Weak Base Equilibria: They accept H from H2O

1

• Weak Base Equilibria They accept H (from H2O)
• In general, B(aq) H2O(l) ? BH(aq)
  OH(aq)
• ex. Find the pH of a 0.20 M solution of CH3NH2,
  methylamine. The Kb 4.38 x 104.
• CH3NH2 H2O ? CH3NH3
  OH
• init. 0.20
  0 0
• change x
  x x
• equil. (0.20x)
  x x
• Ignore the small contribution of water to OH.

2

• Set up the mass-action expression
• Apply the quick dirty approximation, 0.20
  x 0.20
• Solving, x OH 9.4 x 103 M
• pOH log(9.4 x 103) 2.03
• pH 14.00 2.03 11.97

3

• Check the approximation. Is x ltlt 0.20 M?
• a 9.4 x 103 / 0.20 0.047
• 4.7 ionized. Therefore, the approximation
  works.

4
Salts of Acids and Bases

• Salt soluble ionic compound formed by the
  reaction of an acid with a base.
• Note Acid-base reactions always go to
  completion (the limiting reagent is used up).
• Case 1 Neutral Salts
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O
• Strong Acid Strong Base Neutral Salt
• Recall that the counterions of strong acids and
  bases are neutral. Thus, NaCl is neutral.
  Because it is soluble, it exists as Na(aq) and
  Cl(aq) in solution.

5

• Case 2 Basic Salts
• HC2H3O2(aq) NaOH(aq) ? NaC2H3O2(aq) H2O
• Weak Acid Strong Base Salt of Weak Acid
• Now, what does the salt of a WA do?
• Since the salt is soluble, it dissociates first.
• NaC2H3O2(aq) ? Na(aq) C2H3O2(aq)
• 
  Neutral
• We have already seen that the Na cation
  (derived from a strong base) is neutral. What
  about the anion?

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• The conjugate bases of weak acids will behave as
  weak bases in water.
• C2H3O2(aq) H2O ? HC2H3O2(aq) OH(aq)
• Similar behavior for F, OCl, CN, NO2, etc.
• This equilibrium is solved like any other weak
  base equilibrium.
• A problem When we look in the tables for Kb
  values, we do not find ions or salts, only
  molecular weak bases.

7

• A trick multiply the mass-action expression by
  H / H.
• Now regroup the terms
• Notice that this is equal to (Kw / Ka for
  HC2H3O2)

8

• In general, for any weak acid HA and its
  conjugate base A (found in salts)
• Kw Ka (HA) Kb (A)
• Ex. Find the pH of a 0.20 M solution of
  NaC2H3O2.
• We find the Kb of acetate ion as above
• Kb Kw / Ka (1.00 x 1014) / (1.8 x 105)
• 5.6 x 1010 a very weak base.
• Now set up the mass-action expression.

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• C2H3O2(aq) H2O ? HC2H3O2(aq) OH(aq)
• 0.20 0
  0
• x x
  x
• (0.20x) x
  x
• Then,
• Solving, x OH 1.1 x 105 M
• pOH 4.98 and pH 9.02 Basic

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• Case 3 Acidic Salts
• HCl(aq) NH3(aq) ? NH4(aq) Cl(aq)
• Strong Acid Weak Base Salt of a
  Weak Base
• Cl is a neutral ion. NH4 is the conjugate
  acid of a weak base and will behave as a weak
  acid.
• ex. Find the pH of a 0.15 M NH4Cl solution.
• We need the Ka for NH4, which is not in the
  tables.
• KW (Ka of NH4) (Kb of NH3)

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• So, Ka 1.00 x 1014 / 1.8 x 105 5.6 x
  1010
• Thus, NH4 is a very weak acid.
• NH4(aq) ? H(aq)
  NH3(aq)
• initial 0.15 0
  0
• change x x
  x
• equil (0.15x) x
  x
• x H 9.2 x 106 M and pH 5.04

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• Case 4 Salts Derived from Weak Acid and Weak
  Base
• HCN(aq) NH3(aq) ? NH4(aq) CN(aq)
• Weak Acid Weak Base
• The NH4 is a weak acid and the CN is a weak
  base.
• The Ka of NH4 is 5.6 x 1010.
• The Kb of CN is 2.5 x 105.
• So, the solution will be basic.