1 18 Ionic Equilibria: Acids and Bases. 2 Chapter Goals 1.A Review of Strong Electrolytes 2.The...

18 Ionic Equilibria: Acids and Bases

Chapter Goals

1. A Review of Strong Electrolytes

2. The Autoionization of Water

3. The pH and pOH Scales

4. Ionization Constants for Weak Monoprotic Acids and Bases

5. Polyprotic Acids

6. Solvolysis

7. Salts of Strong Bases and Strong Acids

Chapter Goals

8. Salts of Strong Bases and Weak Acids

9. Salts of Weak Bases and Strong Acids

10.Salts of Weak Bases and Weak Acids

11.Salts That Contain Small, Highly Charged Cations

A Review of Strong Electrolytes

• This chapter details the equilibria of weak acids and bases.– We must distinguish weak acids and bases from strong

electrolytes.

• Weak acids and bases ionize or dissociate partially, much less than 100%.– In this chapter we will see that it is often less than 10%!

• Strong electrolytes ionize or dissociate completely.– Strong electrolytes approach 100% dissociation in

aqueous solutions.

• There are three classes of strong electrolytes.

1 Strong Water Soluble AcidsRemember the list of strong acids from Chapter 4.

3(aq)(aq)%100

3(aq)(aq)3100%

)(2)3(

NOHHNO

NOOH OHHNO

3(aq)(aq)%100

3(aq)(aq)3100%

)(2)3(

NOHHNO

NOOH OHHNO

2 Strong Water Soluble BasesThe entire list of these bases was also introduced in

Chapter 4.

100% OH2(s)

-(aq)(aq)

100% OH(s)

OH 2Sr Sr(OH)

OHKKOH

3 Most Water Soluble SaltsThe solubility guidelines from Chapter 4 will help you

remember these salts.

3(aq)2(aq)

100% OHs23

-(aq)(aq)

100% OH(s)

NO 2Ca )Ca(NO

ClNaNaCl

• The calculation of ion concentrations in solutions of strong electrolytes is easy.

• Example 18-1: Calculate the concentrations of ions in 0.050 M nitric acid, HNO3.

0.050 0.050 050.0

NOOHOHHNO 3(aq)(aq)3100%

)(2)3(

• Example 18-2: Calculate the concentrations of ions in 0.020 M strontium hydroxide, Sr(OH)2, solution.

You do it!

0.0202 0.020 020.0

OH 2SrSr(OH) -(aq)

OH2(s)

The Autoionization of Water

• Pure water ionizes very slightly.– The concentration of the ionized water is less than

one-millionth molar at room temperature.

• We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter.

• Because the activity of pure water is 1, the equilibrium constant for this reaction is:

-(aq)(aq)3)(2)(2 OHOHOH OH

Kc H3O+ OH

• Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC.– Note that this is at 25oC, not every temperature!

• We can determine the value of Kc from this information.

Kc H3O+ OH

1.0 x 10-7 1.0 x 10-7 1.0 x10 14

• This particular equilibrium constant is called the ion-product for water and given the symbol Kw.

– Kw is one of the recurring expressions for the remainder of this chapter and Chapters 19 and 20.

Kw H3O+ OH

1.0 x10 14

The Autoionization of Water• Example 18-3: Calculate the concentrations of

H3O+ and OH- in 0.050 M HCl.

HCl + H2O H3O+ Cl

0.050M 0.050M 0.050M

Thus the H3O+ 0.050M.

The H3O+ and Kw will allow us to calculate [OH -].

• Use the [H3O+] and Kw to determine the [OH-].

You do it!

H3O+ OH 1.0 10 14

OH 1.0 10 14

1.0 10 14

5.0 10 2

OH 2.0 10 13M

• The increase in [H3O+] from HCl shifts the equilibrium and decreases the [OH-].– Remember from Chapter 17, increasing the product

concentration, [H3O+], causes the equilibrium to shift to the reactant side.

– This will decrease the [OH-] because it is a product!

. 0.050 10 1.0 0.050 ]O[H overall The

.101.0 is K from ]O[H The

OH OH OH OH

0.050 is HCl from ]O[H The

• Now that we know the [H3O+] we can calculate the [OH-].

You do it!You do it!

