acid base equilibria
DESCRIPTION
Acid Base Equilibria. Complications: Beyond pH of strong acids or strong bases HCl(aq) H + (aq) + Cl - (aq) K a > 1 HF(aq) H + (aq) + F - (aq) K a = 7.2 x 10 -4. Weak conjugate base. Strong acid. Weak acid. stronger conjugate base. Conjugate Bases. Bronsted-Lowry Model - PowerPoint PPT PresentationTRANSCRIPT
Mullis 1
Acid Base Equilibria
Complications: Beyond pH of strong acids or strong bases
HCl(aq) H+(aq) + Cl- (aq) Ka > 1
HF(aq) H+(aq) + F- (aq) Ka = 7.2 x 10-4
Strong acid
Weak acid
Weak conjugate base
stronger conjugate base
Mullis 2
Conjugate Bases• Bronsted-Lowry Model• Conjugate base remains after acid loses its
proton• Strongest conjugate base = best at getting
available protons• Factors affecting conjugate base strength:
1. Electronegativity S- > Cl-
2. Size if acids were binary F- > I-
3. Number of oxygens if acids were oxyacids
ClO- > ClO2-
Mullis 3
pH of Mixture of Weak Acids• Find pH of a solution that contains 1.00M
HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2
(Ka = 4.0 x 10-4).
• Because these are weak acids, dissociation is not complete and [H+] ≠ [HNO2].
• You must use ICE table strategy to solve pH problems with weak acids or bases.
Mullis 4
How to Determine [H+] in an Acid Mixture
• HCN H+ + CN- Ka = 6.2 x 10-10
• HNO2 H+ + NO2- Ka = 4.0 x 10-4
• H2O H+ + OH- Ka = 1.0 x 10-14
• Compare species and find the most likely contributor of protons. Here it is by far HNO2: Ka is 6 orders of magnitude greater than the closest possible H+ contributor.
Mullis 5
Equilibrium Method to Find [H+]
R HNO2 H+ NO2-
I 5.00 M 0 0
C -x +x +x
E 5.00 – x x x• Ka = x2 = 4.0 x 10-4
(5.00 –x)
x2 = 20. x 10-4 x = 4.5 x 10-2
Small: reactants favored
Change must be very small relative to initial conc.: Ignore x in this term
Mullis 6
Accuracy check
• When dealing with weak acids and bases, Ka values are typically valid to a ±5% accuracy. Verify that our assumption in ignoring x in the reduction of [HNO2]0 was valid:
• x x 100% = 4.5 x 10-2 (100%) = 0.90 %
[HNO2] 5.00
• Since 0.90% is less than 5%, the assumption is valid---in other words, our answer for x is good.
Mullis 7
What was the Original Question?
• Find pH of this solution:
• [H+] = 4.5 x 10-2
• pH = -log[H+] = -log(4.5 x 10-2) = 1.35
• Number of decimals in pH = Number of significant figures in concentration (2 here)
Mullis 8
Find [CN-] in that problem.• [H+] = 4.5 x 10-2 M• HCN H+ + CN- Ka = 6.2 x 10-10
• 6.2 x 10-10 = [4.5 x 10-2 ][CN-][HCN]
Remember our assumption that since Ka is small, we ignore the small reduction in initial concentration of product? Be consistent and make that assumption here, too. So:
• 6.2 x 10-10 = [4.5 x 10-2 ][CN-]1.00
[CN-] = 1.4 x 10-8 M (That’s a small number!)
Mullis 9
% DissociationEx: HNO2 H+ + NO2
-
[H+] x 100% = % dissociation
[HNO2]
Or…
[NO2-] x 100% = % dissociation
[HNO2]
If you know % dissociation and original concentration of acid [HNO2], can solve for [H+] or [NO2
-] (or x, the change in concentration).
Mullis 10
% Dissociation and DilutionMore Concentrated Diluted
Less % dissociation More % dissociation(more chance of
recombination)
If Q = K for this reaction Q < Ka for this one, so equilibrium shifts
right
Mullis 11
Ka from % DissociationA 0.100 M solution of lactic acid (HC3H5O3) is 3.7% dissociated. Calculate the value of Ka.
