Download Chapter 8 - Activity

Acids and Bases!
 Acids
and Bases (and calculations involving them)
are essential to all areas of analytical chemistry!
What are Acids/Bases?
Aqueous
Definition
Brønsted-Lowry Definition
Lewis Definition
Conjugate Acids and Bases
 These
are the products of acid/base reactions!
CH3C(O)OH + CH3NH2 
CH3C(O)O- + CH3NH3+
Acid (H+donor)
conjugate
base (lost H+)
base (H+ acceptor)
conjugate
acid (gained H+)
Acids, Bases, and Equilibrium
 We
can write the autoprotolysis (self-ionization) of
water in this way:
H20
H+ + OHFor this reaction, the measured K (Kw) is 1.0 x 1014 at 25° C.
You can calculate initial and final concentrations
of [H+] and [OH-] from this equation.
Remember: K is temperature dependent!
pH!
is another way to express [H+] in solution
 pH = -log [H+], or log 1/[H+]
 Also, pOH = -log[OH-] or pOH =log 1/[OH-]
 A useful relation:
 pH

pH + pOH = 14 at 25°C
Strong/Weak acids and bases
 Strong
acids and bases:
H+ + Cl-
 HCl
 Weak
acids and bases:
 HA +
 HA
H2O
H3 O + + A H+ + A -
Polyprotic Acids/Bases
 These
compounds have the ability to donate or
accept more than one proton.
 Example, PO43- + H20 - can accept 3 protons
 There’s a K for each reaction!
Example problem 1
 What
HCl?
is the pH of a 1.8*10-3 M solution of
Example 2
 What
is the pH of the resulting solution when 0.500
moles of acetic acid are dissolved in water and
diluted to 1.00 L?
CH3COOH(aq)  H+(aq) + CH3COO-(aq)
Ka = 1.8x10-5 at 25oC.
Example 3
 What
is the pH of the resulting solution
when 0.500 moles of trichloroacetic acid
are dissolved in water and diluted to 1.00
L?
CCl3COOH(aq)

H+(aq) + CCl3COO-(aq)
Ka = 1.3x10-1 at 25oC.

Example 4
A 0.100 M solution of the weak acid
HA has a pH of 2.36. Calculate pKa
for HA.