100.2]OH[

] [0.050

101]OH[

. 0.050 ]O[H Since

101][OH-]O[HK

The pH and pOH scales

• A convenient way to express the acidity and basicity of a solution is the pH and pOH scales.

• The pH of an aqueous solution is defined as:

pH = -log H3O+

• In general, a lower case p before a symbol is read as the ‘negative logarithm of’ the symbol.

• Thus we can write the following notations.

pOH = -log OH - pAg = -log Ag+

and so forth for other quantities.

• If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated.

• Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M.

pH = -log H3O+

pH log 3.0 10 2 pH 1.52

• Example 18-5: The pH of a solution is 4.597. What is the concentration of H3O+?

4.597-3

1053.2]O[H

10]O[H

-4.597]Olog[H

]O-log[H4.597

]O-log[HpH

• A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C.

• Taking the logarithm of both sides of this equation gives:

[H3O][OH ] 1.0 10 14

log H3O log OH 14.00

• Multiplying both sides of this equation by -1 gives:

• Which can be rearranged to this form:

-log H3O log OH 14.00

pH pOH 14.00

• Remember these two expressions!! – They are key to the next three chapters!

H3O OH 1.0 10 14

pH pOH 14.00

• The usual range for the pH scale is 0 to 14.

• And for pOH the scale is also 0 to 14 but inverted from pH.– pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0.

H3O 1.0 M to H3O

1.0 10 14M

pH 0 to pH 14.00

OH 1.0 10 14M up to OH 1.0M

pOH 14.00 pOH 0

• Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution.

– Is HNO3 a weak or strong acid?

– What is the [H3O+] ?

70.1pH

100.2-logpH

100.2OH

0.020 0.020 020.0

NOOHOHHNO

100%23

• Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution.

Kw H3O OH 1.0 10 14

OH 1.0 10 14

1.0 10 14

2.0 10 2 5.0 10 13M

pOH log 5.0 10 13 12.30

The pH and pOH scales• To help develop familiarity with the pH and pOH scale we

can look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.

[H3O+] [OH-] pH pOH

1.0 M 1.0 x 10-14 M 0.00 14.00

1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00

1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00

2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30

1.0 x 10-14 M 1.0 M 14.00 0.00

Ionization Constants for Weak Monoprotic Acids and Bases• Let’s look at the dissolution of acetic acid, a

weak acid, in water as an example.• The equation for the ionization of acetic acid is:

• The equilibrium constant for this ionization is expressed as:

-3323 COOCHOHOH COOHCH

Kc H3O

CH3COO CH3COOH H2O

Ionization Constants for Weak Monoprotic Acids and Bases• The water concentration in dilute aqueous

solutions is very high.• 1 L of water contains 55.5 moles of water.• Thus in dilute aqueous solutions:

M5.55OH2

Ionization Constants for Weak Monoprotic Acids and Bases• The water concentration is many orders of

magnitude greater than the ion concentrations.• Thus the water concentration is essentially that of

pure water.– Recall that the activity of pure water is 1.

Kc H2O H3O CH3COO

CH3COOH

CH3COO CH3COOH

Ionization Constants for Weak Monoprotic Acids and Bases• We can define a new equilibrium constant for weak

acid equilibria that uses the previous definition.– This equilibrium constant is called the acid ionization

constant.

– The symbol for the ionization constant is Ka.

Ka H3O

CH3COO CH3COOH 1.8 10 5

for acetic acid

Ionization Constants for Weak Monoprotic Acids and Bases• In simplified formsimplified form the dissociation equation and

acid ionization expression are written as:

CH3COOH H CH3COO -

Ka H CH3COO

CH3COOH 1.8 10 5

Ionization Constants for Weak Monoprotic Acids and Bases• The ionization constant values for several acids

are given below.– Which acid is the strongest?

Acid Formula Ka value

Acetic CH3COOH 1.8 x 10-5

Nitrous HNO2 4.5 x 10-4

Hydrofluoric HF 7.2 x 10-4

Hypochlorous HClO 3.5 x 10-8

Hydrocyanic HCN 4.0 x 10-10

Ionization Constants for Weak Monoprotic Acids and Bases• From the above table we see that the order of

increasing acid strength for these weak acids is:

• The order of increasing base strength of the anions (conjugate bases) of these acids is:

HCN>HClO>COOHCH>HNO>HF 32

- CN<ClO<COOCH<NO<F

Ionization Constants for Weak Monoprotic Acids and Bases• Example 18-8: Write the equation for the

ionization of the weak acid HCN and the expression for its ionization constant.

