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Chapter 18 Equilibria Involving Acids & bases. ARRHENIUS THEORY for ACIDS and BASES. ACIDS : produce hydrogen ions (protons), H + , in solution BASES : produce hydroxide ions, OH - ,in solution, NEUTRALIZATION : H + + OH -  H 2 O Problems with Arrhenius Theory

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chapter 18 equilibria involving acids bases
Chapter 18

Equilibria Involving Acids & bases

arrhenius theory for acids and bases

ARRHENIUS THEORY for ACIDS and BASES

ACIDS: produce hydrogen ions (protons), H+, in solution

BASES: produce hydroxide ions, OH-,in solution,

NEUTRALIZATION: H+ + OH- H2O

Problems with Arrhenius Theory

* H3O+: Hydronium ion rather than H+

* OH(H2O)3- present in solution, not OH-

* Other substances also have acidic or basic properties

bronsted lowery theory of acids and bases

Bronsted-Lowery Theory of Acids and Bases

Acid– any substance donating a proton, H+

Base– any substance accepting a proton

Conjugate Acid-Base Pairs:

e.g. HF + NH3NH4+ + F-

acid 1 base 2 acid 2 base 1

AMPHOTERIC substances have both acidic and basic properties.

Mono-, di-, tri-,………. to polyprotic acids.

Acidic versus nonacidic H atoms in compounds.

slide6

For each of the following reactions, identify the acid, the base, the conjugate base, and the conjugate acid

H2O + H2O  H3O+ + OH-

H2PO4- + H2PO4- H3PO4 + HPO42-

H2SO4 + H2O  H3O+ + HSO4-

H2PO4- + H2O  H3PO4 + OH-

CO2 + 2H2O  HCO3- + H3O+

H2PO4- + H2O  HPO42- + H3O+

Fe(H2O)63+ + H2O  Fe(H2O)5(OH) 2+ + H3O+

HCN + CO32- CN- + HCO3-

slide7

Graphic representations of strong and weak acid equilibria

Strong Acid:

100% Dissociation into ions

HA(aq) + H2O(l) H3O+(aq) + A- (aq)

Equilibrium Favors undissociated acid

Weak Acid:

Very little Dissociation into ions

bronsted lowery theory acid and base strengths

Bronsted-Lowery Theory: Acid and Base Strengths

Proton transfers occur from a

Strong acid to a strong base

e.g. HCl + NaOH  H2O + NaCl

Weak acid to strong base

e.g. CH3COOH + NaOH  ??

Weak base to a stronger base

e.g. HSO41- + HSO31- ???

Will a Reaction occur between…..

a. HS1- and F 1-?? b. HCl and ClO2 1- ??

c. HCl and ClO4 1- ?? d. HCl and HNO3??

relative strengths of some bronsted lowry acids and their conjugate bases

Relative strengths of some Bronsted-Lowry acids and their conjugate bases

AcidBase

StrongestHClO4 ClO4-Weakest

Acids H2SO4 HSO4-bases

HI I-

HBr Br-

HCl Cl-

HNO3 NO3-

H3O+ H2O

HSO4- SO42-

H2SO3 HSO3-

H3PO4 H2PO4-

HNO2 NO2-

HF F-

CH3CO2H CH3CO2-

H2CO3 HCO3-

H2S HS-

NH4+ NH3

HCN CN-

HCO3- CO32-

HS- S2-

H2O OH-

Weakest NH3 NH2- Strongest

Acid OH- O2-bases

bronsted lowery theory acid and base strengths11

Bronsted-Lowery Theory: Acid and Base Strengths

LEVELING EFFECT of SOLVENTS:

The strongest acid in a solvent is the conjugate acid of the solvent. The strongest base is the conjugate base.

Acid H3O+ in water

Base OH- in water

slide12

H2O(l)

H2O(l)

OH-(aq)

H3O+(aq)

Autoionization of Water and the pH Scale

+

+

autoionization of water

Autoionization of Water

H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)

Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25oC

At equilibrium

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

Kw changes with temperature

but [H3O+] = [OH-]

ph and poh scales

pH and pOH SCALES

pH = - log [H3O+]

pOH = - log [OH-]

ph poh calculations

pH, pOH CALCULATIONS

pH = - log [H3O+]

pOH = - log [OH-]

Kw = [H3O+] [OH-] = 1.0 x 10-14

so p[H3O+] + p[OH-] = 14.00

or pH = 14.00 – pOH

CALULATE SOME pH and pOH VALUES

slide16

(Household

Cleaner)

Blood

[H+] pH

10-14 14 1 M NaOH

10-13 13

Basic 10-12 12 Ammonia 10-11 11

10-10 10

10-9 9

10-8 8

Neutral 10-7 7 Pure Water

10-6 6

10-5 5

10-4 4

10-3 3

Acidic 10-2 2

10-1 1

1 0

Milk

Vinegar

Lemon juice

Stomach acid

1 M HCl

slide17

Calculate the pH of each solution:

a. [H+] = 1.4 x 10-3 M e. [OH-] = 8 x 10-11 M

b. [H+] = 2.5 x 10-10 M f. [OH-] = 5.0 M

c,. [H+] = 6.1 M g. pOH = 10.5

d. [OH-] = 3.5 x 10-2 M h. pOH = 2.3

Calculate [H+] and [OH-] for each solution:

a. pH = 7.41 (the normal pH of blood)

b. pH = 15.3

c. pH = -1.0 e. pOH = 5.0

d. pH = 3.2 f. pOH = 9.6

How many significant figures are there in the numbers: 10.78, 6.78, 0.78? If these were pH values, to how many significant figures can you express the [H+]? Explain any discrepancies between your answers to the two questions.

