Chapter 16: Equilibria in Solutions of Weak Acids and Bases.

Chapter 16: Equilibria in Solutions of Weak Acids and Bases

Weak Acids

• All weak acids behave the same way in aqueous solution: they partially ionize

AOHOHHA 32

][

]][[

][

]][[ 3

HA

AH

HA

AOHKa

Ka is called the acid ionization constant

• Table 18.1 and Appendix C list the Ka and pKa for a number of acids

A “large” pKa,means a “small” value of Ka and only a “small” fraction of the acid molecules ionize

aa KK logp

Weak bases in water

][

]][[

isconstant ionization base the2

B

OHHBK

OHHBOHB

b

Consider the following:

ABconsider

B

OHHBK

OHBHOHB

b ][

]][[

2

HF + H2O ↔ H3O+ + F-

F- + H2O ↔ HF + OH- Solve Ka*Kb=

w

ba

b

a

KOHOH

A

OHHA

HA

AOHKK

A

OHHAKOHHAOHA

HA

AOHKAOHOHHA

]][[

][

]][[

][

]][[

isproduct the

][

]][[

:base conjugate for the

][

]][[

:acid weak For the

3

3

2

332

Problem Solving

• Calculate the pH of a 0.0200 M solution of a weak monoprotic acid which is 3% ionized at 25°C? What is Ka for the acid?

Problem Solving

• Calculate the pH of a 0.0300 M solution of a weak base that is 3% ionized at 25°C? What is Kb for the base

If Ka for a weak acid is 2.9E-5, what is its pKa value?a) 4.54 b) 4.82 c) 5.29 d) 6.82 e) 7.89

16.1. Ionization constants can be defined for weak acids and bases 18

What is the value of Kb for the following weak conjugate bases?

• NaF

• NaCN


What is the value of Kb for the following weak conjugate bases?

• NaF – Kw=Ka×Kb Ka for HF =6.8×10-4

– 1.47×10-11

• NaCN – Kw=Ka×Kb Ka for HCN =6.2×10-10

– 1.61×10-5


What is the value of Ka for the following weak conjugate acids?

• NH4Cl

• C6H5NH3NO3


What is the value of Ka for the following weak conjugate acids?

• NH4Cl

– Kw=Ka×Kb Kb for NH3 =1.8×10-5

– 5.56×10-10

• C6H5NH3NO3

– Kw=Ka×Kb Ka =4.1×10-10

– 2.44×10-5

Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the autoionization of water.

• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for x w/ simplifying assumption. [A]>400*K If approximation is not valid, solve for x exactly

4. Calculate concentrations of all species and/or pH of the solution.

15.5

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]

= 7.1 x 10-4

Solved next page



= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka << 1

15.5



= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50= 7.1 x 10-4

0.50 – x 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M15.5

16.2. Calculations can involve finding or using Ka and Kb 26

Determine the pH of 0.1M solutions of:

• HC2H3O2

• Ka=1.8×10-5

• HCN

• Ka=6.2×10-10

I

C

E

I

C

E

0.1M N/A 0 0

0.1M N/A 0 0


Determine the pH of 0.1M solutions of:

• HC2H3O2

• Ka=1.8×10-5

• HCN

• Ka=6.2×10-10

HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2

-(aq)

I

C

E

HCN + H2O↔ H3O+(aq) + CN-

(aq)

I

C

E

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

0.1M N/A 0 0

-x +x +x

(0.1-x)≈0.1 N/A x

-x

x

52

108.10.1x

pH=2.87

102

102.60.1

x

pH=5.10

X=1.34(10-3)M

X=7.87(10-6)M


Determine the pH of 0.1M solutions of:• N2H4

• Kb=1.7×10-6

• NH3

• Kb=1.8×10-5

I

C

E

I

C

E

0.1M N/A 0 0

0.1M N/A 0 0


Determine the pH of 0.1M solutions of:• N2H4

• Kb=1.7×10-6

• NH3

• Kb=1.8×10-5

N2H4 + H2O ↔ OH-(aq) + N2H5

+(aq)

I

C

E

NH3 + H2O↔ H3O+(aq) + CN-

(aq)

I

C

E

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

62

107.10.1x

pOH=3.38

52

108.10.1

x

pH=11.13

pH=10.62

pOH=2.87

X=4.12(10-4)M

X=1.34(10-3)M

16.2. Calculations can involve finding or using Ka and Kb 5.20 30

What is the pH of a 0.30 M solution of phenol (C6H5OH), an ingredient in some older mouthwashes? Ka=1.3×10-10?

