1. Acid-Base Concepts -- Chapter 15
1. Arrhenius Acid-Base Concept (last semester)
Acid: H+ supplier Base: OH– supplier
2. Brønsted-Lowry Acid-Base Concept (more general)
(a) Definition (H+ transfer)
Acid: H+ donor Base: H+ Acceptor
Conjugate Acid-Base Pairs: Base Acid
+ H+
– H+

2. Conjugate Acid-Base Pairs
e.g.,
conjugate pair
NH3(g) +
H2O(l)
NH4+(aq) +
OH–(aq)
base
acid
acid
base
conjugate pair
more examples:
conjugate acids
NH2–
NH3
NH4+
OH–
H2O
H3O+
O2–
HSO4–
H2SO4
H3SO4+
CH3–
CH4
“CH5+”
conjugate bases

3. Amphoteric Substances
Molecules or ions that can function as both acids and bases (e.g. H2O itself!)
e.g. the bicarbonate ion, HCO3–
HCO3– + OH– --> H2O + CO32–
acid base
HCO3– + HCl --> H2CO3 + Cl–
base acid

4. Autoionization of Water
Water undergoes auto-ionization to a slight extent:
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
H3O+ = hydronium ion OH– = hydroxide ion
or, simply,
H2O(l) H+(aq) + OH–(aq)
equilibrium constant:
Kc = [H+][OH–]/[H2O]
But, [H2O] ~ constant ~ 55.6 mol/L at 25 °C
so, instead, use the “ion product” for water = Kw
Kw = [H+][OH–] = 1.0 x (at 25 °C)
In pure water: [H+] = [OH–] = 1.0 x 10-7 M

5. The pH Scale Kw = [H+][OH–] = 1.0 x 10-14
The pH scale: pH = – log [H+]
In general: pX = – log X
e.g. pOH = – log [OH–]
and, in reverse: [H+] = 10–pH mole/L
[OH–] = 10–pOH mole/L
Since Kw = [H+][OH–] = 1.0 x 10–14
pKw = pH + pOH =
Notice the sig figs!
two sig figs
two sig figs!

6. Relative Acidity of Solutions
neutral solution [H+] = [OH–] = 1.0 x 10–7 M
pH = pOH = 7.00
acidic solution [H+] > 10–7
(i.e. more H+ than in pure water)
pH < [OH–] < 10–7 and pOH > 7.00
e.g. if [H+] = 1.00 x 10–3 M
then pH = and pOH =
Basic solution [H+] < 10–7
(i.e. less H+ than in pure water)
pH > [OH–] > 10–7 and pOH < 7.00
e.g. if [OH–] = 1.00 x 10–3 M
then pOH = and pH =

7. Example Problem
The water in a soil sample was found to have [OH–] equal to 1.47 x 10–9 mole/L. Determine [H+], pH, and pOH.
Answer:
[H+] = Kw/[OH–] = (1.00 x 10–14)/(1.47 x 10–9) = 6.80 x 10–6
pH = – log [H+] = – log (6.80 x 10–6) = (acidic!)
pOH = – pH = – = 8.833
{or, pOH = – log [OH–] = – log (1.47 x 10–9) = 8.833}

8. Strong Acids and Bases
Strong Acids (e.g. HCl, HNO3, etc) ~ 100% ionized
HNO3(aq) + H2O H3O+(aq) + NO3–(aq)
or, in simplified form
HNO3(aq) H+(aq) + NO3–(aq)
[H+] = initial M of HNO3
e.g. in a M HNO3 solution:
[H+] = and pH = – log (0.050) = 1.30
Strong Bases (metal hydroxides) % ionized
NaOH(aq) Na+(aq) + OH–(aq)
[OH–] = initial M of NaOH
100%
100%
Memorize!
Table 15.3 p665
Table 15.7 p 682
100%

9. Example Problem
What mass of Ba(OH)2 ( g/mole) is required to prepare
250 mL of a solution with a pH of 12.50?
First: Write the reaction.
Ba(OH)2(aq) Ba2+(aq) OH–(aq)
so, [OH–] = 2 x M of Ba(OH)2 solution (2:1 ratio)
100%
Second: Determine [OH–].
pOH = – pH = – = 1.50
[OH–] = 10–1.50 = M
Third: How much Ba(OH)2 is needed for that much OH–?
0.032 mol OH–
250 mL soln x
= mol OH–
1000 mL soln
1 mole Ba(OH)2
g Ba(OH)2 x
= 0.68 g Ba(OH)2
mol OH– x
2 mol OH–
1 mol Ba(OH)2

10. Sample Problem (a) Determine the pH of a 1.75 M solution of HNO3.
(b) Now suppose that mL of a 1.50 M Ba(OH)2 solution is added to 1.00 L of the above HNO3 solution. Determine the pH of the resulting solution.
(c) What additional volume (in mL) of 1.50 M Ba(OH)2 must be added to the mixture in part (b) to bring the pH to 7.00?

