1. 16.2 Acids - Base Proton Transfer
Dr. Fred Omega Garces
Chemistry 201
Miramar College
Important Notes:
Ka when H3O+ is produced,
Kb when OH- is produced

2. Brønsted-Lowry Acids-Bases
Brønsted - Lowry definition
Acid - Proton H+ (H3O+) donor.
Base - Proton H+ (H3O+) acceptor.
example:
acids: HCl (aq) g H+(aq) + Cl- (aq)
Bases: NH3 (aq) D NH4+ (aq)
HCl (aq) + NH3 (aq) g NH4+ (aq) + Cl- (aq
In an acid - base reaction, H+ & OH- always combine together to form water and an ionic compound (a salt):
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)

3. Consider a Neutralization Reaction
Neutralization reaction is a reaction between acids and base
NH3 + HCl  NH Cl-
Base
NH3 : Bronsted-Lowery: proton acceptor (Conjugate Base1)
Cl-: Bronsted-Lowery: proton accept (Conjugate Base2)
Acid
HCl: Bronsted-Lowery: proton donor (Conjugate Acid2)
NH4+: Bronsted-Lowery: proton donor (Conjugate Acid1)
Conjugates differ by a proton (H+)
Base1
Acid2
Conj Acid1
Conj Base2

4. … more Conjugate Acid-Base pairs
Other Examples of Conjugate Pairs:
1. HCO3- (aq) + H2O(l)  H3O+(aq) CO3 2- (aq)
Acid Base Conj.. acid Conj. base
of H2O of HCO3-
2. HCO3- (aq) + HS-(aq)  H2S(aq) + CO3-2 (aq)
Acid Base Conj. acid Conj. base
of HS- of HCO3-
3. CH3CO2H(aq) + CN-(aq)  HCN(aq) + CH3CO2- (aq)
of CN- of CH3CO2H
4. HSO3-(aq) NH3 (aq)  NH4+(aq) SO3-2(aq)
Acid Base Conj. acid Conj. base of NH3 of HSO3-

5. HNO2 + H2O D NO2- + H3O+ HPO42- + NH4+ D H2PO4- + NH3
Conjugate (Example)
Q: Give the conjugates of NH3, IO-, C2H3O2-, HAsO42-
A: NH4+ , HIO, HC2H3O2 , H2AsO4-, AsO4-3
Q: Identify the acids and bases on both sides of the following equations and show the conjugate acid-base pairs.
HNO2 + H2O D NO H3O+ A:
HPO42- + NH4+ D H2PO4- + NH3 B:
ConjAcid2
Acid1
Base2
ConjBase1ß
Base1
Acid2
ConjAcid1
ConjBase2

6. Relative Strengths and Net Directions
The net direction of an acid-base reaction depends on the relative strengths of the acids and bases involved. A reaction proceeds to a greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base.
The strength of the conjugate acid-base pairs are shown as the stronger the acid, the weaker the conjugate base. The strongest acid appears on the top left and the strongest base on the bottom right. When an acid reacts with a base farther down the list, the reaction proceeds in the forward (right) direction.

7. …more Acids and Bases conjugate pairs
Conjugate Acid Conjugate Base
Name Formula Formula Name
Perchloric acid HClO4 ClO4 - Perchlorate ion
Sulfuric acid H2SO4 HSO4 - Hydrogen sulfate ion
Hydrochloric acid HCl Cl - Chloride ion
Nitric acid HNO3 NO3 - Nitrate ion
Hydronium ion H3O + H2O Water
Hydrogen sulfate ion HSO4 - SO4 2- Sulfate ion
Phosphoric acid H3PO4 H2PO4 - Dihydrogen phosphate ion
Acetic acid CH3CO2H CH3CO2 - Acetate ion
Carbonic acid H2CO3 HCO3 - Hydrogen carbonate ion
Hydrogen sulfide H2S HS - Hydrogen sulfide ion
Dihydrogen phosphate ion H2PO4 - HPO4 2- Hydrogen phosphate ion
Ammonium ion NH4 + NH3 Ammonia
Hydrogen cyanide HCN CN- Cyanide ion
Hydrogen carbonate ion HCO3 - CO3 2- Carbonate ion
Water H2O OH - Hydroxide ion
Ethanol C2H5OH C2H5O - Ethoxide ion
Ammonia NH3 NH2 - Amide ion
Hydrogen H2 H - Hydride ion
Methane CH4 CH3 - Methide ion
Strong Acids
Strong Base