A. 9.2

B. 0.52

C. 9.4

D. none of these


Determine the % ionization of 0.2M solution of HC2H3O2

• Ka=1.8×10-5

I

C

E

0.2M N/A 0 0



• Ka=1.8×10-5


-(aq)

I

C

E

0.2M N/A 0 0

-x +x +x

(0.2-x) ≈ 0.2 N/A x

-x

x

52

108.10.2

x

x=1.90×10-3M

%1000.2

x 100

literper moles available

literper ionized molesionization %

0.95 % ionized



• Ka=1.8×10-5


-(aq)

I

C

E

0.1M N/A 0 0

-x +x +x

(0.1-x) ≈ 0.1 N/A x

-x

x

52

108.10.1x

x=1.34 x 10-3M

%1000.1x 100

literper moles availableliterper ionized moles

ionization %

1.3 % ionized



• Ka=1.8×10-5

I

C

E

0.1M N/A 0 0


What is the % ionization of 0.10M HOCl? (Ka=3.5×10-8)

A. 0.6 %

B. 0.06%

C. 3.5×10-6 %

D. none of these

16.2. Calculations can involve finding or using Ka and Kb 0.3% 36

What is the % ionization of 0.50M HOCl? (Ka=3.5×10-8)

A. 0.13 %

B. 7.0×10-6 %

C. 0.06 %

D. none of these


The pH of a 0.50 M solution of an acidic drug, HD, is 3.5. What is the value of the Ka for the drug?

HD + H2O ↔

H3O+ +

D-

I 0.50 } 0 0

C -x -x +x +x

E 0.50-x

} x x


The pH of a 0.50 M solution of an acidic drug, HD, is 3.5. What is the value of the Ka for the drug?

HD + H2O ↔

H3O+ +

D-

I 0.50 } 0 0

C -x -x +x +x

E 0.50-x

} x x

Given the pH, we find the value of x=10-pH

Ka

Kx

xa

4

24

2

1016.35.0

1016.350.0

Ka=2.0×10-7

x=10-3.5 =3.16×10-4

What is the pH of a 0.5 M HF solution (at 250C)? Ka=7.1E-4 at this temperature.

When can I use the approximation?

0.50 – x 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.0190.019 M0.50 M

x 100% = 3.8%Less than 5%

Approximation ok.

Ka x2

0.05= 7.1 x 10-4 x = 0.006 M

0.006 M0.05 M

x 100% = 12%More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or method of successive approximation. 15.5


The method of successive approximations

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

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

Solved next page



Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka x2

0.122= 5.7 x 10-4

0.122 – x 0.122Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M0.122 M

x 100% = 6.8%More than 5%

Approximation not ok.

15.5

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

x = 0.0081 x = - 0.0081


Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

15.5

Few substances are more effective in relieving intense pain than morphine. Morphine is an alkaloid – an alkali-like compound obtained from plants – and alkaloids are all weak bases.

In 0.010 M morphine, the pH is 10.10. Calculate the K b and pK b

Nicotinic acid, HC2H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4e-5. What is the [H+], pH and percent ionization of a 0.050 M solution of nicotinic acid?

Pyridine, C5H5N, is a bad smelling liquid for which Kb = 1.5e-9. What is the pH and percent ionization of a 0.010 M aqueous solution of pyridine?

Phenol is an organic compound that in water has Ka = 1.3e-10. What is the pH and percent ionization of a 0.15M solution of phenol in water?

Codeine, a cough suppressant extracted from crude opium, is a weak base with a pKb of 5.79. What will be the pH of a 0.020 M solution of codeine?

What is [H3O+] in a 1.1 E-1 M solution of HCN?Ka for HCN is 4.0E-10. a) 4.0E-10 Mb) 3.6E-9 Mc) 6.0E-5 Md) 6.6E-6 Me) 1.1E-1 M

What is the [OH-] concentration of a 0.19 M solution of NH3 in water?Kb for NH3 is 1.8E-5.

a) 1.8E-5 Mb) 1.8E-3 M c) 0.19 Md) 1.54 Me) 12.45 M

What is the pH of a 0.125 M pyruvic acid? (Ka = 3.2E-3)

Phenol, also known as carbolic acid, is sometimes used as a disinfectant. What are the concentrations of all of the substances in a 0.050 M solution of phenol, HC6H5O?

