Demonstrate understanding of
equilibrium principles in
aqueous Systems
A.S. 91392
Chemistry 3.6
5 external credits
Equilibrium Systems
• N2(g) + 3H2(g)
2NH3(g)
-∆H
• What is the formula for the equilibrium constant Kc?
Don’t include
solids or the
solvent in Kc
[C]c [D]d
K eq 
a
b
[A] [B]
• What can be said about the concentrations if Kc is very small?
• The concentrations of reactants are larger than products so the position
of equilibrium is far to the left
• What happens to the equilibrium and concentration of NH3 if heat is
added?
• It shifts in favour of the endothermic (reverse) reaction in order to
minimise the effect of the change therefore the concentration of the NH3
decreases
Le Chatelier Principle
• H2(g) + I2(g)
2HI(g)
• Kc = [HI]2
[H2][I2]
• At equilibrium the concentration of
species doesn’t change
• Explain the changes
• [H2] is increased so forward reaction is
favoured to use up H2 and re-establish Kc,
therefore the [I2] decreases and [HI]
increases
• What would the graph look like if the
pressure was increased?
• What if a catalyst was added?
ESA pg216 Q5
Solubility Definitions
• Solvent – a liquid that dissolves
something
• Solute – gets dissolved (by the
solvent)
• Saturated – no more solute can
dissolve
• Solubility – how much solute will
dissolve in a volume (molL-1 or
gL-1)
Dissolving
When ionic solids are added to water an
aqueous solution of ions is formed.
Remember:
• When things dissolve in water there will
always be OH- and H3O+ present
• 2H2O
H3O+ + OH• If the solution is acidic then [H3O+]>[OH-]
• Kw = [H3O+ ][OH-] = 10-14 therefore low
level of dissociation
• In neutral solutions [H3O+ ] = [OH-] = 10-7
• These species dissolve but
don’t react with water:
• Cl-, Br- and ions in groups 1
and 2
• Anions combined with
oxygen eg SO42-, NO3• Glucose, ethanol
Calculating Ks
• At the point of saturation an equilibrium is
formed.
• Adding Ag2CrO4 to water will create an
equilibrium. Write the equation.
• Ag2CrO4(s)
2Ag+(aq) + CrO42-(aq)
• Ag2CrO4 has a solubility of 6.5 x 10-5 molL-1
• This is a special type of
equilibrium constant
• What is the concentration of CrO42-?
6.5 x 10-5 molL-1
• What is the concentration of Ag+?
1.3 x 10-4 molL-1
•
•
•
•
What is the value of Ks?
Ks = [Ag+]2[CrO42-] therefore
Ks = (1.3 x 10-4)2 x (6.5 x10-5)
Ks = 1.1 x 10-12
• The bigger the value of
Ks, the more likely the
substance will dissolve
Special Equilibrium Constants
Ks
Ks = [A] x [B] for AB salt (NaCl), giving s =
s = solubility (molL-1)
Ks= [A][B]2 for AB2 salt (MgCl2)
Ks =
[A]2[B]
for A2B salt (Na2S), giving s =
3
Ks
4
Plus Kw, Ka and Kb
Use these to calculate solubility if you are
given Ks
Memorise
these!
Solubility calculations
• Write equilibrium and then Ks
expressions for:
• Zn(OH)2 and FeS dissolving in
water
• Now find the solubility of
•
•
•
•
A) FeS where Ks = 6.3 x 10 -18
2.51 x 10 -9 molL-1
B) Zn(OH)2 where Ks = 2.0 x 10 -17
1.71 x 10 -6 molL-1
• Don’t forget the equilibrium
arrow
• Don’t include solids or solvents
as their concentration doesn’t
vary
Does a Precipitate form?
•
•
•
•
•
Known as the Ionic Product
Use the Ks expression
If IP>Ks a precipitate will form
Ks(AgCl) = 1.6 x 10 -10
If a mixture of ions is formed with a
[Ag+] = 2.1 x 10 -4 molL-1 and [Cl-] =
3.4 x 10-5 will a precipitate form?
• An I.P. of 7.14 x 10 -9 > Ks therefore
a precipitate will form.
• ESA pg 229 activity 19E
Ionic Product calculations
• Will a precipitate form if we mix 50ml of
0.01molL-1 AgNO3(aq) with 50ml of
0.002molL-1 NaCl(aq)? Ks(AgCl) = 1.8 x
10-10
• Write equilibrium for the formation of
AgCl(s)
• AgCl(s)
Ag+(aq) + Cl-(aq)
• Write Ks expression
• Ks = [Ag+][Cl-]
• Calculate I.P. by substituting values for
the concentrations of ions I.P. = 5.0 x 10-6
• If you increase the volume you will
decrease the concentration
• Use this formula to work out the
new concentrations
• [new] = original vol x [original]
•
Total vol
• E.g. for Ag+ new concentration is:
• (50/100) x 0.01 = 0.005molL-1
• Comment on the size of I.P.
compared to Ks and therefore
whether a precipitate will form.