10 x 4.0HCN

CN H HCN

Ionization Constants for Weak Monoprotic Acids and Bases• Example 18-9: In a 0.12 M solution of a weak

monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid.

HY H + Y

Ionization Constants for Weak Monoprotic Acids and Bases• Since the weak acid is 5.0% ionized, it is also 95% unionized.• Calculate the concentration of all species in solution.

11.0)12.0(95.0HY

100.6YH

0060.0)12.0(05.0YH3+

Ionization Constants for Weak Monoprotic Acids and Bases

• Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.

103.3K

100.6 100.6K

Ionization Constants for Weak Monoprotic Acids and Bases• Example 18-10: The pH of a 0.10 M solution of a

weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant?

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

101.1H

Ionization Constants for Weak Monoprotic Acids and Bases• Use the [H3O+] and the ionization reaction to

determine concentrations of all species.

HA H A

Equil. []'s 0.10 -1.1 10 1.1 10 1.1 10

-3 -3 -3

Ionization Constants for Weak Monoprotic Acids and Bases• Calculate the ionization constant from this

information.

102.1K

101.1101.1

Ionization Constants for Weak Monoprotic Acids and Bases• Example 18-11: Calculate the concentrations of the

various species in 0.15 M acetic acid, CH3COOH, solution.

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

108.1COOHCH

COOCHOHK

COOCHOH OHCOOHCH

Ionization Constants for Weak Monoprotic Acids and Bases• Next, combine the basic chemical concepts with

some algebra to solve the problem.

M0.15 [] Initial

COOCH OH OHCOOHCH -3323

Ionization Constants for Weak Monoprotic Acids and Bases• Next we combine the basic chemical concepts

with some algebra to solve the problem

xMxMxM

- Change

0.15 [] Initial

Ionization Constants for Weak Monoprotic Acids and Bases• Next we combine the basic chemical concepts

with some algebra to solve the problem

xMxM-x)M.(

xMxMxM

150 [] mEquilibriu

- Change

0.15 [] Initial

• Substitute these algebraic quantities into the ionization expression.

Ka H3O

CH3COO CH3COOH

0.15 x 1.8 10 5

Ionization Constants for Weak Monoprotic Acids and Bases• Solve the algebraic equation, using a simplifying assumption that is

appropriate for all weak acid and base ionizations.

108.115.0

• Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.

108.115.0

[]. tocompared ignore enough to small is

.assumption thismake then 10 K If

108.115.0

• Complete the algebra and solve for the concentrations of the species.

x 2 2.7 10 6

x 1.6 10 3M H3O CH3COO

CH3COOH 0.15 1.6 10 3 M 0.15M

Ionization Constants for Weak Monoprotic Acids and Bases• Note that the properly applied simplifying assumption gives

the same result as solving the quadratic equation does.

0107.2108.1

108.115.0

101.6- and 106.1

107.214108.1108.1

Ionization Constants for Weak Monoprotic Acids and Bases• Let us now calculate the percent ionization for the

0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is

%1.1%10015.0

106.1ionization %

%100COOHCH

COOHCH= ionization %

original3

ionized3

Ionization Constants for Weak Monoprotic Acids and Bases• Example 18-12: Calculate the

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

Ka= 4.0 x 10-10 for HCN

MxMxMx

15.0 15.0HCN

CNH107.7

100.415.0

-0.15 mEquilibriu

+ + - Change

0.15 Initial

CN OH OH HCN

Ionization Constants for Weak Monoprotic Acids and Bases• The percent ionization of 0.15 M HCN solution is

calculated as in the previous example.

%0051.0%10015.0

107.7ionization %

%100HCN

HCN = ionization %

original

ionized

Ionization Constants for Weak Monoprotic Acids and Bases• Let’s look at the percent ionization of two weak acids

as a function of their ionization constants. Examples 18-11 and 18-12 will suffice.

• Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.