What percentage of phenol is ionized, for this acid Ka = 1.3E-10?

Nicotinic acid is a monoprotic acid and another name for niacin which is a vitamin. Minute quantities are found in all living cells.

When 0.10 mole of nicotinic acid HC6H4NO2 is dissolved in enough water to make 1.50 L of solution the pH of the solution is found to be 2.92. What is the Ka of nicotinic acid?

Propionic acid which occurs in dairy products in small amounts is a weak monoprotic acid CH3CH2COOH. If 0.10 mole of the acid is mixed with sufficient water to make 250 mL of solution, calculate the pH of the solution. Ka = 1.13E-5

Aspirin is acetylsalicylic acid HC9H7O4 (HAsp). It is a moderately weak monoprotic acid. If you have a 0.0075 M solution of the acid, what is the pH of the solution?

What is the concentration of HAsp at equilibrium? Ka for HAsp is 3.3e-4.

Hydrazine, N2H4 is used as a rocket fuel and it is a weak base with a Kb of 8.5e-5. If 0.15 moles of hydrazine are dissolved in enough water to make 750.0 mL of solution, what is the pH of the solution?

Example: Calculate the pH of a 0.0010 M solution of dimethylamine, a base, for which Kb=9.6x10-4.

Example: Calculate the pH of a 0.0010 M solution of dimethylamine for which Kb=9.6x10-4.

0.0010 E

)0010.0( C

][

]][[ 0 0 0.0010 I

106.9 )()()(

2

42

xxx

x

xxxx

B

OHBH

KaqOHaqBHOHaqB-

b

Put in standard form

Solve for x and the equilibrium concentrations

7422

2

2

106.9106.90010.00

)0010.0(

)0010.0(

xxKxKx

xKx

x

xK

bb

b

b

MMxB

MxOHBH

M

x

4

4

4

7244

109.3)0010.0(][

101.6][][

and 101.6

)1(2

)106.9)(1(4)106.9(106.9

so allowed, are solutions positiveOnly

A 0.17 M solution of HAsO3 was found to have a pH of 1.09 at 25oC.What is the Ka value for this acid?

a) 1.2E-2 b) 3.2E-2 c) 7.5E-2 d) 1.7E-1 e) 4.8E-1

A 0.53 M solution of a weak base, R3N was found to have a pH of 10.01What is the Kb value for this base?

a) 1.8E-20 b) 7.5E-19 c) 2.0E-8 d) 3.2E-8 e) 2.0E20

Codeine, a cough suppressant extracted from crude opium, is a weak base with a pKb of 5.79. What will be the pH of a 0.020 M solution of codeine?



Solved next page

16.3. Salt solutions are not neutral if the ions are weak acids or bases 65

Learning Check 0.1M solutions of the following are acid/ base/ neutral or amphoteric? …

• HCl• NaCl• NaCN• HCN• Na2S• Na3PO4

• NH4Cl

•acid•neutral•base•acid•base•base•acid

The salt, KC2H3O2, dissociates in water as follows:KC2H3O2 K+ + C2H3O2

-

What pH value should it produce?

a) less than 7b) greater than 7c) about 7

The salt, NaHSO4, dissociates in water as follows:NaHSO4 Na+ + HSO4

-

What pH value should it produce? a) less than 7b) greater than 7c) about 7

The salt, Na2CO3, dissociates in water as follows:Na2O3 2Na+ + CO3

2-


The salt, NaClO4, dissociates in water as follows:NaClO4 Na+ + ClO4

2-


A basic saltIf 0.10 mole of sodium carbonate, Na2CO3, is

added to enough water to make 1.0 L of solution, what is the pH of the solution? Kb1 of CO3

- is 2.1e-4.

• Solved on next pages . . .

Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .

