John A. Schreifels

1
Chapter 17

• Acid-Base Equilibria

2
Overview

• Solutions of a Weak Acid or Base
• Acid ionization equilibria
• Polyprotic acids
• Base ionization equilbria
• Acid-Base properties of Salts
• Solutions of a Weak Acid or Base with Another
  Solute
• Common Ion Effect
• Buffers
• Acid-Base Titration Curves

3
Acid Ionization Equilibria

• Weak acids and weak bases only partially
  dissociate their strengths are experimentally
  determined in the same way as strong acids and
  bases by determining the electrical conductivity.
  
• The reaction of a weak acid (or base) with water
  is the same as discussed in previous section.
• Consider the reaction
• Hydronium ion concentration must be determined
  from the equilibrium expression.
• Relative strengths of weak acids can be
  determined from the value of the equilibrium
  constant.
• Large equilibrium constant means strong acid
• Small equilibrium constant means weak acid
• E.g. determine which acid is the strongest and
  which the weakest.
• Acid Ka
• HCN 4.9x10?10
• HCOOH 1.8x10?4
• CH3COOH 1.8x10?5
• HF 3.5x10?4

4
Determining K from pH

• Ka determined if pH and CHA known.
• Use the equilibrium expression for the acid.
• E.g. Determine the equilibrium constant of acetic
  acid if the pH of a 0.260 M solution was 2.68.
  Determine H3O HA and A?.
• Strategy
• Calculate the H3O from pH this is x in the
  table above.
• The rest of the quantities are obtained from the
  bottom row.

5
Calculating Equilibrium Concentrations in
Weakacid Solutions

• pH determined if Ka and Ca known for the
  dissociation of acetic acid
• H3Ototal H3OCH3COOH H3OH2O.
• H3Ototal ? H3OCH3COOH.
• The total hydronium ion concentration is often
  equal to the contribution from the weak acid
  which is usually a lot stronger acid than water.
• The total hydronium ion concentration is needed
  for the equilibrium calculation.

6
pH from Ka and Ca

• E.g. Calculate the pH of 0.100M acetic acid.
  Given pKa 4.76
• Method I
•  Substitute into equilibrium equation to get ?
• x2 1.75x10?5x ? 1.75x10?6 0.
• Solve using quadratic equation (see book).
• Method 2
• Assume x ltlt CHA. Then x (KaCHA)1/2.
• Check (confirm assumption to be correct)
• Analytical concentration should be Ca
  100xH3O
• Method 3 method of successive approximations.
• As in Method 2 then
• x (Ka(CHA ? x1))1/2 repeat if necessary.
• E.g. Calculate pH of 0.0200M lactic acid if its
  Ka 8.4x10?4M.

7
Dissociated (also called Ionized) Weak Acids

• ionization a useful way of expressing the
  strength of an acid or base.
• 100 ionized? a strong acid.
• Only partial ionization occurs with weak acids.
• E.g. determine the ionization for 0.100 M,
  0.0100 M, 0.00100M HCN if Ka 4.9x10?10.
• Solution determine x for each and sub into
  definition above. Check assumptions.
• Notice ionization increases with dilution.

8
Polyprotic Acids

• Some acids can donate more than one proton to the
  solution. Thus a diprotic acid has two protons
  such as H2S and H2SO4, while a common triprotic
  acid has three acidic protons that can be donated
  (H3PO4).
• First proton easily removed others much more
  difficult.

Treat Polyprotic acids as if they were monoprotic
acids Use Ka1.

• The equilibrium constant for removal of each
  successive proton is about 10?5 times the
  equilibrium constant for removal of the
  preceeding proton.
• E.g. determine the pH of 0.100 M H2SO3. Then
  determine .

9
EquilibriaWeak bases (WB) (proton acceptor)

• Treat bases just like we did the weak acid
  except you are calculating OH?.
• The general equation that describes the behavior
  of a base in solution is
• Set up the equilibrium table as before for the
  acids and substitute values for all the
  quanitities in the equilibrium expression.
• Since usually CB is supplied, we have one unknown
  which we can evaluate using standard equil.
  equation for weak base.
• Remember that x OH? and not H3O.
• E.g. Calculate the pH of 0.10M NH3(aq).
• Hint Expect pH gt 7 when with weak base.

10
EquilibriaWeak bases Structure

• Many nitrogen containing compounds are basic the
  amine most important.
• Most of the amines have a lone pair of electrons
  that are available for bonding with an acidic
  proton (Brønsted-Lowry base).
• Amines usually have a carbon residue in place of
  a hydrogen.

