1. Acids and Bases
Operational definitions are based on observed properties. Compounds can be
Classified as acid or base by observing these sets of properties.

2. Properties of Acids
Taste sour (acere – Latin for sour) (Lemons, vinegar)
Cause certain organic dyes to change colour (Turns blue litmus paper to red – BAR)
Acid properties are destroyed by Bases (React with bases to form a salt and water)
Acid solutions are Electrolytes (substance in solution that conduct an electric current – Acids can be strong or weak electrolytes)
Acids react (corrode) with active metals (Group I and II as well as Zn and Aluminum) (Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g))
Acids react with carbonates (CO32-) and hydrogen carbonates (HCO31-) to produce carbon dioxide gas {2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + H2O(l) + CO2(g)}
Certain nonmetal oxides will dissolve to produce acid solutions. (SO3(g) + H2O → H2SO4(aq) (SO3(g) is the acid anhydride – without water)

3. Properties of Bases Bases taste bitter; mustard and soap
Bases cause weak organic acids (dyes) to change colour (red litmus paper to blue {BB} Basic Blue
Acids destroy base properties - react with acids to form salts and water
Bases are electrolytes {strong or weak}
Feel soapy, slippery
Bases are formed when the oxide of some metals dissolve in water (CaO(s) + H2O → Ca(OH)2(aq) {CaO is the base anhydride}

4. Acid/Base definitions
Definition #1: Arrhenius (traditional)
Acids are compounds with ionizable hydrogen– produce H+ ions (or hydronium ions H3O+) in solution
Bases are compounds that produce OH- ions in solution (problem: some bases don’t have hydroxide ions!)
The reaction between an acid and a base: H+(aq) + OH-(aq) → H2O (l)

5. Arrhenius acid is a substance that produces H+ (H3O+) in water.
The HCl molecule is ionized. (ionization)
Arrhenius base is a substance that produces OH- in water.
The ions are dissociated. (dissociation)

6. Some acids have more than one ionizable hydrogen
H2SO4 → H+(aq) + HSO41-(aq
HSO41- → H+(aq) + SO42-(aq) H2SO4 is diprotic
H3PO4(aq) → H+(aq) + H2PO41-(aq)
H2PO41-(aq) → H+(aq) + HPO42-(aq)
HPO42-(aq) → H+(aq) + PO43-(aq)
Phosphoric acid is a triprotic acid.

7. Water self-ionization
H2O ↔ H+(aq) + OH-(aq)
[H+] = [OH-] = 10-7M at SATP
Keq = [H+][ OH-]
[H2O(l)]
Kw = [H+][ OH-] = 10-7 x 10-7 (at 25ºC)
Kw = at SATP

8. H2O ↔ H+(aq) + OH-(aq)
What happens to this equilibrium if HCl(g) dissolves in the water?
HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)
Increasing Decreasing
H2O ↔ H3O+(aq) + OH-(aq)
[H+] > [OH-] = acidic
What happens when sodium hydroxide dissolves? NaOH(s) + H2O → Na+(aq) + OH-(aq)
Decreasing Increasing
[H+] < [OH-] = basic (alkaline solution)
If [H+] = 10-7 then [OH-] = 10-7 solution is neutral (SATP)

9. pH and logs
[H+] is important in the study of acid-base chemistry. pH is the widely used scale to show [H+].
pH = -log[H+] or pH =
log[H+]
[H+] = 10 – pH (the antilog)
A logarithm is the power to which ten must be raised to get a number.
log1000 = log(103) = 3

10. pH calculations For a neutral solution pH = -log[H+] pH = -log [10-7]
pH = 7 at SATP
Example:
[H+] = 5 x 10-3
pH = -log [5 x 10-3]
pH = -log [0.005]
pH = - (-2.3) = 2.3

11. pH and pOH pOH = - log [OH-] or [OH-] = 10 - pOH
Kw = [H+] x [OH-] = 1 x (at 25ºC)
pKw= pH + pOH
14= pH + pOH
Example:
If pH = (2.3) what is the [OH-]?
pH + pOH = 14
pOH = 14 – pH
pOH = 14 – 2.3
pOH = 11.7
pOH = -log [OH-]
[OH-] = inverse log or ( )
[OH-] = 2.0 x 10-12

12. [H3O+], [OH-] and pH What is the pH of the 0.0010 M NaOH solution?
[OH-] = (or 1.0 X 10-3 M)
pOH = - log
pOH = 3
pH + pOH = 14
pH = 14 – 3 = 11
OR Kw = [H3O+] [OH-]
1.0 x10-14 = [H3O+] x 1.0 X 10-3
[H3O+] = 1.0 x M
pH = - log (1.0 x 10-11) = 11.00

13. Problem 1: The pH of rainwater collected in a certain region of the northeastern New Brunswick on a particular day was What is the H+ ion concentration of the rainwater?
[H+] = 1.51 x 10-5
Problem 2: The OH- ion concentration of a blood sample is 2.5 x 10-7M. What is the pH of the blood?
pOH = 6.6 pH = 7.4
Problem 3: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) M. Calculate the [H3O+], pH, [OH-], and pOH of the two solutions at 25°C.
a) [H3O+] = [3.0], pH = , pOH = 14.48, [OH-] = 3.3 x 10-15
b) [H3O+] = [2.4x10-3], pH = 2.62, pOH = 11.38, [OH-] = 4.2 x 10-12
Problem 4: What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?
[H3O+] = 2.14 x10-4, pOH = 10.33, [OH-] = 4.68x It is an acid.
Problem 5: Problem #4 with pH = 8.05?
[H3O+] = 8.92 x10-9, pOH = 5.95, [OH-] = 1.12x 10-6 It is an base.