NaNa++ + H + H22O O neutral neutral

COCO332-2- ++ HH22OO HCOHCO33

-- ++ OHOH--

basebase acidacid acidacid basebase

KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 1.Step 1. Set up concentration tableSet up concentration table

[CO[CO332-2-]] [HCO[HCO33

--]] [OH[OH--]]

initialinitial

changechange

equilibequilib

Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts


NaNa++ + H + H22O ---> neutralO ---> neutral


-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression

Assume 0.10 - x = 0.10, because 100•KAssume 0.10 - x = 0.10, because 100•Kbb < C < Coo

x = [HCOx = [HCO33--] = [OH] = [OH--] = 0.0046 M ] = 0.0046 M


Kb = 2.1 x 10-4 = [HCO3

- ][OH- ]

[CO32 ]

x2

0.10 - xKb = 2.1 x 10-4 =

[HCO3- ][OH- ]

[CO32 ]

x2

0.10 - x


NaNa++ + H + H22O ---> neutralO ---> neutral


-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 3.Step 3. Calculate the pHCalculate the pH

[OH[OH--] = 0.0046 M] = 0.0046 M

pOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34

pH + pOH = 14, pH + pOH = 14,

so so pH = 11.66pH = 11.66, and the solution is ________., and the solution is ________.


Problem Solving

Calculate the pH of 0.40 M KNO2

Problem Solving

Calculate the pH of 0.20 M NaCN

Problem Solving

Calculate the number of grams of NH4Br that have to be dissolved in 1.00 L of water at 25C to have a solution with a pH of 5.16.

What is the pH of 1.00 L of pure water at What is the pH of 1.00 L of pure water at 2525ºC?ºC?

What is the pH of 1.00 L of pure water after What is the pH of 1.00 L of pure water after adding 1 mL of 1.00 M HCl?adding 1 mL of 1.00 M HCl?

What is the pH of 1.00 L of pure water at What is the pH of 1.00 L of pure water at 2525ºC?ºC?

What is the pH of 1.00 L of pure water after What is the pH of 1.00 L of pure water after adding 1 mL of 1.00 M HCl?adding 1 mL of 1.00 M HCl?

unbufferedunbuffered

QUESTION: What is the pH of 0.25 M NH3(aq)?

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

How does the pH change if we add NH4Cl to the system at equilibrium? We are adding an ion COMMON to the equilibrium.

Le Chatelier** predicts that the equilibrium will shift to the __________.

The pH will go _____________.

After all, NH4+ is an acid!

QUESTION: What is the pH of 0.25 M NH3(aq)?

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

How does the pH change if we add NH4Cl to the system at equilibrium? We are adding an ion COMMON to the equilibrium.

Le Chatelier** predicts that the equilibrium will shift to the __________.

The pH will go _____________.

After all, NH4+ is an acid!

Buffers introduced

What is the pH of the 0.25 M NHWhat is the pH of the 0.25 M NH33(aq) when (aq) when

mixed so that it also contains 0.10 M NHmixed so that it also contains 0.10 M NH44Cl?Cl?

NHNH33(aq) + H(aq) + H22O O NH NH44++(aq) + OH(aq) + OH--(aq)(aq)

What is the pH of the 0.25 M NHWhat is the pH of the 0.25 M NH33(aq) when (aq) when

mixed so that it also contains 0.10 M NHmixed so that it also contains 0.10 M NH44Cl?Cl?


The Common Ion EffectThe Common Ion Effect

Problem:Problem: What is the pH of a solution with 0.10 M NH What is the pH of a solution with 0.10 M NH44Cl Cl

and 0.25 M NHand 0.25 M NH33(aq)?(aq)?


[OH[OH--] = x = (0.25 / 0.10)K] = x = (0.25 / 0.10)Kbb = 4.5 x 10 = 4.5 x 10-5-5 M M

This gives pOH = 4.35 and pH = 9.65This gives pOH = 4.35 and pH = 9.65

pH drops from 11.33 to 9.65 on adding a pH drops from 11.33 to 9.65 on adding a common ion.common ion.

Problem:Problem: What is the pH of a solution with 0.10 M NH What is the pH of a solution with 0.10 M NH44Cl Cl

and 0.25 M NHand 0.25 M NH33(aq)?(aq)?


[OH[OH--] = x = (0.25 / 0.10)K] = x = (0.25 / 0.10)Kbb = 4.5 x 10 = 4.5 x 10-5-5 M M

This gives pOH = 4.35 and pH = 9.65This gives pOH = 4.35 and pH = 9.65

pH drops from 11.33 to 9.65 on adding a pH drops from 11.33 to 9.65 on adding a common ion.common ion.

