Transcript Metathesis Problems (and Some Solutions) Identified

Chapter 17: Acid-base equilibria
Chemistry 1062: Principles of Chemistry II
Andy Aspaas, Instructor
Acid-ionization equilibria
• Weak acids such as acetic acid involve the following
equilibrium when dissolved in water:
HC2H3O2(aq) + H2O(aq) = H3O+(aq) + C2H3O2-(aq)
• Since acetic acid is a weak acid, only a small amount of its
molecules are ionized, and this equilibrium lies far to the
left
• General weak acid equilibrium: HA + H2O = H3O+ + A• Ka: acid-dissociation
constant
[H3O  ][A ]
Kc 
– Related to equilibrium
[HA][H2 O]
constant (Kc)
[H3O  ][A ]
K a  [H2 O]K c 
[HA]
Calculating acid-dissociation constant
• Most Ka values were obtained by recording the pH
of a weak acid solution at equilibrium
• To solve the problem, the molarity of the weak acid
solution must be known (this refers to how the
solution was prepared, and can be considered the
initial value on an ICE table)
• x = [H3O+] = [A-] = antilog(-pH)
• If x is sufficiently small, an approximation can be
used in the Ka equation, (Ca – x)  ca
• Degree of ionization: percentage of solute that’s in
ionized form: ([A-]/Ca) x 100%
Calculating Ka
• A 0.025 M solution of lactic acid has a pH of 2.75.
– Calculate Ka
– Calculate the degree of ionization.
• Construct ICE table
• Construct Ka equation
• Calculate [H3O+] from pH
• Calculate Ka
• Calculate degree of ionization
Using Ka in calculations
• What is the pH of a 0.10 M acetic acid solution?
Ka = 1.7 x 10-5
• Follow same procedure as last slide, in constructing
ICE table and Ka expression
• If Ca / Ka > 100, you can simplify Ka expression to
just Ka = (x2 / Ca)
Solving for pH when the approximation is not valid
• What is the pH of an aqueous 0.0030 M solution of
pyruvic acid, Ka = 1.4 x 10-4
• If Ca / Ka < 100, the approximation is not valid, and Ka =
x2 / (Ca – x) must be solved with the quadratic equation
ax  bx  c  0
2
b  b  4ac
x
2a
2
Polyprotic acids
• Polyprotic acids have more than one acidic proton
• Separate Ka values for each acidic proton
• Ka for the most acidic proton, Ka1 is generally much
larger than Ka2
– Proton concentration as a result of Ka2
equilibrium is negligible, can be ignored in pH
calculations
• For triprotic acids like H3PO4, Ka3 is even smaller
than Ka2
Polyprotic acid calculations
• What is the pH of a 0.25 M solution of sulfurous acid
(H2SO3)? What is the concentration of sulfite ion,
SO32- at equilibrium?
Ka1 = 1.3 x 10-2, Ka2 = 6.3 x 10-8
• Solve for pH by just using Ka1 reaction
• Solve for [SO32-] by using equilibrium values from
Ka1 reaction as initial values in Ka2 equilibrium
Base-ionization equilibria
• Kb (the base-ionization constant) can be found by a method
similar to that of Ka
• B + H2O = HB+ + OH[HB ][OH ]
Kc 
• Weak bases are nearly
[B][H2 O]
all amines
– All contain nitrogen
[HB ][OH ]
K b  [H2 O]K c 
with lone pair that accepts
[B]
proton from water
Weak base pH calculation
• What is the pH of a 0.20 M solution of ammonia in
water? Kb = 1.8 x 10-5
• Use an ICE table and construct a Kb equation to
solve for [OH-]
• Convert to [H3O+] using Kw, and then to pH
– Or, convert to pOH and subtract from 14 to get
pH
Acid-base properties of salt solutions
• A salt is an ionic compound resulting from an acidbase reaction
• NaCl (aq) is obtained by neutralizing NaOH and HCl
• Consider the individual ions produced in a salt to
determine whether a salt solution is acidic or basic
• Hydrolysis of an ion: acid-base reaction of an ion
with water
Predicting whether a salt is an acid or base
• Is an aqueous solution of NH4Cl acidic or basic?