11
Relation between Ka and Kb

• Ka and Kb are always inversely related to each
  other in aqueous solutions.
• Inverse relationship explains why conjugate base
  of very weak acid is relatively strong.
• E.g. given the Kas of the following acid list
  their conjugate bases in terms of relative
  strength.

Acid Ka HF 3.5x10?4 HCOOH 1.8x10?4 HOCl 3.5x10?8 H
CN 4.9x10?10
12
Salts of WA and WB

• Salt an ionic substance formed as a result of an
  acidbase neutralization reaction.
• Salt of an acid(base) obtained by its
  neutralization with acid if it is a base and base
  if it is an acid.
• E.g. NaCl is a salt from the reaction of HCl
  with NaOH.
• The properties of the salt will depend upon the
  strengths of the acid and base that formed the
  salt.
• E.g.1 determine the acidbase reaction that
  would produce CH3COONa, NaCN, NH4Cl, (NH4)2CO3.
• Salts are usually soluble in water because of
  their ionic character.
• When they dissolve, they affect the pH of the
  solution. Depends upon relative strengths of the
  conjugate acid and base.

13
Salt of Strong Acid and Strong Base

• Neutral solution results if the salt is from the
  reaction of a SA SB.
• E.g. NaCl
• Other cations and anions producing neutral
  solutions Li, Na, K, Ca2, Sr2, Ba2 and
  Cl?, Br?, I?, , ).
• E.g. what is the approximate pH of the
  following. NaCl, KCl, LiClO4, etc.?
• Salt of WA SB (basic) and Salt of WB SA
  (acidic).
• Ignore cation (or anion) from SA (base).
• Conjugate of WA is WB ? basic solution.
• Conjugate of WB is WA ? acidic solution.

SA SB ? Neutral (very WA WB) SA WB ? Acidic
(WA) WA SB ? Basic (WB) where SA Strong Acid
SB Strong Base WA Weak Acid WB Weak Base
14
Calculating the pH of Salt of WA or WB (other ion
from SA(SB))

• Salt of WA Use Kb of the conjugate base and
  treat it as a weak base
• A?(aq) H2O(l) ? HA(aq) OH?(aq)
• E.g. determine the pH of 0.100M NaCH3COO. Ka
  (CH3COOH) 1.75x10?5.
• E.g. determine the pH of 0.200 M NaCN. Ka(HCN)
  4.9x10?10.
• Salt of WB Use Ka of conjugate acid and treat as
  a weak acid
• E.g. determine pH of 0.250M NH4Cl. Kb
  1.8x10?5.
• E.g. determine pH of 0.100 M N2H5Br. Kb
  1.1x10?8.

15
Salt of WA WB

• Determine Ka and Kb of acidic and basic portions
  of salt.
• Largest K dominates to make solution either
  acidic or basic.
• E.g. determine if 0.100 M NH4CN is acidic or
  basic. E.g. 2 predict if 0.100 M C6H5NH3F is
  acidic or basic.

16
The Common Ion Effect

• CommonIon Effect the change in the equilibrium
  that results from the addition of an ion that is
  involved in the equilibrium.
• E.g. NaOCl is added to 0.100 M HOCl is
  added to NH3.
• Setting up the standard equilibrium table can
  show the effect.
• E.g. determine the pH of a solution prepared by
  mixing 50.0 mL of 0.100 M HOCl with 50.0 mL of
  0.100 M NaOCl (Ka 3.5x10?8).
• Set up equilibrium table after calculating the
  concentrations of each in the final mixture.
• Initial concentrations change slightly as a
  result of a change reaction.
• Solve using either approximations or quadratic
  equation.
• Shifts equilibrium towards the basic side.

17
Buffers

• Buffer solution a mixture of conjugate acid and
  base that resists pH changes.
• Significant buffering capacity occurs when acid
  base, pH pKa.
• An example of the common ion effect.
• E.g. Calculate pH of solution containing 0.040M
  Na2HPO4 and 0.080M KH2PO4. pKa27.20.
• Set up equilibrium table.
• Ignore the value of x compared to the
  concentrations of the common ion.
• pH in buffering region related to the relative
  amount of conjugate acid and base.
• Let then the equilibrium
  equation is

18
Addition of Acid or Base to a Buffer

• Upon addition of a SB to the buffer we have
• Addition of either acid or base changes ratio of
  acidic and basic forms.
• Big changes in pH occur only when nearly all of
  one species is consumed.
• E.g. determine r after addition of 5.00 mL of
  0.100 M NaOH to 10.00 mL of 0.100 M HOCl.
  Determine pH if Ka 3.5x10?8.
• E.g. Determine pH of 50.00 mL of phosphate
  buffer containing equilmolar concentrations
  (0.200M) of acid/base forms, after 10.00 mL
  0.100 M NaOH or 10.00 mL of HCl. pKa2 7.20
• Changes in volume don't affect pH.