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x(0.10 + x)

0.25 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x(0.10 + x)0.25 - x

The Common Ion Effect

Buffers “work” because the weak acid can react with added base and the weak base can react with added acid

Consider the general buffer made so that both HA and A- are present in solution

• When base (OH-) is added:

• When acid (H+) is added:

– Net result: small changes in pH

OHaqAaqOHaqHA 2)()()(

)()()( aqHAaqHaqA

What is the pH after adding 1.00 mL of 1.00 M HCl to a What is the pH after adding 1.00 mL of 1.00 M HCl to a solution which is 0.10 M NHsolution which is 0.10 M NH44Cl and 0.25 M NHCl and 0.25 M NH33(aq) (aq)

[initial pH=9.65][initial pH=9.65]


What is the pH after adding 1.00 mL of 1.00 M HCl to a What is the pH after adding 1.00 mL of 1.00 M HCl to a solution which is 0.10 M NHsolution which is 0.10 M NH44Cl and 0.25 M NHCl and 0.25 M NH33(aq) (aq)

[initial pH=9.65][initial pH=9.65]


The Common Ion Effect

What is the [H3O+] in a solution that is 0.16 M in HC2H3O2, WITHOUT any added NaC2H3O2?

a) 1.3E-3 Mb) 1.1E-2 M c) 3.3E-2 Md) 2.2E-1 Me) 0.16 M

What is the [H3O+] in a solution that is 0.16 M in HC2H3O2, but, WITH NaC2H3O2 added to make [C2H3O2

-] = 0.88M

a) 3.3E-6 M b) 3.1E-5 Mc) 1.1E-2 M d) 0.16 Me) 0.88 M

For the general weak acid HA:

– Thus both the value of Ka and the ratio of the molarities (or the ratio of moles) affect the pH

– These last two relations are often expressed in logarithmic form

mol

mol][or

][

][][

gives grearrangin

][

]][[ )()()(

A

HAKH

A

HAKH

HA

AHKaqAaqHaqHA

aa

a

• The first is called the Henderson-Hasselbalch equation, and is frequently encountered in biology courses

• When preparing a buffer, the concentration ratio is usually near 1, so the pH is mostly determined by the pKa of the acid

][

][logppH

or ][

][logppH

acid

saltK

HA

AK

a

initial

initiala

Henderson-Hasselbalch Equation

This shows that the pH is determined largely This shows that the pH is determined largely by the pKby the pKaa of the acid and then adjusted by of the acid and then adjusted by

the ratio of acid and conjugate base.the ratio of acid and conjugate base.

pH pKa + log[Conj. base]

[Acid]pH pKa + log

[Conj. base][Acid]

• Typically, the weak acid is selected so the the desired pH is within one unit of the pKa

• A buffer’s capacity is determined by the magnitudes of the molarities of its components

• Generally, the pH change in an experiment must be limited to about

1ppH:buffer useful aFor aK

unit pH 1.0

Adding an Acid to a Buffer

Problem:Problem: What is the pH when 1.00 mL of 1.00 What is the pH when 1.00 mL of 1.00 M HCl is added toM HCl is added to

a)a) 1.00 L of pure water (before HCl, pH = 7.00)1.00 L of pure water (before HCl, pH = 7.00)

b)b) 1.00 L of buffer that has [HOAc] = 0.700 M 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M Ka(HOAc)=1.8E-5and [OAc-] = 0.600 M Ka(HOAc)=1.8E-5

Adding an Acid to a BufferProblem:Problem: What is the pH when 1.00 mL of 1.00 M HCl What is the pH when 1.00 mL of 1.00 M HCl

is added tois added toa)a) 1.00 L of pure water (before HCl, pH = 7.00)1.00 L of pure water (before HCl, pH = 7.00)b)b) 1.00 L of buffer that has [HOAc] = 0.700 M and 1.00 L of buffer that has [HOAc] = 0.700 M and

[OAc-] = 0.600 M Ka(HOAc)=1.8E-5[OAc-] = 0.600 M Ka(HOAc)=1.8E-5

Solution to Part (a)Solution to Part (a)Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of

waterwater MM11•V•V11 = M = M22 • V • V22

MM22 = 1.00 x 10 = 1.00 x 10-3-3 M M

pH = 3.00pH = 3.00(pH = 4.68)(pH = 4.68)

A buffer made from 0.10 mol HA (pKa=7.20) and 0.15 mol NaA in 2.0 L has 0.02 mol of HCl added to it with no volume change. What is the pH change?ANALYSIS: This buffer problem is “best” solved in terms of moles.

