Transcript 12th Grade IB Chemistry

IB Chemistry Power Points
Topic 18
Acids and Bases
www.pedagogics.ca
Buffers
and Salts
Buffer Solutions
 DEFINITION: A buffer solution contains a weak acid mixed
with its conjugate base (or weak base and conjugate acid)
 Buffers resist changes in pH when a small amount of a
strong acid or base is added to it.
HA ∏ H+ + A-
 If a small amount of a strong acid (H+) is added eqm
shifts to the left as [H+] increases so system adjusts to
increase [HA] and reduce [H+] again.
HA ∏ H+ + A-
 A small amount of a strong base will react with H+ to
form H2O and eqm will shift to the right to increase
[H+] again.
HA ∏ H+ + A-
Making Buffer Solutions
 An example of a weak acid is ethanoic acid. This could be
mixed with sodium ethanoate which will provide ethanoate
ions (conjugate base).
CH3COOH(aq) ∏ CH3COO-(aq)+H+(aq)
 An example of a weak base is ammonia. This could be mixed
with ammonium chloride to provide ammonium ions
(conjugate acid).
NH3(aq) + H2O(aq) ∏ NH4+(aq) +OH-(aq)
In order for a buffer to work well the concentration of the acid/base and
its salt must be much higher than the strong acid/base added.
Optimum Buffer
 A buffer is most effective when the concentration
of weak acid and its salt (the conjugate base) are
equal and the pH is equal to pKa.
 In practice it will work reasonably well with similar
concentrations and the effective buffer range of
any weak acid/ base is pKa 1.
Blood has buffering capacity
 Blood must maintain a pH of 7.4 so its enzymes can
work. If 0.01 mol of H+ or OH- is added to blood it
only changes pH by 0.1 unit.
 The eqm is:
CO2(aq) + H2O(l) ) ∏ H+(aq) + HCO3-(aq)
Buffer Calculation #1
Calculate the pH of a 1.00 dm3 buffer solution made by
dissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3
ethanoic acid solution.
pH = 5.6
Identify
the
weak
acidthe
/ conjugate
OR weak base
Write
In thisthe
case,
equilibrium
assume
equation
equilibrium
andbase
expression
concentrations
of /the
conjugate
Determine
concentrations.
weak acidacid
and pair.
the salt
anion aretheir
the same
as the given
+
CH3COOH
∏ H +
CH3COO
information (very
little change
when
equilibrium is
weakestablished.
acid
[CH3COOH] = 0.075 mol dm-3
1.
2.3.

[
H
][0.50]
4.76
conjugate base
[CH3COO-] =
0.50 mol dm-3


ka  [ H ][CH COO
 10 ]
3
ka  [0.075]
 10 pka
[CH3COOH ]
4.76


0.075

10

pH   log[ H ]   log 

0.50


Buffer Calculation #2
Calculate the mass of ammonium chloride that would need to be
dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer
with a pH of 9.00. Assume no change in overall volume.
mass = 9.50 g
In thisthe
Identify
Write
case,
the
equilibrium
weak
assume
acid
the
equation
/ conjugate
equilibrium
andbase
expression
concentrations
OR weak base
of the
/
conjugate
weak
base acid
is thepair.
sameDetermine
as the given
their
information
concentrations.
(very little
NH3+ H2O is∏established.
NH4+ + OHCalculate
change when equilibrium
[OH-]
weakfrom
basedesired pH [NH3] = 0.100 mol dm-3
1.
2.
3.
conjugate acid
[NH4+] = ?
 (14  pH )
[ NH 4 ][10
]  ] 4.75

[
NH
][
OH
4
kb 
 10  10 pkb
kb[0.100]

[ NH3 ]
4.75
0.100

10
-3
[ NH 4 ] 

0.1778
mol
dm
10 (149)
Buffer Calculation #3
A buffer can also be made by mixing excess weak acid/base with a lesser
amount of strong base/acid. For example, calculate the pH of a buffer
formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150
mol dm-3 ammonia solution.
1. In
2.
3.
Dothis
Write
thethe
case,
stoichiometry
equilibrium
assume the
to
equation
equilibrium
determine
and what
expression
concentrations
the concentrations
of the
are AFTER
weak
base neutralization.
and cation are the same as determined by
+
NH3+little
H2Ochange
∏ NHwhen
stoichiometry (very
equilibrium is
4 + OH
established.
HCl + NH3  NH4+ + Cl-
pH= 9.6

[0.02885][
OH ]
(0.025
 0.075)