8. Concentration Calculations for Strong Acids/Bases
Since strong acids or base undergo 100% ionization, the concentration of H+ or OH- is the analytical concentration of these acid or base solution.
That is [H2SO4] = [H3O+] yields pH of solution
[HNO3] = [H3O+] yields pH of solution
and [KOH] = [OH-] yields pOH then pH
[Ca(OH)2] = 2 [OH-] yields pOH then pH
Consider
1. Acid: HClO4 + H2O  H3O ClO4- M
Since this proceeds 100%, then [H3O+ ] also equals M
The pH of the solution = - log[0.0010] = 3.00
2. Base: Ca(OH)2  2OH Ca +2
Since this proceeds 100%, then [OH-] equals 2 • M
The pOH of the solution is - log[0.0020] = and pH = 11.30

9. Weak Acids / Bases Both of these are an equilibrium Problem !!!
Weak acids and bases ionizes less than 100%?
Dissociation properties-
Electrolyte - Substances which dissociate in water.
Strong electrolyte completely dissociates to ions
Weak electrolyte undergoes partial dissociation.
Acid Examples :
H2S + H2O  H3O HS- Less 100% Dissociation (Weak Acid)
Base Examples :
NH3 + H2O  OH- + NH4+ Less 100% Dissociation (Weak Base)
Concentration of Acid / Base depends on the percent dissociation.
Both of these are an equilibrium Problem !!!

10. Weak Acids and Bases, Ka values (revisited)
The % ionization is dependent of the Keq constant.
For acids, this is called K-acid or Ka.

11. Solving Weak-Acid Equilibria
As in equilibrium type problems, there are also two type of acid base equilibrium problems.
• Given the equilibrium concentrations, find Ka or Kb
• Given the Ka (or Kb) and concentrations information, find the other equilibrium concentrations.
Problem Solving Strategy:
• Balance eqn & Mass action Expression
• Construct iCe table
• define x as the unknown
• Make simplifying assumptions
• Solve Mass action expression for x
• Check assumption using the 5% rule.
…do second iteration if error is greater than 5%.
Assumptions
• [H30+] from water, ~ 1•10-7 is negligible compare to x.
[H3O+]eq = 1•10-7 M + x = x
• A weak acid has a very small Ka, therefore x is very small compared to the original acid concentration.
[HA]eq = [HA]int - x = [HA]int

12. Weak Acids : pH Calculation
Consider HF:
HF(aq) + H2O (l)  H3O+ (aq) + F- (aq)
100 molc’ Ka = 6.8•10-4
100 molc’ HF  97molc’ HF & 3F- & 3H+
(Analytical Conc) (Solution at Equilb)
HF(aq) + H2O (l)  H3O+ (aq) + F- (aq)
i 0.10M lots •
C -x -x x x
e x lots •10-7+x x
Amount Dissociation:
Determine conc. Ka = [H3O+] [F- ] << 1
by Ka [HF]
Since Ka << 1 , The major component in solution is the HF

13. Weak Acids : pH comparison with strong acid
The pH value of weak acids is higher than those of strong acids for solutions with the same concentration.
Example: Determine the pH of 0.10 M HF and of 0.10 M HCl

14. Find Ka, Given Concentrations (Try on your own)
Type problem: HA + H2O (l)  H3O+ (aq) + A- (aq)
Mass Action Expression: Ka = [H3O+] [A- ]
[HA]

15. Find Concentrations, Given Ka
Type problem: Concentrations provided for reaction
HPr + H2O (l)  H3O+ (aq) + Pr- (aq)
What is the [H3O+] of 0.10M, Propanoic acid, Ka = 1.3•10-5
Use iCe table to express equilibrium
concentrations in terms of x
Solve Mass action expression and
solve for concentrations.

