Jackson BettisMichael Martzahn

DefinitionsAcids 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 reactionConjugate Bases accept protons in the

forward chemical reaction

IdentificationAcids have an H in front usuallyAcids have a pH of less than 7Bases have an OH sometimesBases 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 acidStrong Acids dissociate completely in water

Therefore, they give up more protons than weak acids

The Six Strong AcidsHClHNO3

H2SO4

HClO4

HIHBr

Acid dissociation reaction in waterH2O <-> H+ + OH-

Therefore, water can act as a base or an acid

KwKw = 1.0 * 10-14

Kw / [OH-] = [H+]

Kw / [H+] = [OH-]

-log[H+] = pH

-log[OH-] = pOH

pH + pOH = 14

Writing Ka expressionsKa = [H+][A-] / [HA]

Kb = [OH-][HB+] / [B]

Ka * Kb = Kw

Calculating pHFor 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 buffersEx.) 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 :DCalculate the pH of a 0.20 M solution of HF

(Ka = 7.2 * 10-4)

Another Sample Problem20. 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 problem12. 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