Solution Ka [H+] pH % ionization

0.15 M acetic acid

1.8 x 10-5 1.6 x 10-3 2.80 1.1

0.15 M

4.0 x 10-10 7.7 x 10-6 5.11 0.0051

Ionization Constants for Weak Monoprotic Acids and Bases• All of the calculations and understanding we have

at present can be applied to weak acids and weak bases!

• One example of a weak base ionization is ammonia ionizing in water.

Ionization Constants for Weak Monoprotic Acids and Bases• All of the calculations and understanding we have at

present can be applied to weak acids and weak bases!

• Example 18-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia.

xMxM-x)M.(

xMxMxM

150 [] mEquilibriu

- Change

0.15 [] Initial

OH NH OH NH -423

xMxM-x)M.(

xMxMxM

15.0101.615.0NH

101.6OHNH

101.6 and 107.2

108.1)15.0(

108.115.0

0.15 x - 0.15 thus0.15x

valid.is assumption gsimplifyin The

108.115.0

OH NHK

150 [] mEquilibriu

- Change

0.15 [] Initial

OH NH OH NH

Ionization Constants for Weak Monoprotic Acids and Bases• The percent ionization for weak bases is

calculated exactly as for weak acids.

% ionization NH3 ionized

NH3 original

1.6 10 3M

0.15M100%

Ionization Constants for Weak Monoprotic Acids and Bases• Example 18-14: The pH of an aqueous

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

363.2pOH-

103.2NH

103.21010OH

2.63 11.37-14.00pH - 14.00=pOH

pOH. thederivecan we14.00, = pOH + pH From

11.37=pH

Ionization Constants for Weak Monoprotic Acids and Bases• Use the ionization equation and some algebra to

get the equilibrium concentration.

NH 3 H2O NH 4 OH -

Initial[] xM

Change - 2.310-3 +2.310-3 +2.310-3

Equilibrium[] x - 2.310-3 M +2.310-3 +2.310-3

Ionization Constants for Weak Monoprotic Acids and Bases• Substitute these values into the ionization

constant expression.

Kb NH4

OH NH3 1.8 10 5

1.8 10 5 2.310 3 2.310 3

x 2.310 3

Ionization Constants for Weak Monoprotic Acids and Bases• Examination of the last equation suggests that

our simplifying assumption can be applied.• In other words (x-2.3x10-3) x.

– Making this assumption simplifies the calculation.

2.310 3 2

x1.8 10 5

x 0.30M NH3

Polyprotic Acids• Many weak acids contain two or more acidic hydrogens.

– Examples include H3PO4 and H3AsO4.

• The calculation of equilibria for polyprotic acids is done in a stepwise fashion.– There is an ionization constant for each step.

• Consider arsenic acid, H3AsO4, which has three ionization constants.1 Ka1 = 2.5 x 10-4

2 Ka2 = 5.6 x 10-8

3 Ka3 = 3.0 x 10-13

Polyprotic Acids

• The first ionization step for arsenic acid is:

H3AsO4 H H2AsO4

Ka1 H H2AsO4

H3AsO4 2.5 10 4

Polyprotic Acids

• The second ionization step for arsenic acid is:

H2AsO41- H HAsO4

Ka2 H HAsO4

2 H2AsO4

1- 5.6 10 8

Polyprotic Acids

• The third ionization step for arsenic acid is:

HAsO42- H AsO 4

Ka3 H AsO4

3 HAsO4

2- 3.0 10 13

Polyprotic Acids

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

• This is a general relationship.– For weak polyprotic acids the Ka1 is always > Ka2,

a3a2a1 KKK

Polyprotic Acids

• Example 18-15: Calculate the concentration of all species in 0.100 M arsenic acid, H3AsO4, solution.

