ch 17 3-15 acid-base calculations

Calculation of Ionization Constants

In a 0.12 In a 0.12 MM solution, a weak monoprotic acid, HY, solution, a weak monoprotic acid, HY, is 5.0% ionized. Calculate the ionization constant is 5.0% ionized. Calculate the ionization constant for the weak acid.for the weak acid.

HY H + Y

KH Y

HY

+ -

a

+ -

Since the weak acid is 5.0% ionized, it is also 95% Since the weak acid is 5.0% ionized, it is also 95% unionized.unionized.

Calculate the concentrations of all species in Calculate the concentrations of all species in solution.solution.

MM

M

MM

114.0)12.0(95.0

100.6

0060.0)12.0(05.03

HY

YH+


Substitute into the ionization constant expression Substitute into the ionization constant expression to get the value of Kto get the value of Kaa

4

33

102.3

114.0

100.6100.6

HY

Y HKa


The pH of a 0.10 The pH of a 0.10 MM solution of a weak monoprotic solution of a weak monoprotic acid, HA, is found to be 2.97. What is the value acid, HA, is found to be 2.97. What is the value for its ionization constant?for its ionization constant?

pH = 2.97 so [HpH = 2.97 so [H++]= 10]= 10-pH-pH

H O3

10

11 10

2 97

3

.

. M


Use the [HUse the [H33OO++] and the ionization reaction to ] and the ionization reaction to

determine concentrations of all species. determine concentrations of all species.


5a

3-3-

a

3-3-3-

-+

102.1K

0.10

101.1101.1

HA

AHK

0.10

101.1 101.1 101.1-0.10 s]' [ Equil.

A H HA

Calculate the concentrations of the various species in a Calculate the concentrations of the various species in a 0.15 0.15 MM acetic acid, CH acetic acid, CH33COOH, solution.COOH, solution.

It is always a good idea to write down the ionization It is always a good idea to write down the ionization reaction and the ionization constant expression.reaction and the ionization constant expression.

5

3

-33

a

-3323

108.1COOHCH

COOCHOHK

COOCHOH OHCOOHCH


Calculate the concentrations of the various species in a 0.15 Calculate the concentrations of the various species in a 0.15 MM acetic acid, CH acetic acid, CH33COOH, COOH,

solution.solution. Next we combine the basic chemical concepts with some algebra to solve the problemNext we combine the basic chemical concepts with some algebra to solve the problem

xM xM x)M-(0.15 ] [ mEquilibriu

xM xM xM- Change

M00M0.15M ] [ Initial

COOCH OH OHCOOHCH -3323


Substitute these algebraic quantities into the Substitute these algebraic quantities into the ionization expression.ionization expression.

5

3

33a

108.115.0

COOHCH

COOCH OHK

x

xx


Solve the algebraic equation, using a simplifying assumption or Solve the algebraic equation, using a simplifying assumption or using the quadratic.using the quadratic.

52

52

108.115.0

108.115.0

xx

x

x


2a

4acbb

c b a

0107.2108.1

2

652

x

xx

3-3

6255

101.6- and 106.1

12

107.214108.1108.1

x

x


Pick the algebraic answer that makes chemical sense. [H3O+] = x = 1.6 x 10-3

Solve the algebraic equation, using simplifying Solve the algebraic equation, using simplifying assumption. x may be small enough to ignore . . .assumption. x may be small enough to ignore . . .

Sometimes called the 5% rule . . Sometimes called the 5% rule . .

5252

52

108.115.0108.115.0

%5%110015.0

0016.01484.00016.015.0

108.115.0

xxx

x

x

x


Complete the algebra and solve for concentrations.Complete the algebra and solve for concentrations.

MM

Mx

x

x

15.0106.115.0COOHCH

COOCHOH106.1

107.2

108.115.0

33

333

6

52


Calculate the percent ionization for the 0.15 Calculate the percent ionization for the 0.15 MM acetic acid. acetic acid.

