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Arrhenius Definition Acids produce H + in water Bases produce OH - in water

Acid-Base Properties. Arrhenius Definition Acids produce H + in water Bases produce OH - in water Applies only to aqueous solutions Allows for only one kind of base (OH - ) 2) Bronsted-Lowery Definition Acid is an H + donor Base is an H + acceptor 3) General Acid Equation

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Arrhenius Definition Acids produce H + in water Bases produce OH - in water

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  1. Acid-Base Properties • Arrhenius Definition • Acids produce H+ in water • Bases produce OH- in water • Applies only to aqueous solutions • Allows for only one kind of base (OH-) 2) Bronsted-Lowery Definition • Acid is an H+ donor • Base is an H+ acceptor 3) General Acid Equation HA + H2O H3O+ + A- • Conjugate base = what is left after H+ leaves acid • Conjugate acid = base + H+ • Conjugate acid-base pair are related by loss/gain of H+ base acid conjugate acid conjugate base

  2. Ka = acid dissociation constant • General Base Equation (Kb = base dissociation constant) B: + H2O BH+ + OH- 6 ) Bases require N: or O: lone pair • Strong Acid = equilibrium lies far to the right ([H+] = [HA]0) HCl H+ + Cl- K = large • Strong Base = equilibrium lies far to the right ([OH-] ≈ [B]0) NaOH Na+ + OH- K = large • Weak Acid = equilibrium lies far to the left ([H+] << [HA]0) HC2H3O2 H+ + C2H3O2- K = 1.8 x 10-5 • Weak Base: equilibrium lies far to the left ([OH-] << [B]0) NH3 + H2O NH4+ + OH- K = 1.8 x 10-5 conjugate base acid conjugate acid base

  3. Water as an Acid and Base • An amphoteric substance can behave as an acid or a base (water) • Autoionization of water (reaction with itself) H2O + H2O H3O+ + OH- • Ionization constant for water = KW = [H3O+][OH-] = [H+][OH-] • For any water solution at 25 oC, [OH-] x [H+] = KW = 1 x 10-14 • Neutral solutions (pure water) have [OH-] = [H+] = 1 x 10-7 • Acidic solutions: [H+] > [OH-] • Basic solutions: [OH-] > [H+] • Example: If [H+] = 1 x 10-5, what is [OH-]? 4) Relationship of pH and pOH • KW = [H+][OH-] • -logKW = -log[H+] – log[OH-] • -log(1 x 10-14) = pH + pOH = 14 • Example: Find [H+], [OH-], and pOH for sample of pH = 7.41

  4. The pH Scale • pH = -log[H+] (simplifies working with small numbers) • If [H+] = 1.0 x 10-7, pH = -log(1 x 10-7) = -(-7.00) = 7.00 3) Number of decimal places in a log: The number of sig. fig’s in the number is how many decimal places you keep when you take the log • Other p Scales: • pOH = -log[OH-] • pKa = -logKa • pH changes by 1 unit for every power of 10 change in [H+] • pH = 3 [H+] = 10 times the [H+] at pH = 4 • pH decreases as [H+] increases (pH = 2 more acidic than pH = 3)

  5. Percent Dissociation = amount dissociated / initial conc. x 100% • For 1.00 M HF we found [H+] = 2.7 x 10-2 Percent Dissociated = (2.7 x 10-2 / 1.00) x 100% = 2.7% • Strong acids are 100% dissociated always • For weak acids, percent dissociation increases as acid is diluted HA + H2O H3O+ + A-Le Chatelier’s Principle Finding Ka or Kb from the pH • Example: an unknown acid HA at 0.40 M has a pH = 2.6. What is the Ka? • Example: an unknown base at 0.5 M has a pH = 11.3. What is the Kb?

  6. Experimental Details • You can skip Procedure 1, the standardization of the pH meter (it has been done) • In Procedure 2, use clean pipet to blow breath through solution (broken glass box) -get your own tap and distilled water -vials of other solutions up front; take to your pH meter and then bring back • In Procedure 3, take 10 ml of the stock solution and dilute it to 100 ml -use volumetric pipet and volumetric flask • Procedure 4 and 5: Use graduated cylinder and be as careful as you can -use pH meter readings to find [H+], find other concentrations from it -substitute concentrations into Ka/Kb expression to find Ka/Kb • Skip the Indicator Procedure Part 6 and PreLab Question #4

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