1 Write the first ionization step and represent the concentrations.Approach this problem exactly as previously done.

xMxMMx 100.0

AsOHHAsOH 4243

Polyprotic Acids

2 Substitute the algebraic quantities into the expression for Ka1.

apply.not does assumption gsimplifyin thecase, In this

0105.2105.2

105.210.0

105.2AsOH

AsOH HK

Polyprotic Acids

3 Use the quadratic equation to solve for x, and obtain both values of x.

x 2.5 10 4 2.5 10 4 2

4 1 2.5 10 5 2 1

x 5.110 3M and x 4.9 10 3M

H H2AsO4 xM 4.9 10 3M

H3AsO4 0.100 x M 0.095M

Polyprotic Acids

4 Next, write the equation for the second step ionization and represent the concentrations.

yMyMMy

)10(4.9lly algebraica

)10(4.9 step1st ]from [

HAsO + H AsOH

Polyprotic Acids

5 Substitute the algebraic expressions into the second step ionization expression.

Ka2 =H3O

HAsO 42

H2AsO4 5.6 10 8

Ka2 =4.9 10-3 y y

4.9 10-3 y

For this step we can apply the assumption.

Polyprotic Acids

Ka2 =H3O

HAsO42

H2AsO4 5.6 10 8

Ka2 =4.9 10-3 y y

4.9 10-3 y

In this case the assumption can be applied.

y 4.9 10-3

Thus, 4.9 10-3 y 4.9 10-3

Ka2 =4.9 10-3 y

4.9 10-3 5.6 10 8

y 5.6 10 8M H 2nd

HAsO42

Note that H 1st H 2nd

Polyprotic Acids

6 Finally, repeat the entire procedure for the third ionization step.

MzMzMz

105.6 changes ] [ of tionsrepresenta algebraic

105.6104.9 105.6 sionization 2 and 1 from s]' [

AsO H HAsO

8-3-8-ndst

Polyprotic Acids

7 Substitute the algebraic representations into the third ionization expression.

Ka3 =H3O

HAsO42 3.0 10 13

Ka3 =4.9 10 3 5.6 10 8 z z

5.6 10 8 z The assumption can be applied, z 5.6 10-8.

Polyprotic Acids

• Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.

H OH 1.0 10 14

OH 1.0 10 14

H 1.0 10 14

4.9 10 3

OH 2.0 10 12M

4.9 10 3 z 5.6 10 8 3.0 10 13

z 3.4 10 18M H 3rd

Polyprotic Acids• A comparison of the various species in 0.100 M

H3AsO4 solution follows.

Species Concentration

H3AsO4 0.095 M

H+ 0.0049 M

H2AsO4- 0.0049 M

HAsO42- 5.6 x 10-8 M

AsO43- 3.4 x 10-18 M

OH- 2.0 x 10-12 M

Solvolysis• This reaction process is the most difficult concept in this chapter.• Solvolysis is the reaction of a substance with the solvent in which

it is dissolved.• Hydrolysis refers to the reaction of a substance with water or its

• Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.

Solvolysis

• Hydrolysis refers to the reaction of a substance with water or its ions.– Hydrolysis is solvolysis in aqueous solutions.

• The combination of a weak acid’s anion with H3O+ ions, from water, to form nonionized weak acid molecules is a form of hydrolysis.

A H O HA H O

recall H O + H O H O OH

Solvolysis

• The reaction of the anion of a weak monoprotic acid with water is commonly represented as:

mequilibriu water the

upsets OH of removal The

OH HA OH A+

Solvolysis

• Recall that at 25oC• in neutral solutions:

[H3O+] = 1.0 x 10-7 M = [OH-]

• in basic solutions:[H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M

• in acidic solutions:[OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M

Solvolysis

• Remember from BrØnsted-Lowry acid-base theory:• The conjugate base of a strong acid is a very weak base.• The conjugate base of a weak acid is a stronger base.• Hydrochloric acid, a typical strong acid, is essentially

completely ionized in dilute aqueous solutions.

HCl H O H O Cl2 3 ~100%

Solvolysis

• The conjugate base of HCl, the Cl- ion, is a very weak base.– The chloride ion is such a weak base that it will not react with the hydronium ion.

• This fact is true for all strong acids and their anions.

Cl H O No rxn. in dilute aqueous solutions3

Solvolysis

• HF, a weak acid, is only slightly ionized in dilute aqueous solutions.• Its conjugate base, the F- ion, is a much stronger base than the Cl-

• The F- ions combine with H3O+ ions to form nonionized HF.– Two competing equilibria are established.

HF + H O H O F

only slightly

F + H O HF + H O

nearly completely

2 3+ -

Solvolysis

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

1. Salts of Strong Bases and Strong Acids

2. Salts of Strong Bases and Weak Acids

3. Salts of Weak Bases and Strong Acids

4. Salts of Weak Bases and Weak Acids

Salts of Strong Bases and Weak Acids• Salts made from strong acids and strong soluble

bases form neutral aqueous solutions.

