• syllabus update, posted now
  • includes assigned problems (not to be collected) for chapters 13-14-15
  • also slightly modifies the coverage on exam 2
• calendar-- we have exam 2 next Wednesday
  • equilibrium, acid base and Ksp (part of chapter 16)
    • we will not include anything from chapter 20
  • you might look over the relevant parts of Ch 16 now
  • we've already included such problems in earlier classes
• Useful Information
  • I'll try to have some samples, study guide on Friday
  • I'll post last year's exam II (same disclaimers)
  • I'll host a review / problem session 6 pm Sunday
  • I'll try to post answers (by Friday) for the problem set due Sept 20

Acid base Equilibrium

Acid Base Equilibrium

• we started with last two pages of Monday's posted lecture notes

pH, pKa, pOH and logs

• log10 (N) ... enter N, press log key

• log compresses large numerical range
  • pH 0 to 14
  • [H+] = 100 (1) to 10-14
    • the price is small change in log or pH
    • [H+] = 1x10-5 or 2 x 10-5
    • pH = 5.00 or 4.70 (a fairly small change)
• the inverse of log is 10^N
  • pH = 3.21 so [H+] = 10 ^-3.21 = 6.2 x 10^-4
    • many calculators have a log button with 10^x above it
    • enter 3.21 (and - key)
    • press (2nd) or (function) key
    • and the button computes 10-3.21
  • or enter 10 and press the xy button followed by 3.21
    • and (-) (equals) to find 10-3.21
• We can, by analogy, write pAnything
  • -log10 [OH-] = pOH
  • -log10 (Ka) = pKa
  • -log10(Kw) = pKw = 14.0
  • these do turn out to be useful

• log(a) + log(b) = log (a*b)
• log(c) - log(d) = log (c/d)
• n log (a) = log (an)
• -log(a) = log (1/a)

• Kw = [H+] [OH-]
  • log(a) + log(b) = log (a*b)
  • log Kw = log [H+] + log [OH-]
  • or pKw = pH + pOH
• for a weak acid HA ---> H+ + A-
  • Ka = [H+] [A-]/[HA]
    • pKa = pH - log ([A-])/[HA] ) or
    • pH = pKa - log ([A-])/[HA] )
  • If we would like a solution with pH = 3.75
    • we pick an acid with pKa 3.75 +/- 2
    • and adjust ratio of salt/acid to make up the difference
      • (not a good idea to go more than 1.5 pH units)
      • (2 pH units needs 100:1 ratio, so one component is quite weak)
        • we'll see that this is a prescription for a pH buffer

What is the pH of a 0.17 M Ca(OH)2 solution?

• [OH-] = 2 * 0.17 = 0.34 M
  • pOH = -log(0.34) = 0.468
  • pH = 14.0 - pOH = 13.532
• we could also say [H+] = 1x10^-14/0.34 =2.94 x10^-14
  • pH = - log10 ( ) = 13.532

One of my working rules--

• If you mix two acids, the pH (and [H+] ) are fixed by the stronger acid
• Try 0.010 M HCl and 0.1M acetic acid
  • HCl = 100% dissociated
    • my rule says [H+] = 0.010 M
    • (ignoring any H from aetic acid)
  • 1.8 x 10^-5 = [H+] [Ac-] / [HAc]
    • let's assume x = amount of HAc that dissociates
    • 1.8x 10^-5 = [0.010 +x] [x] / [0.10 -x]
      • cheat, almost
      • 1.8x10-5 = 0.010 * x / 0.10
      • x = 1.8 x 10-6 M
    • yeh, ignoring x in 0.10-x is justified
• if we are interested in [Acetate] or degree of dissociation of HAc, we need to evaluate x

Classic Questions or Acid Base problems

• what is the pH of a __ M solution of HCl (or NaOH)
• what is the pH of a dilute weak acid solution
  • works equally well for salts like NH4+ , since it is an acid
• what is the pH of a __ M solution of Ammonia
  • (or other weak base)
    • works equally well for most anions associated with a weak acid (the anions are bases)
    • acetate, cyanide, nitrite, carbonate ion.
• what is the pH of a specified mixture of sodium acetate and acetic acid
  • or any other weak acid and the conjugate base (anion)
• we might also have a reaction before equilibrium
  • if we add 3.0 ml of 0.10 M NaOH to 10 ml of 0.15M HCl, what is the pH ?
  • or, if we add 3.0 ml of 0.10 M NaOH to 10 ml of 0.15M Acetic Acid, what is the pH ?
    • (These last two are equivalent to one point in a titration)
• if the pH of a 0.05M solution of a weak acid is 6.23, what is Ka ?

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