Ch. 16: Equilibrium in Acid-Base Systems

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Ch. 16 Equilibrium in Acid-Base Systems

• 16.3a Acid-Base strength and equilibrium law

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Definitions

• Arrhenius
• A produce H in aqueous solution
• B produces OH- in aqueous solution
• very limited
• Bronsted-Lowry
• A H donor
• B H acceptor
• more general

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Acid ionization constant

• equilibrium expression where H is removed to
  form conjugate base
• so for HA H2O lt--gt H3O A-

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Strength

• determined by equilibrium position of
  dissociation reaction
• strong acid
• lies far to right, almost all HA is dissociated
• large Ka values
• creates weak conjugate base
• weak acid
• lies far to left, almost all HA stays as HA
• small Ka values
• creates strong conjugate base

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Water is a stronger base than the CB of a strong
acid but a weaker base than the CB of a weak
acid Water is a stronger acid than the CA of a
strong base but a weaker acid than the CA of a
weak base
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H2O, pH and Kw

• conc. of liquid water is omitted from the Ka
  expression
• we assume that this conc. will remain constant in
  aqueous soln that are not highly concentrated
• pH -logH
• pOH -logOH-
• 14.00 pH pOH

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Example 1

• The OH- of a solution at 25oC is 1.0x10-5 M.
  Determine the H, pH and pOH.
• Kw 1.0x10-14 OH- x H
• H 1.0x10-9
• pH -log(1.0x10-9) 9.00
• pOH -log(1.0x10-5) 5.00
• acidic or basic?
• basic

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Approximations

• If K is very small, we can assume that the change
  (x) is going to be negligible
• rule of thumb is if initial conc. of the acid
  is gt1000 times its Ka value then cancel x
• this makes the answer true to /- 5 and why Ka
  values are given to 2 sig. digs

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Calculating Weak Acids

1. Write major species
2. Decide on which can provide H ions
3. Make ICE table
4. Put equilibrium values in Ka expression
5. Check validity of assumption (x must be less than
   5 of initial conc)
6. Find pH

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Example 2

• Calculate the pH of 1.00 M solution of HF (Ka
  7.2 x 10-4)
• HF, H2O
• HF ? H F- Ka 7.2x10-4
• H2O ? H OH- Kw 1.0 x 10-14
• HF will provide much more H than H2O ignore
  H2O

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Example 2
HF ? H F- HF ? H F- HF ? H F- HF ? H F-
I 1.00 M 0 0
C -x x x
E 1.00 -x x x
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Example 2

• Check assumption

• pH -log(0.027) 1.57

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Example 3

• Find pH of 0.100 M solution of HOCl (Ka
  3.5x10-8)
• HOCl, H2O
• HOCl will provide much more H than H2O, so we
  ignore H2O

HOCl ? H OCl- HOCl ? H OCl- HOCl ? H OCl- HOCl ? H OCl-
I 0.100 M 0 0
C -x x x
E 0.100 -x x x
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Example 3

• Check assumption

• pH -log(5.9x10-5) 4.23

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Homework

• Textbook p743 2a,c,e 5,7,9
• LSM 16.3A and 16.3D

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Ch. 16 Equilibrium in Acid-Base Systems

• 16.3b Base strength and equilibrium law

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Base Strength and Kb

• follows same standard rules as for calculating Ka
  for acids
• Kb is used with weak bases that react only
  partially with water (lt50)

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Bases

• Kb
• base ionization constant
• refers to reaction of base with water to make
  conjugate acid and OH-
• liquid water is again ignored like in Ka
• B(aq) H2O (l) ? BH (aq) OH- (aq)

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Example 4

• Find the pH for 15.0 M solution of NH3 (Kb
  1.8x10-5)
• NH3 will create more OH- than water so self-
  ionization (H2O) can be ignored

Soln

• NH3 H2O lt--gt NH4 OH-

NH3 H2O ? NH4 OH- NH3 H2O ? NH4 OH- NH3 H2O ? NH4 OH- NH3 H2O ? NH4 OH-
I 15.0 0 0
C -x x x
E 15.0-x x x
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Example 4 cont
ion.
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Example 5

• Codeine (C18H21NO3) is a weak organic base. A
  5.0x10-3 M solution of codeine has a pH of 9.95.
• Calculate the Kb for this substance.
• Soln
• What is chemical reaction?
• C18H21NO3 H2O lt--gt HC18H21NO3 OH-
• Find OH- using given pH
• pOH 14.00-9.95 4.05
• OH- 10-4.05 8.9x10-5

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Example 5 cont
C18H21NO3 H2Olt--gt HC18H21NO3 OH- C18H21NO3 H2Olt--gt HC18H21NO3 OH- C18H21NO3 H2Olt--gt HC18H21NO3 OH- C18H21NO3 H2Olt--gt HC18H21NO3 OH-
I 5.0x10-3 0 0
C -x x x
E 5.0x10-3 - x x x

• x OH- 8.9x10-5

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Ka - Kb relationship for conjugate pairs

• Acid-Base strength tables do not give Kb values
• we use the Ka - Kb relationship to solve this
  problem
• this can be used with any conjugate acid-base pair

Kw Ka x Kb
Kw 1.0 x 10-14
or
Kb Kw / Ka
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Example 6

• What is the Kb value for the weak base present
  when sodium cyanide dissociates into an aqueous
  solution?

Soln

• NaCN --gt Na CN- (complete dissociation)
• the CN- is the weak base so we write the
  equilibrium equation with water to determine its
  conjugate acid
• CN- H2O lt--gt HCN OH-
• the conjugate acid is found to be HCN and its Ka
  is 6.2 x 10-10
• Using Kb Kw / Ka find Kb for CN-
• Kb 1 x 10-14 / 6.2 x 10-10 1.61 x 10-5

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Homework

• Textbook p746 11,12
• Textbook p750 2,6,9
• LSM 16.3 A,B D