HCl is a strong acid. The H+ it contributes to solution increases the amount of HA present at the expense of A-.

SOLUTION: The pH before addition of HCl was:

A buffer made from 0.10 mol HA (pKa=7.20) and 0.15 mol NaA in 2.0 L has 0.02 mol of HCl added to it with no volume change. What is the pH change?ANALYSIS: This buffer problem is “best” solved in terms of moles.

HCl is a strong acid. The H+ it contributes to solution increases the amount of HA present at the expense of A-.

SOLUTION: The pH before addition of HCl was:

38.7mol 10.0

mol 15.0log20.7

][

][logppH

HA

AKa

After the HCl ionizes and reacts:

• The pH change was greater than –0.1 unit. The buffer effectively resisted the pH change, however, because if the HCl had been added to pure water, the pH change would have been much larger:

15.038.723.7pHpHpH

is change pH theand ,23.7mol 0.12

mol 0.13log7.20pH

:issolution new theof pH the

mol 0.12mol )02.010.0(][

mol 0.13mol )02.015.0(][

initialfinial

final

final

HA

A

00.500.70.2

02.0logpH

What is the pH of a dihydrogen phosphate buffer in which [H2PO4

-] = 0.31 M, and [HPO4

2-] = 0.35 M?a) 5.3E-2 b) 7.15 c) 7.20 d) 7.26 e) 7.31

What is the pH of a monohydrogen phosphate buffer in which [HPO4

2-] = 0.55 M, and [PO4

3-] = 0.51 M?a) 12.25 b) 12.30 c) 12.35 d) 12.45 e) 12.50

• Acids that can donate more than one H+ to solution are called polyprotic acids

• Table 18.3 (and Appendix C) list the ionization constants of a number of polyprotic acids

• The ionization constants for these acids are numbered to keep tract of the degree of ionization

• Note that, for a given polyprotic acid, the magnitudes of the ionization constants are always: Ka1 > Ka2 (> Ka3, if applicable)

Suppose some H3PO4 is added to water, using the constants in Table 18.3:

13

24

34

334

24

8

42

24

22442

3

43

4214243

105.4

][

]][[ )()(

103.6

][

]][[ )()(

101.7

][

]][[ )()(

HPO

POHKPOaqHaqHPO

POH

HPOHKHPOaqHaqPOH

POH

POHHKPOHaqHaqPOH

a

a

a

– The total [H+] is then

– This is generally true when any polyprotic acid is added to water

• This greatly simplifies the determination of the pH in solutions of polyprotic acids

stepfirst total

step thirdstep secondstepfirst

321

step thirdstep secondstepfirst total

][][

that so ][][][

that followsit ,

Since

][][][][

HH

HHH

KKK

HHHH

aaa

Example: What is the pH and [CO32-] in 0.10 M

carbonic acid (H2CO3)?

ANALYSIS: Carbonic acid is a diprotic acid. The pH will depend on the [H+] generated from the first ionization. Ionization constants can be obtained from Table 18.3

SOLUTION: Solve the first ionization first, then substitute the results into the second reaction.

7

32

31332

103.4

][

]][[ )()()(

COH

HCOHKaqHCOaqHaqCOH a

– Applying the usual procedures:

– The pH is 3.68. Substituting these results into the second ionization equation:

][][101.2 and 10.0

thatso 400

34

2

1

1

HCOHMxx

K

Kx

a

a

11232

3

3

23

2233

107.4][

that so ][][but

][

]][[ )()()(

COK

HCOH

HCO

COHKaqCOaqHaqHCO

a

a

• Titration of a strong acid by a strong base

Titration curve for the titration of 25.00 mL of 0.2000 M HCl (a strong acid) with the 0.2000 M NaOH (a strong base). The equivalence point occurs at 25.00 mL added base with a pH of 7.0 (data from Table 18.4).