19
Henderson-Hasselbalch Equation

• The effect of r (A?/HA) on pH is better
  understood by taking log of both sides of
  equation between K and conc. To give
• Called Henderson-Hasselbach equation.
• Allows us to predict pH when HA/A? mixed.
• When A? /HA 1 (i.e. HAA?), pH pKa
• E.g. Calculate pH of solution containing 0.040M
  Na2HPO4 and 0.080M KH2PO4. pKa27.20.
• E.g.2 determine the ratio of the concentration of
  the conjugate acid to concentration of the
  conjugate base for a weak acid in which the pH
  was 5.45 and pKa was 5.75.
• E.g. determine the pH of a solution consisting of
  0.100 M NH3 and 0.150 M NH4Cl.

20
Neutralization Reactions

• Neutralization Reaction the reaction of an acid
  with a base to produce water.
• Extent of reaction nearly quantitative (except if
  both acid and base are weak.
• SASB
• E.g. HNO3 NaOH ? NaNO3 H2O
• SA produces H3O
• SB produces OH?
• Overall reaction
• WASB thought of as two step reaction.
• E.g. HOCl NaOH ? NaOCl H2O K ?
• Large equilibrium constant means reaction nearly
  quantitative.

21
Neutralization Reactions WB SA and WA WB

• WB SA
• SA produces H3O ions use base as is.
• E.g. NH3 HCl ? Cl? or
• Conclusion Quantitatively generate product
  (nearly).
• WA WB initially undissociated species
  dominates.
•  
• Conclusion Reaction will sometimes, but not
  always, be quantitative.
• E.g. determine the extent of reaction when di
  methyl amine (Kb 5.4x10?4) reacts with either
  HF (Ka 3.5x10?4) or HOCl (Ka 3.5x10?8).

22
pH Titration Curves

• Titration curve plot of pH of the solution as a
  function of the volume of base (acid) added to an
  acid (base).
• Sharp rise in curve is equivalence point.
• pH at equivalence point is 7.0 for SA but higher
  for WA.

• Equivalence point can be used to determine the
  concentration of the titrant.
• E.g. the equivalence point for 15.00 mL of an
  acid occurred when 25.00 mL of 0.075 M NaOH was
  added. What was the molarity of the acid?

23
SASB Titrations

• Base removes some acid and pH increases.
• Let nb moles of base added
• na,r moles of acid remaining
• na,r na ? nb CaVa ? CbVb
• Moles of hydronium ion same as moles of acid
  remaining. nH3O na,r
• Valid until very close to equivalence point.
• Equivalence point(EP) pH 7.00
• Beyond EP pH due only to base added (i.e. excess
  base). Use total volume.
• E.g. Determine pH of 10.0 mL of 0.100M HCl after
  addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH.

24
Titration of SB with SA

• Acid removes some of the base and pH is changed
  by amount of base removed.
• Let na moles of acid added
• nb,r moles of base remaining
• nb,r CbVb ? CaVa
• Moles of hydroxide ion same as moles of base
  remaining.
• nOH? nb,r
• Valid until EP.
• EP pH 7.00
• Beyond EP pH due only to excess acid. Use total
  volume.
• E.g. Determine pH of 10.0 mL of 0.100M NaOH
  after addition of 5.00, 10.0 and 15.0mL of 0.100M
  HCl.

25
WA with SB Titration

• As above base removes some of the acid and pH is
  changed by amount of acid removed.
• Let nb moles of base added
• nHA moles of acid remaining
• nHA CHAVHA ? CbVb
• nA? nb CbVb
• Up to equivalence point moles of hydronium ions
  must be determined from equilibrium expression.
• Equivalence point pH pH of salt of WA
• Beyond Equivalence point Use amount of excess
  base to determine pH.
• E.g. determine pH of 10.0 mL of 0.100M HA after
  addition of 5.00, 10.0 and 15.0mL of 0.100M NaOH.
  Ka 1.75x10?5.

26
WBSA Titrations

• Acid removes some of the base and decreases the
  pH.
• Let na moles of acid added
• nb,r moles of base remaining
• nb,r CbVb ? CaVa
• nBH na CaVa
• Moles of hydroxide ions must be determined from
  equilibrium expression. Valid until EP.
• EP pH pH of salt of weak base.
• Beyond EP pH due only to presence of acid added
  after endpoint (i.e. excess acid) as seen for
  strong base. Volume correction needed as above
  (total volume).
• E.g. Determine pH of 10.0 mL of 0.100M B after
  addition of 5.00, 10.0 and 15.0mL of 0.100M HCl.
  Kb 1.75x10?5.