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Acids and Bases Part 1. 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 - PowerPoint PPT Presentation
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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