair of electrons to

donate for bond formationii. A metal atom or ion which has sufficient number of empty d-

orbitals for bond formationiii. Metal atoms with small size and high nuclear charge (high

Charge to volume ratio) are more effective in accepting electrons and result in the formation of stable complex compound . 151

2.3 Classification of Ligands

Any atom, ion or molecule that is capable of donating a pair of electrons to the central atom is called a coordinating group or ligand. In a ligand, the particular atom that actually donates the electron pair is called the donor atom.

Ligands are classified according to the number of pairs

of electrons that they can share with the central ion. If a ligand contains only one donor atom, i.e., it is capable of forming only one coordinate bond to the metal atom, then the ligand is classified as mono- or unidentate ligand. Examples of monodentate ligands are: F-, Cl-, Br-, I-, CN-, SCN-, NO2

-, NH3, H2O, N(CH2CH3)3, CH3COCH3, etc 152

When a ligand contains more than one donor atom that can simultaneously be coordinated to a metal atom, the ligand is referred to as a multi or poly dentate ligand. Depending upon the number of donor sites, these ligands may be referred to as bidentate (two donor atoms) (example NH2CH2CH2NH2, C2O4

2), tridentate (three donor atoms), terdentate (or quadridentate, four donor atoms), pentadentate (or quinqui dentate, five donor atoms), or hexadentate (sexadentate, six donor atoms) ligands

153

2.4 Chelation Chelating ligand is a multidentate ligand that on

coordination with the central atom results in ring formation. The complex produced is called a chelate and the process is called chelation. A chelate is formed if two or more donor atoms of a ligand coordinate by the simultaneous use of two or more electron pairs to the same metal atom.

Example: Mn+ + Y4- (MY)⇌ (n-4)

where Y4- represents:

and MY(n-4) represents

All types of bidentate, tridentate, tetradentate, pentadentate and hexadentate ligands can act as chelating ligands and their complexes with metals are known as chelates. Normally chelated complexes are more stable than similar non-chelated complexes. 154

Some special features of chelates: Chelating ligand form far more stable complexes

than their monodentate analogous. The enhanced stability of complexes containing chelating ligands is known as chelate effect.

The stability of chelates depends on the number of atoms in the ring. Generally, chelates that do not contain double bonds such as ethyldiamine that form five member rings are very stable compounds. Ligands that contain double bonds (Example CH3COCH=C(OH)CH3 (acetylacetonate) form stable complexes containing six member rings.

Because of sterric reasons, ligands with large groups form less stable compounds than the analogous ligands with smaller groups

155

2.5 Importance of Chelates

Chelates find application both in industry and laboratory where fixing of metal ions is required.

In analytical chemistry, chelates are used in both qualitative and quantitative analysis. For example, Ni2+, Mg2+, and Cu2+ are quantitatively precipitated by chelating agents as insoluble compounds. In volumetric analysis chelating agents are often used as a reagent (example EDTA) or as indicators for the titration of some metal ions.

EDTA, for example, is added to certain canned foods to remove transition-metal ions that can catalyze the deterioration of the food. The same chelating agent has been used to treat lead poisoning because it binds Pb2+ ions as the chelate, which can then be excreted by the kidneys.

156

2.6 Stability and Instability Constants of Complexes

Several factors including the nature of the ligand and the central metal ion, the chemical nature of the complex and the geometry of the complex affect the stability of the complex. Generally, multidentate ligands that form five- or six-membered rings with central metal ions have unusually high stability.

Stability constant of a complex is the measure of extent of formation of a complex at equilibrium. Thus the stability of a complex depends upon the strength of the linkage between the central metal ion and the ligands. Therefore, the stronger the metal ligand bond the more stable the complex.

157

Metal complexes are formed by replacement of molecules in the solvated shell of a metal ion in aqueous solution with the ligands by stepwise reaction as:

[M(H2O)n] + L [M(H⇋ 2O)n-1 L] + H2O

[M(H2O)n-1L] + L [M(H⇋ 2O)n-2 L2] + H2O

[M(H2O)n-2 L2] + L [M(H⇋ 2O)n-3 L3] + H2O

The overall reaction is:-

[M(H2O)n] + nL [ML⇋ n] + nH2O

158

In the above equations, we can ignore watermolecules as their activity remains practically constant at low concentrations. Then we can write the above equations and their equilibrium constants as:

M + L ML K1 = [ML]/[M][L]⇋ ML + L ML⇋ 2 K2 = [ML2]/[ML][L]

ML2 + L ML⇋ 3 K3 = [ML3]/[ML2][L]

Mn-1 + L ML⇋ n Kn = [MLn]/[MLn-1][L]