• Titration of a weak acid by a strong base– This can be divided into four regions

1) Before the titration begins: this is simple a solution of weak acid

2) During the titration, but before the equivalence point: the solution is a buffer

3) At the equivalence point: the solution contains a salt of the weak acid, and hydrolysis can occur

4) Past the equivalence point: the excess added OH- is used to determine the pH of the solution

– Data for the titration of acetic acid with sodium hydroxide is tabulated in Table 18.5

The titration curve for the titration of 25.00 mL of 0.200 M acetic acid with 0.200 M sodium hydroxide. Due to hydrolysis, the pH at the equivalence point higher than 7.00 (data from Table 18.5).

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?pH of the final solution?

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?pH of the final solution?

Equivalence pointEquivalence pointEquivalence pointEquivalence point

pH of solution of benzoic pH of solution of benzoic acid, a weak acidacid, a weak acid

pH of solution of benzoic pH of solution of benzoic acid, a weak acidacid, a weak acid

Acid-Base ReactionsAcid-Base Reactions

EQUILIBRIUM PORTION

Bz- + H2O HBz + OH- Kb = 1.6 x 10-10

[Bz-] [HBz] [OH-]

equilib 0.020 - x x x

Solving in the usual way, we find

x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25

EQUILIBRIUM PORTION

Bz- + H2O HBz + OH- Kb = 1.6 x 10-10

[Bz-] [HBz] [OH-]

equilib 0.020 - x x x

Solving in the usual way, we find

x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?

Kb 1.6 x 10-10 = x2

0.020 - xKb 1.6 x 10-10 =

x2

0.020 - x

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?point. What is the pH of the final solution?

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?point. What is the pH of the final solution?

Half-way Half-way pointpointHalf-way Half-way pointpoint

Acid-Base ReactionsAcid-Base Reactions

HBzHBz + H+ H22O O H H33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5

[H[H33OO++] = { [HBz] / [Bz] = { [HBz] / [Bz--] } K] } Kaa

At the half-way point, [HBz] = [BzAt the half-way point, [HBz] = [Bz--], so], so

[H[H33OO++] = K] = Kaa = 6.3 x 10 = 6.3 x 10-5-5

pH = 4.20 pH = 4.20 NOTE: pH = pKNOTE: pH = pKaa at 1/2 point at 1/2 point

HBzHBz + H+ H22O O H H33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5

[H[H33OO++] = { [HBz] / [Bz] = { [HBz] / [Bz--] } K] } Kaa

At the half-way point, [HBz] = [BzAt the half-way point, [HBz] = [Bz--], so], so

[H[H33OO++] = K] = Kaa = 6.3 x 10 = 6.3 x 10-5-5

pH = 4.20 pH = 4.20 NOTE: pH = pKNOTE: pH = pKaa at 1/2 point at 1/2 point

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point?What is the pH at the half-way point?

• Titration of a weak base by a strong acid– This is similar to the titration of a weak acid

by strong base– Again dividing into four regions

1) Before the titration begins: this is a solution of a weak base in water

2) During the titration, but before the equivalence point: the solution is a buffer

3) At the equivalence point: the solution contains the salt of the weak base, and hydrolysis can occur

4) Past the equivalence point: excess added H+ determines the pH of the solution

Titration curve for the titration of 25.00 mL of 0.200 M NH3 with 0.200 M HCl. The pH at the equivalence point is below 7.00 because of the hydrolysis of NH4

+.

Problem Solving

• For the titration of 25.00 mL of 0.1000 M ammonia with 0.1000 M HCl, calculate the pH (a) before the addition of any HCl solution, (b) after 10.00 mL of the acid has been added, (c) after half of the NH3 has been neutralized, and (d) at the equivalence point.

Problem Solving• Determine the pH for the titration of 25.00 mL of

0.200 M acetic acid with 0.200 M sodium hydroxide after

the addition of 0.00 mL of NaOHthe addition of 10.00 mL of NaOHthe addition of 24.99 mL of NaOHthe addition of 25.00 mL of NaOHthe addition of 25.01 mL of NaOHthe addition of 26.00 mL of NaOH

This applies to the next 4 questions Exactly 100 mL of 0.24 M HCl is pipetted into a 300 mL flask. What is the pH before any base is added?

a) 0.24 b) 0.48 c) 0.62 d) 0.97 e) 1.62

Exactly 100 mL of 0.24 M HCl is pipetted into a 300 mL flask.