• An example is potassium nitrate, KNO3, made from nitric acid and potassium hydroxide.

neutral. issolution theThus

OHOHupset oreaction t no is There

amounts. equalin present are HNO and KOH The

solution in are that ions The

OH OH OHOH

NOK KNO

HNOKOH

3+OHin %100~

Salts of Strong Bases and Weak Acids• Salts made from strong soluble bases and weak acids hydrolyze to

form basic solutions.

– Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions.

• An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid.

base?or acidstronger theisWhich

solution in ions Notice

OH + OH OH + OH

ClONaNaClO

HClONaOH

-OHin %100~)(

Salts of Strong Bases and Weak Acids

Na ClO Na ClO

H O + H O OH + H O

ClO H O HClO H O

+ - in H O -

• We can combine these last two equations into one single equation that represents the total reaction.

ClO H O HClO OH-2

Salts of Strong Bases and Weak Acids

Kb =HClO OH -

• The equilibrium constant for this reaction, called the hydrolysis constant, is written as:

Salts of Strong Bases and Weak Acids• Algebraic manipulation of the previous expression give

us a very useful form of the expression.• Multiply the expression by one written as [H+]/ [H+].

H+/H+ = 1

Kb =HClO OH -

ClO - H H

Kb =HClO

H ClO - H OH -

Salts of Strong Bases and Weak Acids• Which can be rewritten as:

Kb =HClO

H ClO - H OH -

Ka for HClO

Salts of Strong Bases and Weak Acids• Which can be used to calculate the hydrolysis

constant for the hypochlorite ion:

3.5 10

K =HClO OH

ba for HClO

a for HClO

2 9 10 7.

Salts of Strong Bases and Weak Acids• This same method can be applied to the anion of

any weak monoprotic acid.

A H2O HA OH -

Kb =HA OH

Ka for HA

Salts of Strong Bases and Weak Acids• Example 18-16: Calculate the hydrolysis constants for

the following anions of weak acids.

1. The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4.

F H O HF OH

KHF OH

a for HF

7 2 1014 10

Salts of Strong Bases and Weak Acids• The cyanide ion, CN-, the anion of hydrocyanic acid, HCN.

For HCN, Ka = 4.0 x 10-10.

CN H2O HCN +OH -

Kb HCN OH

Ka for HCN

Kb 1.0 10 14

4.0 10 10 2.5 10 5

Salts of Strong Bases and Weak Acids• Example 18-17: Calculate [OH-], pH and percent

hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

MMM 0.10 0.10 0.10

ClO Na NaClO OHin 100%~(s)

Salts of Strong Bases and Weak Acids • Set up the equation for the hydrolysis and the

algebraic representations of the equilibrium concentrations.

ClO + H O HClO + OH

Initial: 0.10

Change: - + +

At equil: 0.10 -

-M M M

xM xM xM

x M xM xM

Salts of Strong Bases and Weak Acids• Substitute the algebraic expressions into the

hydrolysis constant expression.

Kb HClO OH

ClO 2.9 10 7

Kb x x

0.10 x 2.9 10 7

Salts of Strong Bases and Weak Acids• Substitute the algebraic expressions into the

hydrolysis constant expression.

10.23. pH get the we14.00 pOH pH From

3.77 pOH get the we][OH theFrom

OHClO107.1 becomesWhich

109.2 toreducesequation The

0.10 -0.10 and 0.10

case. in this made becan assumption gsimplifyin The

Salts of Strong Bases and Weak Acids• The percent hydrolysis for the hypochlorite ion

may be represented as:

% hydrolysis =ClO- hydrolyzed

ClO- original

% hydrolysis =1.7 10-4M

0.10M100% 0.17%

Salts of Strong Bases and Weak Acids• If a similar calculation is performed for 0.10 M NaF

solution and the results from 0.10 M sodium fluoride and 0.10 M sodium hypochlorite compared, the following table can be constructed.