• Consider reaction of individual ions with water
NH4+ + H2O = NH3 + H3O+
Cl- + H2O = no reaction
• Ammonium can donate a proton to water, so it will
make the solution acidic
• Chloride does not accept a proton from water, since
HCl is a strong acid
• Therefore, the solution will be acidic
Generalizations about salt pH
• A salt of a strong base and strong acid is neutral
since no ions will react with water
• A salt of a strong base and a weak acid is basic (the
anion is strongly basic and will produce hydroxide)
• A salt of a weak base and a strong acid is acidic (the
cation is strongly acidic and will produce hydronium)
• A salt of a weak base and a weak acid can be either
acidic or basic. If Ka of the cation is larger than Kb
for the anion, the solution is acidic, and vice-versa.
pH of salt solutions
• The Ka and Kb of a conjugate acid-base pair are related
• Ex. for the HCN / CN- conjugate acid-base pair, both
hydrolysis equilibria can be drawn
HCN + H2O = H3O+ + CNKa
CN- + H2O = HCN + OHKb
• The sum of these two equilibria is water’s autoionization
equilibrium, 2H2O = H3O+ + OHKw
• When the two reactions are added, the equilibrium constants
are multiplied, so KaKb = Kw
pH of salt solutions
• What is the pH of a 0.015 M solution of sodium
benzoate? Ka of benzoic acid is 6.3 x 10-5
• Consider the hydrolysis reactions of each of the ions
• Calculate Kb of sodium benzoate from the Ka of its
conjugate acid
• Solve for [OH-], convert to pOH and then pH
Common-ion effect
• Addition of another solute to a weak acid or base equilibrium
can disrupt the equilibrium
• Ex. if HCl is added to a solution of acetic acid:
HC2H3O2 + H2O = C2H3O2- + H3O+
• Since HCl is a strong acid, it will add H3O+ to the system, and
the equilibrium above will shift to the left (Le Chatelier’s
principle)
• The ionization of weak acids is repressed by the addition of
strong acid
• The same applies to the addition of extra conjugate base to
the equilibrium: if acetate were added, the ionization of acetic
acid would be repressed
Common-ion calculation
• What is the concentration of formate ion, CHO2-, in a
solution of 0.10 M HCHO2 and 0.20 M HCl?
Ka = 1.7 x 10-4
• Set up the equilibrium with the weak acid on the left
HCHO2 + H2O = CHO2- + H3O+
• Initial concentrations of HCHO2 and H3O+ will be as
given in the problem
• An approximation can be used if x is sufficiently
smaller than either of the concentrations
Buffers
• Buffer: a solution of a weak acid and its conjugate
base, or a weak base and its conjugate acid
• Ability to resist changes in pH
• In a solution of equimolar amounts of a weak acid
and its conjugate base…
– If acid is added, the conjugate base will accept
the protons to neutralize it
– If base is added, the weak acid will supply
protons to neutralize it
pH of a buffer
• What is the pH of a buffer prepared by adding 30.0
mL of 0.15 M acetic acid to 70.0 mL of 0.20 M
sodium acetate? Ka = 1.7 x 10-5
• Calculate initial molarities
• Use those as initial values in ICE and solve like a
common-ion problem
• Use the approximation if x is sufficiently small
Adding a strong acid or base to a buffer
• First use stoichiometry to determine immediate
effects of adding strong acid/base
• Ex. adding HCl to an acetic acid/sodium acetate
buffer solution
HC2H3O2 + H2O = C2H3O2- + H3O+
• Assume that acetate will be stoichiometrically
reduced in moles by the moles of H3O+ added
• Then allow the solution to equilibrate using new
calculated concentrations
Adding a strong acid or base to a buffer
• Using the buffer in the last example, calculate the
effect of adding 9.5 mL of 0.10 M HCl
• Calculate moles of H3O+ added to solution, subtract
this from the base, and add this to the acid
• Calculate new molarities and solve the equilibrium
equation as before
Henderson-Hasselbalch equation
• In buffer solutions made with weak acids and their
salt, the equilibrium concentrations of HA and Adiffer very little from their initial values
• Using the same assumption that x is very small, the
equation can be rearranged
• pH = pKa + log ( [A-] / [HA] )
– for weak acid / salt buffers only! (the most
common kind)
Acid-base titration curves
• Titration curve: plot of pH vs volume of acid or base
added to a solution
• Equivalence point: point in a titration where a
stoichiometric amount of reactant has been added
Strong acid, strong base
• Strong acid / strong base titration: equivalence point
is at pH 7.0 because the salt of a strong acid and
strong base is neutral
• Calculating points on a titration curve for a strong
acid strong base is a matter of stoichiometry: just
subtract molar amounts of acid and base to find the
concentration of unreacted H3O+
• What is the pH of a solution in which 15 mL of 0.10
M NaOH has been added to 25 mL of 0.10 M HCl?
Titration of a weak acid by a strong base
• What is the pH at the equivalence point when 25 mL
of 0.10 M HF is titrated by 0.15 M NaOH?
Ka = 6.8 x 10-4
• In this case the pH is basic at the equivalence point
• To find the pH of the equivalence point, first find the
number of moles of the salt formed if the acid were
completely deprotonated, and use total volume to
get its concentration
• Then, find the pH just as before.
• Similar calculations can be used for titration of a
weak base by a strong acid.