16. Find Concentrations, pH and Ka, Given percent dissociation or ionization (a)
A M solution of a weak acid HX is 9.40 percent ionized (a).
Using this info., calculate [H+], [X-], [HX] and Ka for HX.
Percent dissociation (a) = [HA]Dissoc = X .
[HA]Ini [HA]Init
* 9.40% of M is M (x = )
HX(aq) + H2O (l)  H3O+ (aq) + X- (aq)
i lots 1•10-7 0 *C
e lots 1•
[e] 0.181M lots M M
Note that the Ka for this unknown acid can be determined:
Ka Determination Ka = [H3O+] [X- ] = (0.0188)2
[HX]
Ka = 1.95•10-3

17. Weak Acids: percent dissociation (a)
Lower concentration of weak acids will dissociate to a greater extent because according to LeChatelier, the system will go in the direction to offset the lower number of particles (concentrations) by breaking up to more particles.*
For a weak acid the percent ionization varies as the inverse square-root of the acid concentration.
A graph illustrating that the percent ionization, a, is inversely proportional to the original acid concentration.
* In the reaction HF D H+ + F-, there are more number of particles in the right side of the equation. Dilution will cause the number of particles in the solution to decrease, in order to attain more particles in the solution more HF must dissociate.
From another perspective, [H][F]/[HF] = Ka, if the concentration of [HF] decrease, then in order for the ratio to maintain Ka, the [H] and [F] concentration must increase; more ionization at lower concentration
Note: % Ionization is inversely related to [HF]o concentration.

18. Behavior of Polyprotic Acids
Many acids have more than one ionizable H atom.
Polyprotic acids are acids with multiple ionizable protons
Consider the following Acids:
HNO H2O  H3O NO2- Ka1 = 7.1• Monoprotic
2. H2SO3 + H2O  H3O HSO3- Ka1 = 1.7•10 -2
HSO H2O D H3O SO Ka2 = 6.4• Diprotic
3. H3PO4 + H2O D H3O H2PO4- Ka1 =7.3•10 -3
H2PO H2O D H3O HPO Ka2 = 6.2•10 -8
HPO H2O D H3O PO Ka3 = 4.2• Triprotic
Note that the proceeding Ka’s are always smaller than the previous. This makes sense in that the removal of the next H+ is from a negatively charge specie. The proceeding Ka’s (for successive H+ ionization) are generally smaller by factors of 1000.

19. Ka’s of Polyprotic Acids
The pH estimate of a polyprotic acid can be estimated by only the pKa1 for polyprotic acids if pKa1 differs by a factor of 103 or greater to pKa2

20. Series Equation for Polyprotic
Consider the ionization of Carbonic acid:
The reaction is never: H2CO3 + H2O  2 H3O CO3 2-
rather it is a two step process:
step i ) H2CO3 + H2O  H3O HCO3- Ka1 =4.3•10 -7
Ka1 = [H3O+] [HCO3- ] / [H2CO3]  [H3O+](1)  pH (1)
step ii) HCO H2O  H3O CO32 - Ka2 = 5.6 •
Ka1 = [H3O+] [CO3-2 ] / [HCO3- ]  [H3O+](2)  pH (2)
Note that when Ka1 >> Ka2 ,
then [H3O+](1) >> [H3O+](2) and therefore
[H3O+](T) = [H3O+](1) + [H3O+](2) = [H3O+](1)
pH soln = pH (1)
When solution can be treated as a mixture of strong acid and weak acid, the contribution from the weaker specie is negligible.
negligible