slide18

Values of Kw as a function of temperature are as follows:

Temp (oC)Kw

0 1.14 x 10-15

25 1.00 x 10-14

35 2.09 X 10-14

40 2.92 x 10-14

50 5.47 x 10-14

a. Is the autoionization of water exothermic or endothermic?

b. What is the pH of pure water at 50oC?

slide19

Values of Kw as a function of temperature are as follows:

Temp (oC)Kw

0 1.14 x 10-15

25 1.00 x 10-14

35 2.09 X 10-14

40 2.92 x 10-14

50 5.47 x 10-14

a. Is the autoionization of water exothermic or endothermic?

b. What is the pH of pure water at 50oC?

c. Restate your answers to water at 50oC. Which of the three criteria for neutrality is most general?

d. From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 37oC, normal physiological temperature.

e. What is the pH of a neutral solution at 37oC?

slide20

-28

-29

Y = -9.2338 – 6870.6x R^2 = 0.999

-30

ln Kw

-31

-32

-33

-34

-35

0.0028

0.0030

0.0032

0.0034

0.0036

1/T

ph measurement

pH MEASUREMENT

Indicators: colored weak acids and bases

pH Meters: Glass membrane with a voltage (potential) difference across the glass.

pH and BODY CHEMISTRY

Normal pH 7.3 to 7.5

Acidosis pH < 7.3

Alkalosis pH > 7.45

Body Chemistry is “buffered” with bicarbonates (HCO3-) dihydrogenphosphates (H2PO4-) and proteins which help to maintain a constant pH

weak acids ionization constants
WEAK ACIDSIONIZATION CONSTANTS

HA(aq) + H2O(l) H3O+(aq) + A- (aq)

Ka values at 25oC are known and tabulated for a large number of weak acids.

slide23

Graphic representations of strong and weak acid equilibria

Strong Acid:

100% Dissociation into ions

HA(aq) + H2O(l) H3O+(aq) + A- (aq)

Equilibrium Favors undissociated acid

Weak Acid:

Very little Dissociation into ions

values of ka for some common monoprotic acids

Values of Ka for Some Common Monoprotic Acids

Formula Name Value of Ka*

HSO4- Hydrogen sulfate ion 1.2 x 10-2

HClO2 Chlorous acid 1.2 x 10-2

HC2H2ClO2 Monochloroacetic acid 1.35 x 10-3

HF Hydrofluoric acid 7.2 x 10-4

HNO2 Nitrous acid 4.0 x 10-4

HC2H3O2 Acetic acid 1.8 x 10-5

[Al(H2O)6]3+ Hydrated aluminum (III) ion 1.4 x10-5

HOCl Hypochlorous acid 3.5 x 10-8

HCN Hydrocyanic acid 6.2 x 10-10

NH 4+ Ammonium ion 5.6 x 10-10

HOC6H5 Phenol 1.6 x 10-10

*The units of Ka are mol/L, but are customarily omitted.

Increasing acid strength

slide25

Write the dissociation reaction and the corresponding equilibrium expression for each of the following acids in water.

a. H3PO4

b. H2PO41-

c. HCO31-

d. HCN

e. Glycine, H2NCH2COOH

f. Acetic acid, CH3COOH (HC2H3O2)

g. Phenol, C6H5OH

h. Benzoic acid, C6H5COOH

slide26

Write the reaction and the corresponding Kb equilibrium expression for each of the following substances acting as bases in water.

a. PO43- g. Glycine, NH2CH2COOH

b. HPO42- h. Ethylamine, CH3CH2NH2

c. H2PO4- I. Aniline, C6H5NH2

d. NH3 j. Dimethylamine, (CH3)2NH

e. CN-

f. Pyridine, C5H5N

weak acid calculations

WEAK ACID CALCULATIONS

HA(aq) + H2O(l)  H3O+(aq) + A-(aq)

2H2O(l)  H3O+(aq) + OH-(aq)

To simplify calculations, if % ionization is < 5%, then CHA [HA] OR [HA] = CHA – [H3O+],

Set up pH equilibria calculations in tables as in previous equilibria problems.

solving weak acid equilibrium problems

Solving Weak Acid Equilibrium Problems

List the major species in the solution

Choose the species that can produce H+, and write balanced equations for the reactions producing H+

Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+

Write the equilibrium expression for the dominant equilibrium.

List the initial concentrations of the species participating in the dominant equilibrium.

Define the change needed to achieve equilibrium; that is, define x..

Write the equilibrium concentrations in terms of x.

Substitute the equilibrium concentrations into the equilibrium expression.