%

..

.

ionization =CH COOH

CH COOH3 ionized

3 original

100%

16 10015

100% 11%3 M

M


Calculate the concentrations of the species in Calculate the concentrations of the species in 0.15 0.15 MM hydrocyanic acid, HCN, solution. hydrocyanic acid, HCN, solution.

KKa a = 4.0 x 10= 4.0 x 10-10-10 for HCN for HCN

You do it!You do it!


MMx

Mx

x

x

xx

xMxMMx

xMxMxM

M

15.015.0

107.7

100.6

100.415.0

6

112

10

HCN

CNH

HCN

CN HK

-0.15 mEquilibriu

+ + - Change

0.15 Initial

CN OH OH HCN

a

-32


The percent ionization of 0.15 The percent ionization of 0.15 MM HCN solution is HCN solution is calculated as in the previous example. calculated as in the previous example.

%

..

.

ionization = HCN

HCNionized

original

100%

7 7 10015

100% 0 0051%6 M

M


All of the calculations and understanding we have at present can be All of the calculations and understanding we have at present can be applied to weak acids applied to weak acids and weak bases!and weak bases!

One example of a weak base ionization is ammonia ionizing in water.One example of a weak base ionization is ammonia ionizing in water. Calculate the % Ionization and the pHCalculate the % Ionization and the pH

-5b

-423 101.8K OHNHOHNH


Determining pH from Kb and Initial [B]–I

Problem: Ammonia is commonly used cleaning agent in households andis a weak base, with a Kb of 1.8 x 10-5. What is the pH of a 1.5 M NH3

Solution and percent ionization?Plan: Ammonia reacts with water to form [OH-] , calculate [H3O+] and the pH. The balanced equation and Kb expression are:

NH3 (aq) + H2O(l) NH4+

(aq) + OH-(aq)

Kb = [NH4

+] [OH-]

[NH3]

Concentration (M) NH3 H2O NH4+ OH-

Initial 1.5 ---- 0 0Change -x ---- +x +xEquilibrium 1.5 - x ---- x x

making the assumption: since Kb is small: 1.5 M - x = 1.5 M

Determining pH from Kb and Initial [B]–II

Substituting into the Kb expression and solving for x:

Kb = = = 1.8 x 10-5[NH4

+] [OH-]

[NH3]

(x)(x)

1.5

x2 = 2.7 x 10-5 = 27 x 10-6

x = 5.20 x 10-3 = [OH-] = [NH4+]

Calculating pH:

[H3O+] = = = 1.92 x 10-12Kw

[OH-]

1.0 x 10-14

5.20 x 10-3

pH = -log[H3O+] = - log (1.92 x 10-12) = 12.000 - 0.283

pH = 11.72

Calculate the percent ionization for weak bases Calculate the percent ionization for weak bases exactly as we did for weak acids.exactly as we did for weak acids.

%35.0

%1005.1

102.5

%100NH

NH

ionization %

3

original3

ionized3

M

M


The pH of an aqueous ammonia solution is The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original 11.37. Calculate the molarity (original concentration) of the aqueous ammonia concentration) of the aqueous ammonia solution.solution.

You do it!You do it!


M

M3

4

363.2pH-

423

103.2NH

103.21010OH

2.63=pOH

14.00 = pOH + pH 11.37;=pH

OHNH OHNH


Use the ionization equation and some algebra to Use the ionization equation and some algebra to get the equilibrium concentration.get the equilibrium concentration.

3-3-3-

3-3-3-

-423

102.3+ 102.3+ 102.3- ] m[Equilibriu

102.3+ 102.3+ 102.3- Change

] Initial[

OH NH OH NH

Mx

xM


Substitute into the ionization constant expression.Substitute into the ionization constant expression.