14. Acid/Base Definitions
Definition #2: Brønsted – Lowry
Acids – proton donor A “proton” is a hydrogen ion (the atom lost it’s electron)
Bases – proton acceptor (accepts a hydrogen ion)
No longer needs to contain the OH- ion

15. A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
acid
conjugate acid
conjugate base
base

16. The Bronsted-Lowry concept+ Cl H O
base
conjugate acid
conjugate base
acid
conjugate acid-base pairs
Acids and bases are identified based on whether they donate or accept H+.
“Conjugate” acids and bases are found on the products side of the equation. A conjugate base is the same as the starting acid minus H+.

17. OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l)
Practice problems
Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs:
CH3COOH(aq) + H2O(l)  CH3COO–(aq) +H3O+(aq)
acid
base
conjugate base
conjugate acid
conjugate acid-base pairs
OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l)
base
acid
conjugate base
conjugate acid
conjugate acid-base pairs

18. Base Conjugate acid
\ \
NH3(g) + H2O(l) ↔ NH4+(aq) + OH-(aq)
/ /
Acid Conjugate Base
HCl(aq) + H2O(l) ↔ H3O+(aq) + Cl-(aq)
Acid Base Conjugate Conjugate
Acid Base
The water has acted as both an acid and a base, depending on what it is mixed with. Substances that can act as both an acid and a base are amphoteric (also called amphiproteric).

19. Strong acid and base :
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq)
At equilibrium the ionic form is favored
Weak acid and base :
At equilibrium the molecular form is favored

20. CH3COOH(aq) + H2O(l) ↔ H+(aq) + CH3COO-(aq)
Keq= [H+] [CH3COO-] .
[CH3COOH] [H2O(l)]
[H2O] is a constant, so collect the constants
(Keq)[H2O(l)] = [H+] [CH3COO-]
[CH3COOH]
 (Keq)[H2O] is represented Ka(ionization constant for an acid)
Ka = [H+]e [CH3COO-]e = x 10-5
[CH3COOH]e
Ka < 1 weak acid
General Formula for the ionization constant of a weak acid.

21. a) What is the pH of an ethanoic acid solution with a concentration of 0.100M?
Ka CH3COOH = 1.82 x 10-5
CH3COOH(aq) ↔ H+(aq) + CH3COO -(aq)
[H+] = x
[CH3COOH]i = 0.100M
Ka = [H+]e [CH3COO-]e [H+]e= [CH3COO-]e 1:1 ratio
[CH3COOH]e
[H+]e = x = [CH3COOH]R ([CH3COOH]R – ionized)
Ka = x2e [CH3COOH]e = [CH3COOH]i - [H+]e = 0.100M
[CH3COOH]e Because it is a very weak acid [CH3COOH]R = 0
x2 = Ka x [CH3COOH]e
x2 = 1.82 x 10-5 x 0.100
x2 = 1.82 x 10-6
[H+] = x = M
pH = -log[H+] pH = -log[ ] pH = 2.87

22. b) What is the percent ionization of this acetic (ethanoic) acid solution?
% ionization = [H+] x 100
[CH3COOH]
= x 100 = 1.35%
0.100
(very low degree of ionization)

23. The value of Ka for phosphoric acid, H3PO4(aq), is 7. 0 x 10-3 at 25C
The value of Ka for phosphoric acid, H3PO4(aq), is 7.0 x 10-3 at 25C a) Calculate the [H3O+] in a 0.10 M solution of H3PO4.
H3PO4 ↔ H+(aq) + H2PO4-(aq)
Ka= [H+] [H2PO4-] [H+] = [H2PO4-] = x [H3PO4]e = [H3PO4]i - x
[H3PO4] 
Ka= x
0.10 – x 
7.0 x 10-3 = x
0.10 – x
x2 = -7.0 x 10-3x x 10-4
x x 10-3 x – 7.0 x 10-4 =0
x = -b± √ b2 – 4ac 2a
x = -7.0 x 10-3 ± √ (7.0 x 10-3)2 – 4 x 1 x – 7.0 x 10-4
2 x 1

24. x = -7.0 x 10-3 ± √ 4.9 x 10-5 – -2.8 x 10-3 2
x = -7.0 x 10-3 ± √ 2.85 x 10-3
x = -7.0 x 10-3 ± x 10-2
x = x 10-2
x = 2.32 x 10-2 = [H+]
pH = - log[2.32x10-2] = 1.63