Solve for x the “easy” way; that is , by assuming that [HA]0-x [HA]0.

Use the 5% rule to verify whether the approximation is valid.

Calculate [H+] and pH.

slide29

Ka Problems

For trichlorophenol (HC6H2Cl3O), Ka = 1 x 10-6, Calculate the concentrations of all species and the pH of a 0.05 M solution of trichlorophenol in water.

A solution is prepared by dissolving 0.56 g benzoic acid (C6H5CO2H), Ka = 6.4 x 10-5) in enough water to make 1.0 L of solution. Calculate [C6H5CO2H]. [C6H5CO2-], [H+], [OH-], and the pH in this solution.

Calculate the pH of a solution containing a mixture of 0.050 M HNO3 and 0.50M HC2H3O2.

weak bases ionization constants

WEAK BASES IONIZATION CONSTANTS

B(aq) + H2O(l) BH+ (aq) + OH- (aq)

Kb values at 25oC are tabulated or may be calculated from Kw and Ka

Kw = (Ka)(Kb) so Kb = Kw/Ka

Where Ka is the conjugate acid constant

slide31

Kb Problems

Thallium (Tl) hydroxide is a strong base used in the synthesis of some organic compounds. Calculate the pH of a solution containing 2.48 g TlOH per liter.

For the reaction of hydrazine (N2H4) in water.

H2NNH2 + H2O  H2NNH3+ + OH-

Kb is 3.0 x 10-6. Calculate the concentrations of all species and the pH of a 2.0 M solution of hydrazine in water.

two weak acids in solution

Two Weak Acids in Solution

A solution of 0.100 M HClO, Ka =3.5 x 10-8 and 0.100 M Formic acid, HCO2H, Ka = 1.8 x 10-4 are mixed in equal proportions. What is the resulting solution pH?

polyprotic acid equilibria

Polyprotic Acid Equilibria

What is the pH of a solution 0.100 M sulfurous acid, H2SO3, Ka1 = 1.5 x 10-2 and Ka2 = 1.0 x 10-7?

bronsted lowry theory

BRONSTED-LOWRY THEORY

OXIDES, HYDROXIDES, ANHYDRIDES

Acid, base reactions

Dehydrations (formation of anhydrides)

Hydration of oxides

acid strengths of weak acids

ACID STRENGTHSof WEAK ACIDS

Oxoacids: Ka  up as the central atom

oxidation state  up.

Ka  up as central atom of same oxidation state moves left to right in the periodic table.

Ka  up as the central atom of the same oxidation state moves UP in the same Group or family.

Polyprotic acids:

Ka decreases by approximately 105 for each successive H+ ionized.

Binary acids: (only H and another element)

Within a period, Ka  up as electronegativity of the other element.

Within a group, Ka  up going down the group to higher mass and larger size.

slide44
SnO2 + ? H2O  ?

?HCrO4-  ?Cr2O72- + ?

?HMnO4-  ? Mn (VI) compound + ?

hydroloysis of ionic salts

HYDROLOYSIS OF IONIC SALTS

The pH of each type salt in solution depends on the Ka or Kb of the hydrolyzing ion(s).

slide46
Salt Derived Ions Undergoing

From: Hydrolysis pH Examples

Strong base Neither Neutral NaCl, KNO3,

strong acid pH = 7 BaCl2, CaBr2

Strong base Anion Basic LiCN, KNO2, CaF2

weak acid pH > 7 NaCH3CO2

Weak base Cation Acidic, NH4Cl, Al(NO3)3,

strong acid pH < 7 (CH3)3NHBr

Weak base, Both Acidic NH4NO2

weak acid if Kb< Ka;

Neutral NH4CH3CO2

if Kb = Ka;

basic NH4CN

if Kb > Ka

slide47

Arrange the following 0.10 M solutions in order from most acidic to most basic

KOH, KBr, KCN, NH4Br, NH4CN, HCN

Given that the Ka value for acetic acid is 1.8 x 10-5 and the Ka value for hypochlorous acid is 3 x 10-8, which is the stronger base, OCl- or C2H3O2-

slide48

Acid-Base Equilibria

What is the pH of a solution of 0.150 M sodium nitrite, NaNO2?

HNO2, Ka = 4.0 x 10-4.

What is the pH of a solution of 0.150 M hydrazinnium chloride, H2NNH3+?

H2NNH3+, Kb = 3.0 x 10-6

slide49

Calculate the pH of each of the following solutions.

a. 0.10 M CH3NH3Cl

b. 0.050 M NaCN

c. 0.20 M Na2CO3 (consider only the reaction )

CO32- + H2O  HCO3- + OH-

Sodium azid (NaN3) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN3. The Ka value for hydrazoic acid (HN3) is 1.9 x 10-5.

lewis theory acids and bases

LEWIS THEORYACIDS and BASES

Lewis base - an electron pair donor

Lewis acid - an electron pair acceptor

Lewis acid + Lewis base  Adduct

(coordination compound)

e.g. Cu2+(aq) + 4CN-(aq)  Cu(CN)42-(aq)

Look at Lewis Dot structures for lone pairs.