3

335

5

3

4b

103.2

103.2 103.2108.1

108.1NH

OH NHK

x


Examination of the equation suggests that Examination of the equation suggests that ((xx-2.3x10-2.3x10-3-3))xx. Making this assumption . Making this assumption

simplifies the calculation and gives simplifies the calculation and gives

3

5

23

NH 30.0

108.1103.2

Mxx


The Relation Between Ka and Kb of a Conjugate Acid-Base Pair

Acid HA + H2O H3O+ + A-

Base A- + H2O HA + OH-

2 H2O H3O+ + OH-

[H3O+] [OH-] = x[H3O+] [A-]

[HA]

[HA] [OH-]

[A-]

Kw = Ka x Kb

For HNO2

Ka = 4.5 x 10-4

Kb = 2.2 x 10-11

Ka x Kb = (4.5 x 10-4)(2.2 x 10-11) = 9.9 x 10-15

or ~ 10 x 10-15 = 1 x 10 -14 = Kw

Calculate the KCalculate the Kbb for the anions of a weak acids. for the anions of a weak acids.

FF--, fluoride ion, the anion of hydrofluoric acid, HF. , fluoride ion, the anion of hydrofluoric acid, HF.

For HF, KFor HF, Kaa=7.2 x 10=7.2 x 10-4-4..

F H O HF OH

KHF OH

F

KK

K

2

bw

a for HF

b

10 10

7 2 1014 10

14

411.

..

The Relation Between Ka and Kb of a Conjugate Acid-Base Pair

The Stepwise Dissociation of Phosphoric Acid

Phosphoric acid is a weak acid, and normally only looses one proton in solution, but it will loose all three when reacted with a strong basewith heat. The ionization constants are given for comparison.

H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+

(aq)

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+

(aq)

HPO42-

(aq) + H2O(l) PO43-

(aq) + H3O+(aq)

H3PO4 (aq) + 3 H2O(l) PO43-

(aq) + 3 H3O+(aq)

Polyprotic Acids

Many weak acids contain two or more acidic hydrogens.Many weak acids contain two or more acidic hydrogens. polyprotic acids ionize stepwisepolyprotic acids ionize stepwise ionization constant for each stepionization constant for each step

Consider arsenic acid, HConsider arsenic acid, H33AsOAsO44, which has three ionization constants, which has three ionization constants1 KKa1a1=2.5=2.51010-4-4

2 KKa2a2=5.6=5.61010-8-8

3 KKa3a3=3.0=3.01010-13-13

The first ionization step isThe first ionization step is

Polyprotic AcidsArsenic Acid

The second ionization step isThe second ionization step is

Polyprotic Acids

The third ionization step isThe third ionization step is

Polyprotic Acids

Notice that the ionization constants vary in the following fashion:Notice that the ionization constants vary in the following fashion:

This is a general relationship.This is a general relationship. a3a2a1 KKK

Polyprotic Acids

Calculate the concentration of all species in 0.100 Calculate the concentration of all species in 0.100 MM arsenic acid, Harsenic acid, H33AsOAsO44, solution., solution.

1 Write the first ionization ionization step and represent Write the first ionization ionization step and represent the concentrations.the concentrations.

xMxMMx 100.0

AsOHHAsOH 4243

Polyprotic Acids

2 Substitute into the expression for KSubstitute into the expression for K11. .

applynot does assumption gsimplifyin

0105.2105.2

105.210.0

K

105.2AsOH

AsOH HK

542

4a1

4

43

42a1

xx

x

xx

Polyprotic Acids

Use the quadratic equation to solve for Use the quadratic equation to solve for xx, and , and obtain two valuesobtain two values

MMx

MxM

MxMx

x

095.0100.0AsOH

109.4AsOHH

109.4 and 101.5

12

105.214105.2105.2

43

342

33

5244

Polyprotic Acids

4 Now we write the equation for the second step Now we write the equation for the second step ionization and represent the concentrations.ionization and represent the concentrations.

yMMyMy

yMyMyM

M

10(4.9 )-10(4.9 mequilibriu

change

M104.9 104.9 step 1st from

HAsO + H AsOH

3-3-

3-3-

-24

+-42

)

Polyprotic Acids

A comparison of the various species in 0.100 A comparison of the various species in 0.100 MM HH33AsOAsO44 solution follows. solution follows.