R HC3H5O3 H+ C3H5O3 -
I 0.100 M 0 0
C -x +x +x
E 0.100 – x x x
% dissociation = x x (100%) Ka = x2
0.10 (0.100 –x)
3.7% (0.10) = x Ka = 0.00372
x = 0.0037 M (0.100 - .0037)
Ka = 1.4 x 10-4
Mullis 12
Organic Acids(Acid = proton donor)
-OH phenol
Compound Formula Example Alcohol R-OH hydroxyl group 1-propanol Alkyl Halide -X X = any halide 1,2-dibromopropane Ether R-O-R’ one oxygen bonded to 2
hydrocarbon groups diethyl ether CH3-CH2 –O--CH2--CH3
Aldehyde O || R-C-H
Carbonyl group attached to end carbon
Ethanal O || CH3—C--H
Ketone O || R-C-R’
Carbonyl group attached to a middle carbon
2-propanone O || CH3—C-- CH3
Carboxylic Acid O || R-C-OH
Carboxyl group ethanoic acid O || CH3—C—OH
Ester O || R-C-O-R’
Carboxyl group without the H
methyl ethanoate O || CH3—C—O-- CH3
Mullis 13
Bases without OH-
• B (aq) + H2O BH+ (aq) + OH- (aq)
• NH3 (aq) + H2O NH4+ (aq) + OH- (aq)
• Kb = [BH+][OH-]
[B]
conjugate baseconjugate acidacidbase
Mullis 14
Amines RxNH(3-x)
• Ammonia with one or more of the H atoms replaced by another group.
• Many organic molecules involve amines.
N. .pyridine
ethylamine
. .N
HH
C2H5HO
HO
OH
CHCH2NHCH3
adrenaline
Mullis 15
Find the pH of 1.0M methylamine (Kb=4.38x10-4).
R CH3NH2 CH3NH3+ OH-
I 1.0 M 0 0
C -x +x +x
E 1.0 – x x x
CH3NH2 + H2O CH3NH3+ + OH-
Kb = x2 = 4.38x10-4 x2 ≈ 4.38x10-4
1.0-x x≈ 2.1x10-2
[OH-] = x = 2.1x10-2 M
pOH=-log[2.1x10-2 ] pOH = 1.68 pH= 14 -1.68= 12.32
Mullis 16
PolyProtic Acids• Lose one proton at a time: H3PO4 H2PO4
- HPO42- PO4
3-
• The reactions written to express this loss have successively smaller Ka values:
• Ka1 > Ka2 > Ka3
• …..makes sense: The loss of a second proton occurs less readily than the loss of the first one.In fact, Ka1 is usually so much larger than the
others, the loss of the first proton is the only reaction
that significantly contributes [H+].
Mullis 17
Sequential Loss of Protons
H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3
H2PO4- H+ + HPO4
2- Ka2 = 6.2 x 10-8
HPO42- H+ + PO4
3- Ka3 = 4.8 x 10-13
Mullis 18
pH of a Polyprotic Acid• Find pH of 5.0M H3PO4 and the equilibrium concentrations of
each species: H3PO4, H2PO4-, HPO4
2-, PO43-
• Ka1 = [H+][H2PO4-] = x2 = 7.5 x 10-3
[H3PO4] 5.0-x (see previous problems’ RICE tables & pH calculations for details)
x2 = (5.0)7.5 x 10-3 x ≈ 0.19 M= [H2PO4-] = [H+] pH = 0.72
[H3PO4] = 5.0 – x = 4.8 M
Ka2 = [H+][HPO42-] = (0.19) [HPO4
2-] = 6.2 x 10-8
[H2PO4-] 0.19
[HPO42-]= 6.2 x 10-8 M
Ka3 = [H+][PO43-] = (0.19) [PO4
3-] = 4.8 x 10-13
[HPO42-] 6.2 x 10-8
[PO43-]= 1.6 x 10-19 M
Mullis 19
Sulfuric acid: Relatively High Conc.
• Find pH of 1.0 M H2SO4.
• H2SO4 H+ + HSO4- Ka1 > 1
• HSO4- H+ + SO4
2-Ka2 = 1.2 x 10-2
• In this case, does the 2nd reaction contribute significantly to [H+]?
• [H+] is at least 1.0 since the first step is total dissociation.
Mullis 20
Sulfuric acid: Relatively High Conc.
