Jackson Bettis
Michael Martzahn
Definitions
 Acids are H+ donors. They give up H+ ions (protons)
 Bases are H+ acceptors. They are compounds that
snatch up H+ ions.
 Conjugate Acids donate protons in the forward
chemical reaction
 Conjugate Bases accept protons in the forward
chemical reaction
Identification
 Acids have an H in front usually
 Acids have a pH of less than 7
 Bases have an OH sometimes
 Bases have a pH of more than 7
 Conjugate Bases of strong acids are terrible bases that
have no effect on pH
 Conjugate Bases of weak acids are weak bases and thus
do affect pH
Identification, cont.
 Conjugate acids of weak bases are weak acids and do
affect pH
What it means to be a strong acid
 Strong Acids dissociate completely in water
 Therefore, they give up more protons than weak acids
The Six Strong Acids
 HCl
 HNO3
 H2SO4
 HClO4
 HI
 HBr
Acid dissociation reaction in water
 H2O <-> H+ + OH Therefore, water can act as a base or an acid
Kw
 Kw = 1.0 * 10-14
 Kw / [OH-] = [H+]
 Kw / [H+] = [OH-]
 -log[H+] = pH
 -log[OH-] = pOH
 pH + pOH = 14
Writing Ka expressions
 Ka = [H+][A-] / [HA]
 Kb = [OH-][HB+] / [B]
 Ka * Kb = Kw
Calculating pH
 For strong acids: -log[H+]
 For strong bases: -log[OH-]
 For weak acids or bases: ICE table
Calculating pH, cont.
 1.) determine major species in solution
 2.) Decide which species in the reaction will control
[H+]
 3.) Set up an ICE table for the reaction to determine
[H+]
Calculating pH of buffers
 Ex.) We add 0.05 mols of NaOH to a 500 mL solution
of 0.25 M HOCl and 0.20 M NaOCl. Assume no
volume change.
Sample problem :D
 Calculate the pH of a 0.20 M solution of HF (Ka = 7.2 *
10-4)
Another Sample Problem
 20. The ionization constant for acetic acid is 1.8 × 10–5; that
for hydrocyanic acid is 4 × 10–10. In 0.1 M solutions of
sodium acetate and sodium cyanide, it is true that
 (a) [H+] equals [OH–] in each solution
 (b) [H+] exceeds [OH–] in each solution
 (c) [H+] of the sodium acetate solution is less than that of
the sodium cyanide solution
 (d) [OH–] of the sodium acetate solution is less than that
of the sodium cyanide solution
 (e) [OH–] for the two solutions is the same
Yet another sample problem
 12. A solution prepared by mixing 10 mL of 1 M HCl
and 10 mL of 1.2 M NaOH has a pH of
 (a) 0
 (b) 1
 (c) 7
 (d) 13
 (e) 14