 NH 4  kb 
HCl
104.75
limiting (why?)
0.065


[0.06346]
[ NH
4 ][OH ]
 pkb

k


10
 NH 4   b0.02885  0.06346 104.75 
NH3 ]
pOH  [log
  (0.025  0.075)

(0.040  0.150)
0.02885


 NH 3  
0.065
 NH 3   0.06346
Salt Hydrolysis
 A soluble salt is an ionic compound made of
cations (ex. Na+) and anions (ex. Cl-) which
completely dissociates into ions in aqueous
solution.
 Salts can affect the pH of a solution because the
cations act as weak acids by bonding with OHand the anions act as weak bases by accepting H+
ions.
Salt Hydrolysis
 If the cation comes from a strong base then it will
have less acidic activity than one from a weak
base. For example Na+ from NaOH is a weaker
acid than NH4+ from NH3
 If the anion comes from a strong acid then it will
have less basic activity than one from a weak acid.
For example Cl- from HCl is a weaker base than
CH3COO- from CH3COOH.
 Salts from a strong acid and strong base ex. NaCl
will form a neutral solution.
Salt Hydrolysis
 Would a solution of sodium ethanoate be
acidic, basic or neutral?
 Would a solution of ammonium chloride be
acidic, basic or neutral?
 Would a solution of potassium chloride be
acidic, basic or neutral?
Salt Hydrolysis
 With salts of weak acids and weak bases the pH of
the solution formed will reflect the relative
strengths of the acid and base. Ex. Ammonium
ethanoate is about neutral.
BE CAREFUL THOUGH . . .
 Acidity of salts also depends on size and charge of
the cation
Salt Hydrolysis
 Salts with small, highly charged cations are
more acidic than large, low charge cations.
 recall Period 3 chloride salts : NaCl, MgCl2 and
AlCl3  which is most acidic
 The aluminum ion and those of transition
metals exist in water in hydrated form ie.
[Al(H2O)6]3+ , [Fe(H2O)6]3+
Salt Hydrolysis
 the e- attracting power of the ion weakens the
O-H bond and stabilizes the resulting OH- ion.
 As a result these ions are quite acidic in water.
[Fe(H2O)6]3+  [Fe(OH)(H2O)5]2+(aq) + H+(aq)
Learning Check – salt hydrolysis
Acid
Base
Example Salt
Strong
Strong
Weak
Weak
ammonium ethanoate
neutral
Strong
Weak
ammonium chloride
acidic
Weak
Strong
sodium ethanoate
basic
NaCl
Solution
neutral
IB Chemistry Power Points
Topic 18
Acids and Bases
www.pedagogics.ca
Buffers
and Salts
Acid Base Titrations
 If a strong acid is added to a strong base gradually the pH
will start off as 1.
 Once enough base is added that it is now in excess the
pH will change very suddenly to about 13.
 The point at which this change is seen is when the
amount of acid = amount of base.
 This is called the equivalence point.
 With this combination it occurs at pH 7 as the acid and
base combine to make a neutral solution.
Strong Acid and Strong Base
 Most indicators will work for this combination.
Weak Acid + Strong Base
 A weak acid will have a pH of 3-5. When a strong base is
added the pH will increase gradually as HA is converted
to A-.
 This is called the buffering region as it’s acting like a
buffer.
 When half the amount of base has been added it is called
the half-neutralization point and [HA] = [A-]
 so [H+] = Ka x [HA]/[A-]
 Ka = [H+] and pKa = pH
Weak Acid + Strong Base
 Ka = [H+] and pKa = pH
 This is the best way to determine the Ka for a weak acid.
 At the equivalence point the pH increases quickly to 13.
 The equivalence occurs when pH >7.
 Most suitable indicator is phenolphthalein
Strong Acid + Weak Base
 When a weak base is added to a strong acid the pH will
remain around 1 until near the equivalence point when
all the base has been converted into its conjugate acid.
 B(aq) + H+(aq)  BH+(aq)
 At the equivalence point pH is <7 and exact pH can be
found from Kb of the base.
 [OH-] = Kb x [B]/[BH+]
 At equivalence point [B] = [BH+] so Kb = [OH-]
 And pKb = pOH = 14 - pH
Strong Acid + Weak Base
 A suitable indicator here is methyl orange.
Weak Acid + Weak Base
 When a weak acid is added to a weak base the change in
pH is gradual from acidic to basic so it is hard to detect
the equivalence point.
 It is hard to find a suitable indicator.
 The volume of base that will neutralize an acid is not
affected by the strength.
 It only depends on the stoichiometric amounts reacting.
Learning Check
 Do exercise 18.4 on packet.
Indicators
 An indicator is a substance (often an organic dye) that has
a different color in acidic and alkaline solutions so it can
be used to show the end point of a titration.
 The color change is seen because the indicator is a weak
acid/base in which the two forms have different colors.
 Ex. HIn(aq)H+(aq) + In-(aq)
 litmus: red
blue
 Hin stands for the indicator and In- is the other form when it
dissociates. Each of these forms will have a different color.
Indicators
 In the presence of an acid the eqm is driven to the left to form




HIn as the H+ ions combine with In- to reduce [H+].
With a base it shifts to the right to form more In- as the OHof the base combines with H+ reducing [H+] so more Hin
will dissociate.
The usual equation for Ka applies:
Ka = [H+][In-] / [HIn]
[HIn] / [In-] = [H+] / Ka
Indicators
 The color depends on the pH and [H+] and also on Ka so
different indicators change color over different pH
ranges.
 When pH = pKa the two colored forms will have equal
concentrations and the indicator will be in the middle of
its color change.
 If one form is in excess by 10 fold then the color will be
of that form.
pH of Indicator
 If [HIn] is 10x [In-] the pH will be:
 [H+] = Ka x [Hin] / [In-]
 = Ka x 10/1
 = 10Ka
 or pH = pKa -1
 If [In-] were 10x [HIn] then pH = pKa + 1.
 Many indicators change over a region of 2 pH units.
 An effective indicator will change color at the
equivalence point of the particular titration.
Choosing an Indicator
 Phenolphthalein
 pKa = 9.6
 pH range: 8.3 -10.0
 Color in acid: colorless
 Color in alkali: pink
 Good for titrations with strong bases
Choosing an Indicator
 Methyl Orange
 pKa = 3.7
 pH range: 3.1 - 4.4
 Color in acid: red
 Color in alkali: yellow
 Good for titrations with strong acids
Learning Check
 Do exercise 18.5 in packet