1.2 x 10-2= [H+][SO42-] = (1.0 + x)(x)
[HSO4-] (1.0 – x)
Assume change is small compared to 1.0, so: 1.0(x) 1.0
X = 1.2 x 10-2, or 1.2% of 1.0 M. If this is added to 1.0,result is 1.0012M, but with 2 sig. figs., [H+] is 1.0 M. (pH = 0.00)
R HSO4- H+ SO4
2-
I 1.0 M 1.0 0
C -x +x +x
E 1.0 – x 1.0 + x x
From 1st disso
ciation step
Mullis 21
0.01 M H2SO4 (Relatively Low Conc.)
1.2 x 10-2= [H+][SO42-] = (.01 + x)(x)[HSO4
-] (.01 – x)If assume change is small compared to .01, .01(x)
.01x = 1.2 x 10-2, or more than 0.01 M! Therefore, cannot ignore the subtraction
or addition of x—use the quadratic to find x = .0045. Total [H+] = 0.01(from 1st dissociation step) + .0045 = 0.0145.
pH = -log(0.0145) = 1.84
R HSO4- H+ SO4
2-
I 0.01 M 0.01 0
C -x +x +x
E 0.01 – x 0.01 + x x
From 1st disso
ciation step
Mullis 22
Salts of Weak Acids or Bases
• Cations of strong bases do not change pH of a solution. – (Ex: Na+ will not attract or contribute H+)
• Salts in which cations do not change pH and the anion is the conjugate base of a weak acid will produce basic solutions.– Ex. KC2H3O2 dissociates to produce C2H3O2
- and OH-
• Salts in which anions are not a base and the cation is a conjugate acid of a weak base will produce acidic solutions. – Ex. NH4Cl dissociates to produce NH3 and H+.
Mullis 23
Finding Kb for Conjugates
• Look up the Ka for the weak acid that makes the conjugate base.
• Ex: NaC2H3O2 (aq)
• C2H3O2- + H2O HC2H3O2 + OH-
• Ka(acetic acid) = 1.8 x 10-5
• KaKb = Kw
• Kb = 1x10-14 = 5.6 x 10-10
1.8 x 10-5
Mullis 24
Salt as a Weak Base• Find pH of a 0.30 M NaF solution.• Na+ does not change pH. • F- is the conjugate base of HF. Water has to
be the proton donor for this solution.
• F- (aq) + H2O (aq) HF (aq) + OH- (aq)
• Ka for HF is 7.2 x 10-4.
• Kb = 1.0 x 10-14 = 1.4 x 10-11
7.2 x 10-4
Mullis 25
Salt as a Weak Base
Kb = 1.4 x 10-11 = [HF][OH-] = x2
[F-] .30-xx2 =.30(1.4 x 10-11) x = 2.0 x 10-6 (<5% of .30)
[OH-] = 2.0x10-6
pOH = 5.69 pH= 14.00 – 5.69 = 8.31
R F- H2O (l) HF OH-
I 0.30 -- 0 ≈ 0
C -x -- +x +x
E 0.30 - x -- + x x
Mullis 26
Salt as a Weak Acid• Find pH of a 0.10 M NH4Cl solution.
• Cl- does not change pH.
• NH4+ produces H+ in water. It is the conjugate
acid of NH3.
• NH4+ (aq) NH3 (aq) + H+ (aq)
• Kb for NH3 is 1.8 x 10-5.
• Ka for NH+ = 1.0 x 10-14 = 5.6 x 10-10
1.8 x 10-5
Mullis 27
Salt as a Weak Acid
Kb = 5.6 x 10-10 = [NH3][H+] = x2
[NH4+] .10-xx2 =.10(5.6 x 10-10 ) x = 7.5 x 10-6 (<5% of .10)
[H+] = 7.5 x 10-6 pH = -log(7.5 x 10-6 ) = 5.13
R NH4+ NH3 + H+
I 0.10 0 ≈0
C -x +x +x
E 0.10 - x + x x
Mullis 28
Acid-Base Properties of SaltsSalt Example pH+ from strong base
- from strong acidNaCl neutral
+ from strong base
- from weak acidNaC2H3O2 Anion acts as
basebasic
+ conj. acid fm. weak base
- from strong acidNH4Cl Cation acts as
acidacidic
+ conj. acid fm. weak base
- conj. base fm. weak acidNH4C2H3O2 Anion acts as
base, cation as acid
Acidic if Ka> Kb
+ from metal ion
- from strong acidFeCl3
Hydrated cation acts as acid
acidic
Mullis 29
Practice1. A 0.15 M solution of a weak acid is 3.0% dissociated.
Calculate Ka.
2. What are the major species present in a 0.150 M NH3 solution? Calculate the [OH-] and the pH of this soln.
3. Calculate the pH of a 5.0 x 10-3 M solution of H2SO4.
4. Sodium azide 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 HN3 (hydrazoic acid) is 1.9 x 10-5.
5. Calculate the pH of 0.10 M CH3NH3Cl. Kb of CH3NH3+
is 4.38 x 10-4.