Species Concentration H3AsO4 0.095 M

H+ 0.0049 M H2AsO4

- 0.0049 M HAsO4

2- 5.6 x 10-8 M AsO4

3- 3.4 x 10-18 M OH- 2.0 x 10-12 M

Polyprotic Acids

Hydrolysis

Hydrolysis refers to the reaction of a Hydrolysis refers to the reaction of a substance with water or its ions.substance with water or its ions.

solution theofbasicity theNote

OHHAOHA

?nextdoAdoeswhat

ANaNaA

2

The conjugate base of a strong acid is a very weak base.The conjugate base of a strong acid is a very weak base. The conjugate base of a weak acid is a stronger base.The conjugate base of a weak acid is a stronger base. Hydrochloric acid, a typical strong acid, is essentially Hydrochloric acid, a typical strong acid, is essentially

completely ionized in dilute aqueous solutions.completely ionized in dilute aqueous solutions.

HCl H O H O Cl2 3 ~100%

Hydrolysis

-32 FOH OHHF

ANaNaA

The conjugate base of HCl, the ClThe conjugate base of HCl, the Cl- - ion, is a very weak base.ion, is a very weak base.

True of all strong acids and their anions.True of all strong acids and their anions.

Cl H O No rxn. in dilute aqueous solutions3

Hydrolysis

HF, a weak acid, is only slightly ionized in dilute aqueous solutions.HF, a weak acid, is only slightly ionized in dilute aqueous solutions. Its conjugate base, the FIts conjugate base, the F-- ion, is a much stronger base than the Cl ion, is a much stronger base than the Cl-- ion. ion. FF-- ions combine with H ions combine with H33OO++ ions to form nonionized HF. ions to form nonionized HF.

HF + H O H O F

only slightly

F + H O HF + H O

nearly completely

2 3+ -

-3

+2

Hydrolysis

Dilute aqueous solutions of salts that contain Dilute aqueous solutions of salts that contain no free acid or base come in four types:no free acid or base come in four types:

1 Salts of strong acids and strong basesSalts of strong acids and strong bases2 Salts of weak acids and strong basesSalts of weak acids and strong bases3 Salts of strong acids and weak basesSalts of strong acids and weak bases4 Salts of weak acids and weak basesSalts of weak acids and weak bases

Hydrolysis

Salts of Strong Bases and Strong Acids Salts made from strong acids and strong bases Salts made from strong acids and strong bases

form form neutral aqueous solutionsneutral aqueous solutions.. An example is potassium nitrate, KNOAn example is potassium nitrate, KNO33, made , made

from nitric acid and potassium hydroxide.from nitric acid and potassium hydroxide.

neutral OHor OH producenot doesn dissolutio

OH+OH OHOH

NOK KNO

-3

+3

-22

3+OHin %100~

)(32

s

Salts made from strong bases and weak acids hydrolyze to Salts made from strong bases and weak acids hydrolyze to form basic solutionsform basic solutions.. Anions of weak acids (strong conjugate bases) react with Anions of weak acids (strong conjugate bases) react with

water to form hydroxide ionswater to form hydroxide ions An example is sodium hypochlorite, NaClO, made from An example is sodium hypochlorite, NaClO, made from

sodium hydroxide and hypochlorous acid. sodium hydroxide and hypochlorous acid.