21. Example: pH Estimate for Polyprotic Acids
Phosphoric acids is a triprotic acid. Calculate the pH of a M solution of phosphoric acid.
H3PO4 + H2O  H3O H2PO4- Ka1 =7.3•10 -3
H2PO H2O  H3O HPO Ka2 = 6.2•10 -8
HPO H2O  H3O PO Ka3 = 4.2•10 -13
Ka1 = [H3O+] [H2PO4-] = x2  7.3 •10 -3 (0.100*) = x2 a x = [H3O+] = 2.70•10 -2 M [H3PO4 ] * check assumption
Ka2 = [2.7•10-2+x] [HPO42-] = x = 6.2•10-8 a x = [H3O+] = 6.2•10 -8 M
[2.7•10-2-x ]
Ka3 = [2.7•10-2+x] [PO43-] = •105 x = 4.2• a x = [H3O+] = 9.64• M
[6.2•10-8-x ]
[H3O+ ] = 2.7•10-2 M •10 -8 M • M ~ 2.7•10 -2 M pH =

22. Take-Home (In-Class) Assignment: pH Estimate for Polyprotic Acids
Sulfuric acid is a diprotic acid.
Its two stages of ionization is shown below.
H2SO4 (aq) g H+ + HSO4-(aq) pKa = ???
HSO4-( (aq)  H SO42- pka = p766 Silberberg
Use your text to find relevant information to determine the following:
Calculate the concentration of HSO4- & SO42- ion in a 0.5 M H2SO4 solution
Determine the pH of the solution.

23. Weak Bases
Weak bases produces OH- at less than 100%?
Base Examples :
NH3 + H2O  OH- + NH4+ HS- + H2O  OH- + H2S
Less 100% Dissociation (Weak Base)
In a base equilibrium process, hydroxides are formed. When the hydroxide formation is not stoichiometric, the strategy of equilibrium analysis must be used.

24. Types of Weak Bases Two type of behavior for hydroxide formation.
1) Molc’ with atom with nonbonding pair of electrons serving as proton acceptor site. i.e., amines such as NH3, CH3NH2, pyridine. (Lone pair)
2) Anions of weak acids (conjugate base of a weak acid). i.e., acetate ion (C2H3O2-) which is the conjugate of acetic acid HC2H3O2. (Lone pair via anion)

25. Assumption Check [React]initial / keq
In general, simplifying assumptions works when the initial concentration of the reactant is high but not when it is low. To summarize, we assume that x ( [React]) can be neglected if Keq is small relative to [React]initial. or [React]initial / keq = ?.
The benchmark in justifying the assumption is-
What is the threshold for
[React]initial / keq so that the assumption is justified.
If [React]initial / keq > 400, the assumption is justified; neglecting x introduces an error of < 5%
If [React]initial / keq < 400, the assumption is not justified;
neglecting x introduces an error of > 5%

26. Example: pH for weak Base soln’
Calculate [OH -] and pH for a M solution of ethylamine, C2H5NH2. (Kb= 6.4•10-4)
C2H5NH H2O D C2H5NH OH -
i M lots 0 1•10-7
C -x -x + x + x
[e] x lots x 1• x
Kb = 6.4• = [OH -] [C2H5NH3+ ] = x2
[C2H5NH2 ]
6.4•10-4 (0.075 ) = x2  x = [OH -] = [C2H5NH3 ] = 6.93•10 -3 M, pOH = 2.16, pH = 11.84
Verify assumption method, % error = 9.2% > 5% necessitate 2nd iteration
Second iteration Kb = 6.4• = x = x
yields x = [OH -] = [C2H5NH3 ] = 6.60•10 -3 M, pOH = 2.18, pH = 11.82
Using the Quadratic Equation, pH = 11.82, confirms second iteration answer.