11. Weak Acids and Bases
(a) Weak Acids -- Less than 100% ionized (equilibrium !)
In general: HA is a weak acid, A– is its conjugate base
HA(aq) + H2O H3O+(aq) + A–(aq)
or, simply, HA(aq) H+(aq) + A–(aq)
Acid Dissociation Constant: Ka
Ka =
[H+][A–]
[HA]
Relative acid strength:
Weak acid: Ka < ~10–3
Moderate acid: Ka ~ 1 to 10–3
Strong acid: Ka > 1

12. Example Problem
Hypochlorous acid, HOCl, has a pKa of What is the pH of an
0.25 M solution of HOCl? What is the percent ionization?
First: Write equation and ICE table.
HOCl(aq) ⇌ H+(aq) + OCl–(aq) I
0.25 M –x +x
0.25 – x x C E
Second: Write equilibrium expression.
pKa = – log Ka, so Ka = 10–pKa = 10–7.52 = 3.0 x 10–8
3.0 x 10–8 =
[H+][Cl–]
[HOCl] =
(x)(x)
(0.25 –x) x2
Third: Solve for [H+] and pH.
Since Ka is very small, assume x << 0.25
x2/(0.25) ≈ 3.0 x 10– x ≈ 8.7 x 10–5 (OK!)
pH = –log (8.7 x 10–5) = (solution is acidic!)

13. Example Problem, cont. Fourth: Determine percent ionization.
% ionization = (amount HA ionized)/(initial) x 100%
= (8.7 x 10–5)/(0.25) x 100% = 0.035%

14. Weak Bases (b) Weak Bases
In general: B is a weak base, HB+ is its conjugate acid
B(aq) + H2O(l) HB+(aq) + OH–(aq)
Kb =
[HB+][OH–]
[B]
Base Dissociation Constant:
e.g. NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
[NH4+][OH–]
Kb =
= 1.8 x 10–5
[NH3]
pKb = -log Kb = 4.74
Note: Since OH– rather than H+ appears here, first find [OH–] or pOH,
and then convert to pH
Example problem: pH of 0.25 M solution of NH3?
Set up conc. table as usual, solve for x = [OH–]
[OH–] = 2.1 x 10–3 pOH = pH = (basic!)

15. Sample Problem
The nitrite ion (NO2–) is a weak base with a pKb value of
(a) Write a balanced net ionic equation for the major equilibrium reaction that is occurring in an aqueous solution of sodium nitrite (NaNO2).
(b) Calculate the pH of a 0.25 M solution of sodium nitrite (NaNO2). Clearly state and justify any assumptions that you make.

16. Salts of Weak Acids and Bases
A) conjugate acid-base pairs (HA and A–)
Ka: HA H+ + A–
Kb: A- + H2O HA + OH–
For any conjugate acid-base pair:
KaKb = Kw
pKa + pKb = 14.00

17. Salt of a Weak Acid (e.g. NaCN) -- basic solution
Anion acts as a weak base:
Kb: CN–(aq) + H2O(l) HCN(aq) + OH–(aq)
[OH–][HCN]
Kb = Kw/Ka =
[CN–]
e.g. Ka for HCN is 6.2 x 10–10
What is pH of a 0.50 M NaCN solution?
Kb = Kw/Ka = (1.0 x 10–14)/(6.2 x 10–10) = 1.6 x 10–5
Use a concentration table based on Kb reaction above:
x = [OH–] = [HCN]
[CN–] = 0.50 – x ≈ (since Kb is small)
Kb = [OH–][HCN]/[CN–] ≈ x2/0.50 ≈ 1.6 x 10–5
x = [OH–] ≈ 2.8 x 10–3
pOH = and pH = (basic!)

18. Salt of a Weak Base (e.g. NH4Cl) -- Acidic Solution
Cation acts as a weak acid:
Ka: NH H+ + NH3
[H+][NH3]
Ka = Kw/Kb =
[NH4+]
e.g. Kb for NH3 is 1.8 x 10–5
What is pH of a 0.50 M NH4Cl solution?
Ka = Kw/Kb = (1.0 x 10–14)/(1.8 x 10–5) = 5.6 x 10–10
Use a concentration table based on Ka reaction above:
x = [H+] = [NH3]
[NH4+] = 0.50 – x ≈ (since Ka is small)
Ka = [H+][NH3]/[NH4+] ≈ x2/0.50 ≈ 5.6 x 10–10
x = [H+] ≈ 1.7 x 10–5 pH = (acidic!)

19. Sample Problem
The pKa value for HCN is What molar concentration of NaCN is required to make a solution with a pH of 11.75?