Solution Ka Kb [OH-] (M) pH%

hydrolysis

NaF 7.2 x 10-4 1.4 x 10-11 1.2 x 10-6 8.08 0.0012

NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23

Salts of Weak Bases and Strong Acids• Salts made from weak bases and strong acids

form acidic aqueous solutions.

• An example is ammonium bromide, NH4Br, made from ammonia and hydrobromic acid.

base?or acidstronger theisWhich

aresolution in Ions

OH OH OHOH

Br NH BrNH

HBrOHNH

100%~OHs4

Salts of Weak Bases and Strong Acids

OH excess generates

OHNHOHNH

OH excess leavingsolution fromit

removingion OH with thereacts

,NH acid, strong relatively The

• The reaction may be more simply represented as:

OHNH OHNH 3324

Salts of Weak Bases and Strong Acids

• The hydrolysis constant expression for this process is:

• Or even more simply as:

Ka NH3 H3O

or Ka NH3 H

HNH NH 34

Salts of Weak Bases and Strong Acids• Multiplication of the hydrolysis constant

expression by [OH-]/ [OH-] gives:

Ka NH3 H3O

OH - OH -

Ka NH3

NH4 OH -

H3O OH - 1

Salts of Weak Bases and Strong Acids• Which we recognize as:

18 105 6 10

Salts of Weak Bases and Strong Acids• In its simplest form for this hydrolysis:

Salts of Weak Bases and Strong Acids• Example 18-18: Calculate [H+], pH, and percent

hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH4Br, solution.

1. Write down the hydrolysis reaction and set up the table as we have done before:

NH H O NH H

Initial[]: 0.10 + +

Change: - + +

Equilibrium[]: 0.10

M xM xM

xM xM xM

x M xM xM

Salts of Weak Bases and Strong Acids2. Substitute the algebraic expressions into the

hydrolysis constant.

0.10 - 0.10 thus0.10

.applicable is assumption The

105.6=10.0

106.5NH

Salts of Weak Bases and Strong Acids3. Complete the algebra and determine the

concentrations and pH.

12.5pH

107.5HNH

107.5=

106.510.0

Salts of Weak Bases and Strong Acids4. The percent hydrolysis of the ammonium ion in

0.10 M NH4Br solution is:

hydrolysis =NH

hydrolysis =7.5 10

hydrolysis = 0.0075%

hydrolized

original

010100%

Salts of Weak Bases and Weak Acids• Salts made from weak acids and weak bases can

form neutral, acidic or basic aqueous solutions.

– The pH of the solution depends on the relative values of the ionization constant of the weak acids and bases.

1. Salts of weak bases and weak acids for which parent Kbase =Kacid make neutral solutions.

• An example is ammonium acetate, NH4CH3COO, made from aqueous ammonia, NH3,and acetic acid, CH3COOH.

Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.

Salts of Weak Bases and Weak Acids• The ammonium ion hydrolyzes to produce H+

ions. Its hydrolysis constant is:

NH NH H

5 6 10 10.

Salts of Weak Bases and Weak Acids• The acetate ion hydrolyzes to produce OH- ions.

Its hydrolysis constant is:

CH COO H O CH COOH OH

KCH COOH OH

CH COO

5 6 10 10.

Salts of Weak Bases and Weak Acids• Because the hydrolysis constants for both ions

are equal, their aqueous solutions are neutral.• Equal numbers of H+ and OH- ions are produced.

solution!in formed are base and acidA weak

aresolution in Ions

OH OH OH OH

COOCHNHCOOCHNH

COOHCHOHNH

34100%~OH

Salts of Weak Bases and Weak Acids2. Salts of weak bases and weak acids for

which parent Kbase > Kacid make basic solutions.

• An example is ammonium hypochlorite, NH4ClO, made from aqueous ammonia, NH3,and hypochlorous acid, HClO.

Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8

Salts of Weak Bases and Weak Acids• The ammonium ion hydrolyzes to produce H+

NH4 NH3 H

Ka NH3 H

NH4 5.6 10 10

Salts of Weak Bases and Weak Acids• The hypochlorite ion hydrolyzes to produce OH-

• Because the Kb for ClO- ions is three orders of magnitude larger than the Ka for NH4

+ ions, OH- ions are produced in excess making the solution basic.

ClO H O HClO OH

KHClO OH

2 9 10 7.