What is the pH after 17 mL of 0.30M NaOH have been added to the flask?a) 0.62 b) 0.80 c) 0.97 d) 1.06 e) 7.00


How many total mL of 0.30 M NaOH must be added to the flask to reach the equivalence point?a) 60 b) 80 c) 75 d) 87 e) 93


What is the pH at the equivalence point in the titration of HCl by NaOH? a) 4 b) 5c) 6d) 7e) 8

• Titration curves for diprotic acids– The features are similar to those for monoprotic

acids, but two equivalence points are reached

The titration of the diprotic acid H2A by a strong base. As each equivalence point is reached, the pH rises sharply.

• A few general comments about indicators can be made– Most dyes that are acid-base indicators are

weak acids, which can be represented as HIn– The color change can be represented as:

color)(another color) (one

form base form acid

][

]][[ )()()(

HIn

InHKaqInaqHaqHIn HIn

– The color change will “appear” to the human eye near the equivalence point of the indicator

– At the equivalence point, the concentration of the acid and base form are equal, so that

– The best indicators have intense color(s) so only a small amount will produce an intense color change that is “easy” to see and won’t consume too much of the titrant

HInKppH point eequivalenc at the

Exactly 100 mL of 0.20 M HC2H3O2 are pipetted into a 300 mL flask. What is the pH before any base is added?a) 0.20 b) 0.70 c) 0.89 d) 1.73 e) 2.72 This applies to next 3 questions also

Exactly 100 mL of 0.20 M HC2H3O2

are pipetted into a 300 mL flask. What is the pH after 23 mL of 0.30M NaOH have been added to the flask?a) 3.48 b) 3,72 c) 4.45 d) 5.97

e) 7.00

Exactly 100 mL of 0.20 M HC2H3O2 are

pipetted into a 300 mL flask.

What is the total mL of 0.30 M NaOH that must be added to the flask to reach the equivalence point in this titration?a) 47 b) 58 c) 61 d) 67 e) 97

Exactly 100 mL of 0.20 M HC2H3O2 are

pipetted into a 300 mL flask.

What is the pH at the equivalence point, when 67 mL of 0.30 M NaOH have been added?a) 6.35 b) 7.00 c) 7.90 d) 8.91 e) 9.12

What is the pH of a 2.0E-1 M solution of NH4+?

Ka for NH4+ is 5.6E-10.

a) 4.74b) 4.97 c) 6.98 d) 8.55 e) 9.25

What is the pH of a 0.06 M solution of NH3 in water?(Kb for NH3 is 1.8E-5.)

a) 1.15 b) 4.74 c) 9.25 d) 11.02 e) 12.85

What is the % ionization in a 1.27E-3 M solution of H2CO3?Ka for H2CO3 is 4.2E-7.

a) 0.033 % b) 1.8% c) 3.0 %d) 3.3 %e) 30%

The pH of a HC2H3O2 solution is 3.37. What is the [C2H3O2

-]/[HC2H3O2]ratio at this pH?Ka for HC2H3O2 is 1.8E-5.a) 1.0 E-2 b) 3.0 E-2 c) 4.2E-2 d) 1.2 E-1 e) 1.23

When 0.0300 moles of a monoprotic acid is dissolved in 500. ml water, the pH is 1.90.What is the Ka for this acid?

a) 3.3E-3 b) 3.6E-3 c) 6.6E-2 d) 1.1 E-2 e) 2.1E-1

The Ka for H2PO4- is 6.2E-8.

What is the Kb value for the conjugate base, HPO4

2-, at 298 K?

a) 6.2E-8 b) 1.6E-7 c) 4.38 d) 7.21 e) 6.2E6

What is the [OH-] of a 0.034 M solution of NaCN in water a) 1.0E-7 M b) 3.3E-6 Mc) 4.3E-6 Md) 8.3E-5 Me) 9.2E-4 M

What is the pH of a 0.031 M solution of NaC2H3O2 in water

a) 6.82 b) 7.89 c) 8.31 d) 8.62e) 9.10

Chapter 16: Equilibria in Solutions of Weak Acids and Bases.

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Transcript of Chapter 16: Equilibria in Solutions of Weak Acids and Bases.