+3

-22

-OHin %100~)(

OH+OH OH + OH

ClONaNaClO 2

s

Salts of Strong Bases and Weak Acids

OHHClOOHClO 23-

OHHClOOHClO

OH+OH OH + OH

ClONaNaClO

23-

+3

-22

-OHin %100~)(

2

s

Combine these equations into one single equation Combine these equations into one single equation that represents the reaction:that represents the reaction:

OHHClOOHClO 2-


Calculate [OHCalculate [OH--], pH and percent hydrolysis for the ], pH and percent hydrolysis for the hypochlorite ion in 0.10 hypochlorite ion in 0.10 MM sodium hypochlorite, sodium hypochlorite, NaClO, solution. NaClO, solution.

Na ClO Na ClO

0.10 0.10 0.10

~100%inH O2 M M M


Set up the equation for the hydrolysis and the algebraic Set up the equation for the hydrolysis and the algebraic representations of the equilibrium concentrations.representations of the equilibrium concentrations.

ClO + H O HClO + OH

Initial: 0.10

Change: - + +

At equil: 0.10 -

-2

-M M M

xM xM xM

x M xM xM

0 0


Substitute the algebraic expressions into the Substitute the algebraic expressions into the hydrolysis constant expression. hydrolysis constant expression.

7b 109.2

ClO

OH HClOK



Kb

x xx010

2 9 10 7

..



10.23pH3.77;pOH

OHHClO

Kb

Mxx

x

xx

482

7

107.1;109.2

109.210.0


The percent hydrolysis for the hypochlorite ion The percent hydrolysis for the hypochlorite ion may be represented as:may be represented as:

%

% .

hydrolysis =ClO

ClO

hydrolysis =1.7 10

0.10

-

hydrolyzed-

original

-4

100%

100% 017%M

M


Salts made from weak bases and strong acids form Salts made from weak bases and strong acids form acidic aqueous solutionsacidic aqueous solutions..

An example is ammonium bromide, NHAn example is ammonium bromide, NH44Br, made Br, made

from ammonia and hydrobromic acid.from ammonia and hydrobromic acid.

acidic issolution OH excess generates

OHNHOHNH

OH OH OHOH

Br NH BrNH

3

23-

4

3-

22

-4

100%~OHs

-4

2

Salts of Weak Bases and Strong Acids

The reaction may be more simply represented as:The reaction may be more simply represented as:

OHNH OHNH 3324

HNH NH 34

Salts of Weak Bases and Strong Acids

Or even more simply as:Or even more simply as:

If Parent KIf Parent Kbasebase > K > Kacidacid

make make basic solutionsbasic solutions An example is ammonium hypochlorite, NHAn example is ammonium hypochlorite, NH44ClO, made ClO, made

from aqueous ammonia, NHfrom aqueous ammonia, NH33,and hypochlorous acid, ,and hypochlorous acid,

HClO.HClO. KKbb for NH for NH33 = 1.8x10 = 1.8x10-5-5

KKaa for HClO = 3.5x10 for HClO = 3.5x10-8-8

Salts of Weak Bases and Weak Acids

63

Because the KBecause the Kbb for ClO for ClO-- ions is three orders ions is three orders

of magnitude larger than the Kof magnitude larger than the Kaa for NH for NH44++

ions, OHions, OH-- ions are produced in excess ions are produced in excess solution is basicsolution is basic


If Parent KIf Parent Kbasebase < K < Kacidacid

make make acidic solutionsacidic solutions An example is trimethylammonium fluoride,An example is trimethylammonium fluoride,

(CH(CH33))33NHF, made from trimethylamine, NHF, made from trimethylamine,

(CH(CH33))33N,and hydrofluoric acid acid, HF.N,and hydrofluoric acid acid, HF.

KKbb for (CH for (CH33))33N = 7.4x10N = 7.4x10-5-5

KKaa for HF = 7.2x10 for HF = 7.2x10-4-4


Because the KBecause the Kaa for (CH for (CH33))33NHNH++ ions is one ions is one

order of magnitude larger than the Korder of magnitude larger than the Kbb for F for F--

ions, Hions, H++ ions are produced in excess. ions are produced in excess. solution is acidicsolution is acidic


ch 17 3-15 acid-base calculations

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