20. Polyprotic Acids
e.g. diprotic acids, H2A, undergo stepwise dissociation:
[HA–][H+]
Ka1 =
H2A HA– + H+
[H2A]
[A2–][H+]
Ka2 =
HA– A2– + H+
[HA–]
Usually, Ka1 >> Ka2 so that:
The 1st equilibrium produces most of the H+ and [HA–]
but the 2nd equilibrium determines [A2–]

21. Example Problem
Ascorbic acid (vitamin C), H2C6H2O6, is an example of a diprotic acid with Ka1 = 7.9 x 10–5 and Ka2 = 1.6 x 10–12. For a 0.10 M solution of ascorbic acid, determine the pH and the concentrations of the mono anion, HC6H2O6–, and the dianion, C6H2O62–.
Based on the first equilibrium:
x = [H+] ≅ [HA–] and [H2A] = 0.10 – x ≅ 0.10
Ka1 = 7.9 x 10–5 ≅ x2 / (0.10)
∴ x ≅ 2.8 x 10–3 so pH = 2.55
Must use the 2nd equilibrium to find [A2–]:
Ka2 = [A2–][H+] / [HA–] but, from above [H+] ≅ [HA–]
∴ Ka2 ≅ [A2–] (a general result for H2A!)
[A2–] ≅ 1.6 x 10–12

22. How to tell if it’s acidic or basic?
Anions
Anion that is conjugate base of a weak acid is itself a weak base. Exception: H2O!!
An anion that is a conjugate base of a strong acid is pH-neutral.
Anion of a polyprotic acid is amphoteric, e.g. H2PO4–.
Cations
Cation that is conjugate acid of a weak base is itself a weak acid.
A cation that is a conjugate acid of a strong base is pH-neutral.
Small, highly charged metal cations form weakly acidic solutions (not Group I or II), e.g.
Al3+(aq) H2O(l)  Al(H2O)63+(aq)
Neutrals
Weak bases: Table 15.8, p683 and Fig , p687
Weak acids: Tables 15.4, p666, and 15.12, p687, and Fig
Mixed Salts
Salts where the cation is acidic and the anion basic form solutions where the pH depends on the relative strengths of the acid and base.
weak acid

23. More Sample Problems
1. Write balanced chemical equations for the important equilibrium that is occurring in an aqueous solution of each of the following.
(a) HClO (b) (NH4)2SO4 (c) KCl
(d) NaCHO2
2. Write the appropriate equilibrium constant expressions (Ka, etc) for the above reactions.
3. Determine the pH of a M solution of each of the above compounds. (Use equilibrium constants from the textbook as needed).

24. Relative Strengths of Brønsted Acids
Binary Acids e.g. HCl, HBr, H2S, etc.
e.g., relative acidity: HCl > H2S (across a period)
HI > HBr > HCl > HF (up in a group)
Periodic Table
Acid Strength Increases

25. Oxo Acids e.g. HNO3, H2SO4, H3PO4, etc. 1. for same central element,
acid strength increases with # of oxygens
acid strength increases
HClO < HClO2 < HClO3 < HClO4
2. for different central element, but same # of oxygens
acid strength increases with electronegativity
e.g. H2SO4 > H2SeO4 > H2TeO4 A c i d S t r e n g h I a s P o T b l

26. Relative Strengths of Conjugate Acid-Base Pairs
For example,
HF H2O H3O F–
acid base acid base
In this case, the equilibrium lies mainly on the reactant side. Therefore, “HF is a weaker acid than H3O+”
In general, weaker Brønsted acids have stronger conjugate bases. (and vice versa)

27. Lewis Acid-Base Concept
(most general)
Definition (electron pair transfer)
Acid: e– pair acceptor
Base: e– pair donor
Lewis acids electron deficient molecules or cations
Lewis bases electron rich molecules or anions.
(have one or more unshared e– pairs)

28. Lewis Acid-Base Reactions
(i.e. all non-redox reactions!)
OH–(aq) NH4+(aq) --> H2O(l) + NH3(aq)
OH–(aq) + CO2(g) --> HCO3–(aq)

29. Yet More Sample Problems
Among the following, which is the strongest acid?
HBrO3 HBrO4 H2SeO3 H2SeO4 H2TeO3 H2TeO4
Among the following, which is the weakest acid?
H2Se NH3 PH3 H2S H2O AsH3
The conjugate base of CH3OH is ______________.
4) The conjugate acid of HF is _________.
5) Write the equation for the reaction of H2Se with AsH3, then answer the questions below.
Label the Bronsted acid, base, conjugate acid, and conjugate base.
If Kc = 0.078, label which is the strongest acid and strongest base in your equation.

30. And Finally…
Consider the reaction of hydroxide ion with the nitrosyl cation, NO2+, to form nitric acid.
OH–(aq) + NO2+(aq) --> HNO3(aq)
Write Lewis electron dot formulas (including formal charges and/or resonance forms if needed) for all three species in this reaction. Clearly indicate which reactant is the Lewis acid and which is the Lewis base. Use arrow(s) to illustrate the formation of any new chemical bond(s) during the reaction.