Chapter 18 Equilibria Involving Acids & bases. ARRHENIUS THEORY for ACIDS and BASES ACIDS: produce...
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Transcript of Chapter 18 Equilibria Involving Acids & bases. ARRHENIUS THEORY for ACIDS and BASES ACIDS: produce...
Chapter 18
Equilibria Involving Acids & bases
ARRHENIUS THEORY for ACIDS and BASESARRHENIUS 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
H+ surrounded by four H-bonded H2O molecules
H9O4+
OH- surrounded by three H-bonded H2O molecules
OH(H2O)3-
Bronsted-Lowery Theory of Acids and BasesBronsted-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.
For each of the following reactions, identify the acid, the For each of the following reactions, identify the acid, the base, the conjugate base, and the conjugate acidbase, the conjugate base, and the conjugate acid
1. H2O + H2O H3O+ + OH-
2. H2PO4- + H2PO4
- H3PO4 + HPO42-
3. H2SO4 + H2O H3O+ + HSO4-
4. H2PO4- + H2O H3PO4 + OH-
5. CO2 + 2H2O HCO3- + H3O+
6. H2PO4- + H2O HPO4
2- + H3O+
7. Fe(H2O)63+ + H2O Fe(H2O)5(OH) 2+ + H3O+
8. HCN + CO32- CN- + HCO3
-
Graphic representations of strong and weak acid equilibria
Strong Acid:
100% Dissociation into ions
Weak Acid:
Very little Dissociation
into ions
HA(aq) + H2O(l) H3O+(aq) + A-
(aq)
Equilibrium Favors undissociated acid
[HA]
][A ]O[H K
-3
a
Acid strength versus conjugate base strength
Bronsted-Lowery Theory: Acid and Base StrengthsBronsted-Lowery Theory: Acid and Base StrengthsProton 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- + HSO3
1- ???
• 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 basesRelative strengths of some Bronsted-Lowry acids and their conjugate basesAcid Base
Strongest HClO4 ClO4- Weakest
Acids H2SO4 HSO4- bases
HI I-
HBr Br-
HCl Cl-
HNO3 NO3-
H3O+ H2OHSO4
- SO42-
H2SO3 HSO3-
H3PO4 H2PO4-
HNO2 NO2-
HF F-
CH3CO2H CH3CO2-
H2CO3 HCO3-
H2S HS-
NH4+ NH3
HCN CN-
HCO3- CO3
2-
HS- S2-
H2O OH-
Weakest NH3 NH2- Strongest
Acid OH- O2- bases
Bronsted-Lowery Theory: Acid and Base StrengthsBronsted-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
Autoionization of Water and the pH Scale
H2O(l) H2O(l)
H3O+(aq) OH-(aq)
+
+
Autoionization of WaterAutoionization 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 SCALESpH and pOH SCALES
pH = - log [H3O+]
pOH = - log [OH-]
pH, pOH CALCULATIONSpH, 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 – pOHCALULATE SOME pH and pOH VALUES
[H+] pH10-14 14 1 M NaOH
10-13 13Basic 10-12 12 Ammonia
10-11 1110-10 1010-9 910-8 8
Neutral 10-7 7 Pure Water
10-6 610-5 510-4 410-3 3
Acidic 10-2 210-1 11 0
(HouseholdCleaner)
Blood
Milk
VinegarLemon juice
Stomach acid
1 M HCl
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.
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?
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?
Y = -9.2338 – 6870.6x R^2 = 0.999
0.0028 0.0030 0.0032 0.0034 0.0036
-35
-34
-33
-32
-31
-30
-29
-28
ln K
w
1/T
pH MEASUREMENTpH MEASUREMENT Indicators: colored weak acids and bases pH Meters: Glass membrane with a voltage (potential)
difference across the glass.
pH and BODY CHEMISTRYNormal 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 ACIDSWEAK ACIDSIONIZATION CONSTANTSIONIZATION CONSTANTS
HA(aq) + H2O(l) H3O+(aq) + A-
(aq)
Ka values at 25oC are known and tabulated for a large number of weak acids.
[HA]
][A ]O[H K
-3
a
Graphic representations of strong and weak acid equilibria
Strong Acid:
100% Dissociation into ions
Weak Acid:
Very little Dissociation
into ions
HA(aq) + H2O(l) H3O+(aq) + A-
(aq)
Equilibrium Favors undissociated acid
[HA]
][A ]O[H K
-3
a
Values of Ka for Some Common Monoprotic AcidsValues 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.
Incr
easi
ng
acid
str
engt
h
a. H3PO4
b. H2PO41-
c. HCO31-
d. HCN
e. Glycine, H2NCH2COOH
f. Acetic acid, CH3COOH (HC2H3O2)
g. Phenol, C6H5OH
h. Benzoic acid, C6H5COOH
Write the dissociation reaction and the corresponding equilibrium expression for each of the following acids in water.
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 CALCULATIONSWEAK 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.
100% x [HA]
]O[H Ionization %
3
Solving Weak Acid Equilibrium ProblemsSolving Weak Acid Equilibrium Problems1. List the major species in the solution
2. Choose the species that can produce H+, and write balanced equations for the reactions producing H+
3. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+
4. Write the equilibrium expression for the dominant equilibrium.
5. List the initial concentrations of the species participating in the dominant equilibrium.
6. Define the change needed to achieve equilibrium; that is, define x..
7. Write the equilibrium concentrations in terms of x.
8. Substitute the equilibrium concentrations into the equilibrium expression.
9. Solve for x the “easy” way; that is , by assuming that [HA]0-x [HA]0.
10. Use the 5% rule to verify whether the approximation is valid.
11. Calculate [H+] and pH.
Ka Problems
1. 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.
2. 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.
3. Calculate the pH of a solution containing a mixture of 0.050 M HNO3 and 0.50M HC2H3O2.
WEAK BASES IONIZATION CONSTANTSWEAK 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
[B]
][OH ][BH K
_
b
Kb Problems
1. 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.
2. 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 SolutionTwo 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 EquilibriaPolyprotic 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?
ACID-BASE PROPERTIES OF THE OXIDES (PART I)
ACID BASE PROPERTIES OF THE OXIDES (PART II)
ACID-BASE PROPERTIES OF THE OXIDES (PART III)
ACID BASE PROPERTIES OF THE OXIDES (PART IV)
ACID-BASE PROPERTIES OF THE OXIDES (PART V)
ACID BASE PROPERTIES OF THE OXIDES PART (VI)
ACID BASE PROPERTIES OF THE OXIDES PART (VII)
ACID-BASE PROPERTIES OF THE OXIDES PART (VIII)`
BRONSTED-LOWRY BRONSTED-LOWRY THEORYTHEORY
OXIDES, HYDROXIDES, ANHYDRIDES
Acid, base reactions
Dehydrations (formation of anhydrides)
Hydration of oxides
ACID STRENGTHSACID STRENGTHSof WEAK ACIDSof WEAK ACIDS
Oxoacids: Ka up as the central atomoxidation 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.
SnO2 + ? H2O ?
?HCrO4- ?Cr2O7
2- + ?
?HMnO4- ? Mn (VI) compound + ?
HYDROLOYSIS OF IONIC SALTSHYDROLOYSIS OF IONIC SALTS
The pH of each type salt in solution depends on the Ka or Kb of the
hydrolyzing ion(s).
Salt Derived Ions UndergoingFrom: 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 NH4CNif Kb > Ka
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
-
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
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 THEORYLEWIS THEORYACIDS and BASESACIDS 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.