Salts of Weak Bases and Weak Acids

3. Salts of weak bases and weak acids for which parent Kbase < Kacid make acidic solutions.

• An example is trimethylammonium fluoride,(CH3)3NHF, made from trimethylamine, (CH3)3N,and hydrofluoric acid acid, HF.

– Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4

Salts of Weak Bases and Weak Acids• Both the cation, (CH3)3NH+, and the anion, F-,

hydrolyze.

CH NH F CH NH F3 3H O~100%

Salts of Weak Bases and Weak Acids• The trimethylammonium ion hydrolyzes to

produce H+ ions. Its hydrolysis constant is:

(CH ) NH CH N H

KCH N H

(CH ) NH

b for CH N

7 4 1014 10

Salts of Weak Bases and Weak Acids• The fluoride ion hydrolyzes to produce OH- ions.

Its hydrolysis constant is:

• Because the Ka for (CH3)3NH+ ions is one order of magnitude larger than the Kb for F- ions, H+ ions are produced in excess making the solution acidic.

F H O HF OH

KHF OH

a for HF

7 2 1014 10

Salts of Weak Bases and Weak Acids• Summary of the major points of

hydrolysis up to now.1. The reactions of anions of weak

monoprotic acids (from a salt) with water to form free molecular acids and OH-.

A + H O HA + OH

Salts of Weak Bases and Weak Acids2. The reactions of anions of weak monoprotic

acids (from a salt) with water to form free molecular acids and OH-.

BH + H O B + H O

K B = weak base

Salts of Weak Bases and Weak Acids• Aqueous solutions of salts of strong acids and

strong bases are neutral.• Aqueous solutions of salts of strong bases

and weak acids are basic. • Aqueous solutions of salts of weak bases and

strong acids are acidic.• Aqueous solutions of salts of weak bases and

weak acids can be neutral, basic or acidic.The values of Ka and Kb determine the pH.

Hydrolysis of Small Highly-Charged Cations

• Cations of insoluble bases (metal hydroxides) become hydrated in solution.– An example is a solution of Be(NO3)3.

– Be2+ ions are thought to be tetrahydrated and sp3 hybridized.

2+ Be(OH2 )4 2

1s 2s 2p

molecules water dcoordinateon pairs e

)Be(OH

2p 2s 1s

hybrids sp form

xxxxxxxx

Hydrolysis of Small Highly-Charged Cations• In condensed form it is represented as:

or, even more simply as:

Be(OH ) H O Be OH OH H O2 4 2 2 3 32

Be H O Be(OH) H2+2

Hydrolysis of Small Highly-Charged Cations• The hydrolysis constant expression for [Be(OH2)4]2+

and its value are:

or, more simply

Ka Be(OH2)3(OH)+ H3O

Be(OH2)4

2 1.0 10 5

Ka Be(OH)+ H

Be2+ 1.0 10 5

Hydrolysis of Small Highly-Charged Cations• Example 18-19: Calculate the pH and percent

hydrolysis in 0.10 M aqueous Be(NO3)2 solution.

1. The equation for the hydrolysis reaction and representations of concentrations of various species are:

xMxMMx 10.0

H Be(OH) OH Be 22

Hydrolysis of Small Highly-Charged Cations2. Algebraic substitution of the expressions into

the hydrolysis constant:

x x 0.10 x 1.0 10 5

The simplifying assumption applies.

x 0.10 and 0.10 - x 0.10

x x 0.10

1.0 10 5

x 2 1.0 10 6

x 1.0 10 3M

H Be2 hydrolyzed

1.0 10 3M

pH 3.00

Hydrolysis of Small Highly-Charged Cations

3 Calculate the percent hydrolysis of Be2+.

% hydrolyzed =Be2+ hydrolyzed

Be2+ original

% hydrolyzed =1.0 10-3

0.10100% 1.0%

Hydrolysis of Small Highly-Charged Cations• This table is a comparison of 0.10 M Be(NO3 )2

solution and 0.10 M CH3COOH solution.

Solution [H3O+] pH

% hydrolysis or

% ionization0.10 M Be(NO3)2 1.0 x 10-3

M3.00 1.0%

0.10 M CH3COOH

1.3 x 10-3 M 2.89 1.3%

Notice that the Be solution is almost as acidic as the acetic acid solution.

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