27. Relationship between Ka and Kb
We note that weak acid have conjugate strong base. This suggest there is a relationship between the ka of the weak acid and the Kb of the strong base.
That is the strength of an acid and the strength of its conjugate base is expressed by: Ka • Kb = Kw
Consider (1) HF + H2O  H3O+ + F- Ka = 3.5•10-4
(2) F- + H2O  OH HF Kb = ?
Plug

28. Amphoteric
Specie which can act as either a base or an acid by donating or accepting protons are considered amphoteric substances. Under certain conditions, chemicals can either give up a proton or accept a proton.
i.e., H2O, HCO3- , HPO42 -, H2PO4-
Consider Hydrogen carbonate (bicarbonate) :
Acid process:
H2CO3 + H2O  H3O HCO3- 1st H+, Ka1 =4.3 •10 -7
HCO H2O  H3O CO nd H+, Ka2 = 5.6 •
Base process:
CO H2O  OH HCO st OH-, Kb1 = 1.79•10 -4
HCO H2O  OH - + H2CO3 2nd OH-, Kb2 =2.33 •10 -8
Note from the above reaction that Ka1 competes against Kb2
these two equilibrium constant are related by Kw: Kb2 = kw / Ka1
To determine acidity or basicitiy of an amphoteric solution,
the Ka’s and Kb’s must be weighted against each other.

29. Amphoteric Behavior of Carbonic Acid
Ka1 = 4.3•10-7
Ka2 = 5.6•10-11
H2CO3 HCO3- CO3-2
Kb2 = 2.33•10-8
Kb1 = 1.79•10-4
Is the solution Acidic or Basic?
Other salts that pose the same problem-
NaHSO3 sodium bisulfite
NaH2PO4 sodium dihydrogen phosphate
K2HPO4 Potassium biphosphate
KHC2O4 Potassium oxalate
KH2C6H5O7 Potassium citrate
NH4HCO3 Ammonium bicarbonate
From polyprotic acids
Is the solution Acidic or Basic?
See work out solution in binder
42.0 g NaHCO3 in 500 mL, What is the pH?
MW = = 84 g/mol
Ka = 5.6• HCO- + H2O D H3O CO32-
Kb = 2.3• HCO- + H2O D OH H2CO3
Ka = 5.6 • = X2/1 ; X = 7.48•10-6 M = H3O+
Kb = • = X2/1 ; X = 1.53•10-6 M = OH-
pOH = 3.84
pH =

30. Amphoteric Behavior of Carbonic Acid
Ka1 = 4.3•10-7
Ka2 = 5.6•10-11
H2CO3 HCO3- CO3-2
Kb2 = 2.33•10-8
Kb1 = 1.79•10-4
From the magnitude of the equilbrium constant for bicarbonate ion -
For the HCO3- : |Kb2| > |Ka2|
In this process, the HCO3- is a better base than HCO3- is an acid.
Therefore is a solution basic or acidic,
if HCO3- is introduced in a solution i.e., NaHCO3 ?
i.e., 42 g NaHCO3 is added to 500 ml of water, What is the pH ?
HCO3-
NaHCO3
Kb2 = 2.33•10-8
Ka2 = 5.6•10-11
OH- H+
From inspection, Kb2 >> Ka2, therefore this solution is basic.
To determine the pH, solve the equilb rxn for both reaction.
Kb2 = [OH -] [H2CO3 ]  [OH-] & Ka2 = [H3O+] [CO32 ]  [H3O+]
[HCO3- ] [HCO3- ]
Excess of OH- and H3O+ will determine the pH of the solution.
See work out solution in binder
42.0 g NaHCO3 in 500 mL, What is the pH?
MW = = 84 g/mol
Ka = 5.6• HCO- + H2O D H3O CO32-
Kb = 2.3• HCO- + H2O D OH H2CO3
Ka = 5.6 • = X2/1 ; X = 7.48•10-6 M = H3O+
Kb = • = X2/1 ; X = 1.53•10-6 M = OH-
pOH = 3.84
pH =

31. Factors Affecting Solubility: Acidic Soln’
Solubility of salts -
influenced by acid, base properties of the salt’s cation & anion.
Idea - LeChatelier’s Principle is obeyed.
Consider Solubility of BaF2 in acidic solution:
BaF2(s)  Ba+2 (aq) F- (aq) Eqn(1)
but
F H2O  OH- (aq) + HF
F- reacts further decreasing the amt. of F - in eqn 1
As HF is produced, more BaF2 dissolves to maintain Ksp equilibrium.
F - can be removed further by addition of H3O+
Solubility increases because H3O+ removes F - and converts it to HF
BaF2(s) Ba+2 (aq) F- (aq)
Acidic Solution,
Solubility increase
BaF2(s) + H3O+ (in acidic conditions, soluble)
 Ba+2 (aq) F - (aq)
+ H3O+  HF