The equilibrium constants, K1, K2, K3, …, Kn are known as stepwise formation constants or stepwise stability constants or consecutive stability constants

159

Example:

Write the stepwise formation constantans and the overall formation constant for the complex, Cu(NH3)4

2+? Consider the simple complexation of copper (II) ion by

the unidentate ligand NH3 in water. The reaction between these two species is

Cu2+ + NH3 ⇌ CuNH32+

where monoaminecopper (II) complex ion is formed (the H2O is omitted for simplicity). In aqueous solution the copper (II) ion is actually hydrated and NH3 replaces H2O. The equilibrium constant for this reaction is the stepwise formation constant, K1, with the following formula:

K1 = [CuNH32+] = 2.0 x 104

[Cu2+][NH3]

160

The overall process for the addition of two NH3 molecules to a Cu2+ ion and the equilibrium constant

for that reaction are given by the following:

Cu2+ + NH3 CuNH⇌ 32+

CuNH32+ + NH3 Cu(NH⇌ 3)2

2+

Cu2+ + 2NH3 ⇌ Cu(NH3)22+

The stepwise and overall formation constant expressions for the complexation of a third and fourth molecule of NH3 to copper (II) are given by the following:

Cu(NH3)22+ + NH3 ⇌ Cu(NH3)3

2+

Cu(NH3)32+ + NH3 ⇌ Cu(NH3)4

2+

Cu2+ + 4NH3 ⇌ Cu(NH3)42+

The values of K3 and K4 are, respectively, 1.0 x 103 and 2.0 x 102. Therefore, the values of 3 and 4 are 1.0 x 1011 and 2.0 x 1013. 161

Example: Calculate the percent dissociation of a 0.10 mol/L Ag(NH3)2

+ solution if its instability constant, Kins = 6.3 x 10-8. A given complex behaves as a weak electrolyte and dissociates to a

small degree. The equilibrium constant for the dissociation of a complex is simply the inverse of its formation constant and is known as the instability constant. For example, the complex ion Ag(NH3)2

+ dissociates according to the equilibrium reaction:

Ag(NH3)2+ ⇌ Ag+ + 2NH3

and its instability constant is given by, In actual practice the dissociation of a complex ion, like the ionization of

a polybasic acid, occurs in steps as shown below:Ag(NH3)2

+ ⇌ Ag(NH3)+ + NH3, Ag(NH3)+ ⇌ Ag+ + NH3,

The overall instability constant Kins = K1.K2

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2.7 Factors Affecting the Stability of Complexes

2.7.1 Nature of the central metal ion Charge and size of the central metal ion

Cations with noble gas configurations The alkali metals, alkaline earth and aluminium

belong to this group that exhibit electrostatic force of attraction in complex formation, so interactions between small ions of high charge are particularly strong and lead to stable complexes. Thus fluoro complexes are particularly stable, water is more strongly bound than

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ammonia that has a smaller dipole moment and cyanide ions have little tendency to form complexes since they only exist

in alkaline solutions where they cannot compete successfully with hydroxyl ions. Cations with completely filled d-subshells

Typical of this group are copper(I), silver(I) and gold(I). These ions have high polarizing power and the bonds formed in their complexes have appreciable covalent character. Complexes are more stable the more noble the metal and the less electronegative the donor atom of the ligand; thus cadmium(II) and mercury(II) form strong complexes with I- and CN- ions, but weak complexes with F-. Transition metal ions with incomplete d-

subshell. The behaviour (stability) of these metals lies between the

former two.

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Summary Complex compounds and double salts are compounds of

higher order. The basic difference between complex compounds and double salts are:-

Double salts are formed from the chemical combination of two simple salts and upon dissociation release the ions of the simple salts.

Complex compounds are formed by the association of two or more chemical species that are capable of independent existence. Once the complex is formed the species lose their identity and form a new complex ion.

Complex compounds have two basic components; the central atom (ion) and the ligand. Central ion accepts electron pairs by acting as Lewis acid; ligand is the coordinating group that donates the electron pairs and acts as Lewis base.

Ligands can be classified based on the number of donor atoms as monodentate and polydentate ligands.

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Polydentate ligands when coordinated to central ion form chelated complexes. The

enhanced stability of chelates over non-chelated complexes is called chelate effect.

Chelates are very useful in industry and laboratory where fixing of metal ions is required.

The overall formation constant of a complex is the product of its stepwise formation constants.

Stability constant of a complex is a measure of the extent of formation of a complex.

Stability of a complex is basically dependent on nature of central metal ion and nature of the ligand.

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2.Acid-base Equilibria Final Ppt 2013

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