32. Factors Affecting Solubility: Basic Soln’
Solubility of salts -
influenced by acid, base properties of the salt’s cation & anion.
Idea - LeChatelier’s Principle is obeyed.
Consider Solubility of BaF2 in basic c solution:
BaF2(s)  Ba+2 (aq) F- (aq) Eqn(1)
but F H2O  OH- (aq) + HF
F- reacts further to produce HF from eqn 1.
BaF2(s) + OH- (in basic conditions, insoluble)
 Ba+2 (aq) F -(aq)
+ H2O  HF (aq)
+ OH-  F - (aq) + H2O
Solubility decreases because F- is generated when HF reacts with excess OH-. Increase F- shift equilibrium back to solid.
According to LeChatelier’s principle, less BaF2 will not dissociate.
BaF2(s) Ba+2 (aq) F- (aq)
Basic Solution, Solubility decrease

33. Acid Base Reaction, which chemical dominates
Consider the following reactions below. Determine which chemical in each equation is the dominant specie in solution. Which direction will the equilibrium favor?
H2S + H2O  H3O HS-
H2PO NH  H3PO NH3
HBrO BrO3-  HBrO BrO4-K
HCH3CO2 + CN-  CH3CO HCN
H2SO3
NaCH3CO2 HF
H2S
NH4+
H3O+
NaCN
HIO
H2C2O4
HBrO3
H3PO4
KH2PO4 -
Strong Acids
HCN
HBrO4
NaHCO3
H2PO4-

34. …more Acids and Bases conjugate pairs
Conjugate Acid Conjugate Base
Name Formula Formula Name
Perchloric acid HClO4 ClO4 - Perchlorate ion
Sulfuric acid H2SO4 HSO4 - Hydrogen sulfate ion
Hydrochloric acid HCl Cl - Chloride ion
Nitric acid HNO3 NO3 - Nitrate ion
Hydronium ion H3O + H2O Water
Hydrogen sulfate ion HSO4 - SO4 2- Sulfate ion
Phosphoric acid H3PO4 H2PO4 - Dihydrogen phosphate ion
Acetic acid CH3CO2H CH3CO2 - Acetate ion
Carbonic acid H2CO3 HCO3 - Hydrogen carbonate ion
Hydrogen sulfide H2S HS - Hydrogen sulfide ion
Dihydrogen phosphate ion H2PO4 - HPO4 2- Hydrogen phosphate ion
Ammonium ion NH4 + NH3 Ammonia
Hydrogen cyanide HCN CN- Cyanide ion
Hydrogen carbonate ion HCO3 - CO3 2- Carbonate ion
Water H2O OH - Hydroxide ion
Ethanol C2H5OH C2H5O - Ethoxide ion
Ammonia NH3 NH2 - Amide ion
Hydrogen H2 H - Hydride ion
Methane CH4 CH3 - Methide ion
Strong Acids
Strong Base

35. Midterm Practice Assignment:
pH of Sodium hydroxide, sodium bisulfite and sodium phosphate NaOH, NaHSO3, Na3PO4
Calculate the pH if 20.0 grams of sodium hydroxide is dissolved in enough water to form 1.00L solution.
pH = 13.
699
Calculate the pH if 52.0 grams of sodium bisulfite is dissolved in enough water to form 1.00L solution.
pH = 2.756
Calculate the pH if 35.5 grams of sodium phosphate is dissolved in enough water to form 1.00L solution
pH =

36. Summary Strength of Acid or Base Keq or Ka, kb value.
The competition for amphoteric substance to give OH- or H3O+
Solubility of salt is directly influence by LeChatelier Principle and how reactant or product reacts with the acid.