• I.P. > Ks so a precipitate will form.
• What would happen if the volume
of AgNO3 were 10ml and NaCl were
40ml?
Common Ion effect
• Adding or removing a common ion will disturb the
solubility equilibrium (Le Chatellier)
• Less AgCl will dissolve in 0.1 M NaCl(aq) compared to in
pure water. (Less Ag+ can dissolve because the solution
already contains Cl-.)
• Given Ks(AgCl) = 1.56 x 10-10 calculate the solubility of AgCl
• As the [Cl-] is 0.1M then Ks = [Ag+] x 0.1 = 1.56 x 10 -10
• So the [Ag+] = 1.56 x 10 -9 molL-1which will be equivalent to
the concentration of AgCl To maintain Ks
A common ion will reduce
the solubility of a substance
[Ag+] must
decrease
adding more Cl- to
the RHS shifts the
equilibrium to the
left
Complex Ions and solubility
•
•
•
•
•
•
Formed if a precipitate disappears when excess reagent is added.
Metal cations with several ligands attached.
Ligands have a pair of non-bonding electrons e.g. H20, NH3, OH-, SCNThe number of ligands is twice the charge on the cation. E.g. Cu2+ forms [Cu(NH3)4]2+
Zn(OH)2
Zn2+ + 2OHAccording to equilibrium principals adding extra OH- will shift the equilibrium in the
reverse direction in order to remove the extra OH- and maintain Kc, therefore more
Zinc Hydroxide will precipitate but…
ESA pg
• Zn(OH)2 + 2OH[Zn(OH)4]2227
2+
• When excess OH is added the zincate ion forms which removes Zn from the
activity
2+
solution thus shifting the equilibrium forwards to produce more Zn and maintain 19C and
Ks, this causes more Zn(OH)2 to dissolve.
D
Formation of a complex ion increases the solubility of a substance
• Can you predict what might happen if the reaction took place in acidic conditions?
Solubility in acidic conditions
• If an excess of H3O+
• Zn(OH)2
Zn2+ + 2OH• 2H2O
H3O+ + OH• Therefore hydroxide ions are
neutralised by the hydronium
thus removing them from the
equilibrium which shifts forward
in order to maintain Ks, so more
Zn(OH)2 can dissolve.
Solubility in basic conditions
Calculate the solubility of Fe(OH)2 in
pH10 solution. Ks(Fe(OH)2) = 8.0 x 10-16
Fe(OH)2(s)
Fe2+ + 2OHKs=[Fe2+ ][OH-]2
We need to work out[OH-]
[H3O+] = 10-pH = 1 x10-10molL-1
Kw =1x10-14=[H3O+][OH-]
So [OH-] =1x10-4molL-1
Substitute this into the Ks expression
• 8.0x10-16 =[Fe2+](1x10-4)2
• [Fe2+] = 8x10-8molL-1
• This is equal to the solubility of
Fe(OH)2 due to stoichiometry.
Solubility in basic conditions
• Basic anions will react with acid
and therefore alter the
equilibrium
• NaOH is added to seawater to
precipitate out Mg2+ as Mg(OH)2.
[Mg2+] = 0.555molL-1
• Calculate the minimum
hydroxide concentration and
hence the pH of the solution
needed for precipitation to
occur. Ks (Mg(OH)2) = 7.1x10-12
• Mg(OH)2
Mg2+ + 2OH• Ks= [Mg2+ ][OH-] 2
• 7.1x10-12 > [Mg2+ ][OH-]2
• 7.1x10-12 > 0.555 x [OH-] 2
• [OH-] > 3.58x10-6molL-1
• pOH = -log[OH-]
• pOH = 5.45
• Minimum pH = 14 – 5.45 = 8.55
Acids and Bases
•
•
•
•
•
•
•
•
You should know:
pH = -log[H3O+]
Kw = [H3O+] x [OH-] = 10 -14
[H3O+] = 10 -pH
pOH + pH = 14
pOH = -log[OH-]
[OH-] = 10 -pOH
Strong acids such as HCl, HBr, HNO3
and H2SO4 all fully dissociate and
therefore are NOT in equilibrium.
• Weak acids only partially dissociate
so if a strong and weak acid have
the same concentration, the strong
acid will have a greater [H3O+] and
therefore the lower pH. (pH is
inversely proportional to the
[H3O+])
Weak Acids
• CH3COOH + H2O
+
• Work out the equation for the equilibrium
constant for the dissociation of weak acids (Ka)
• Ka is the acid dissociation constant or acidity
constant
• Ka =[product]
[reactant]
• [water] remains very high and can be thought of as
constant therefore… Ka = [H3O+][A-]
[HA]
CH3COO-
Also as [H3O+] = 10-pH then…
Ka = 10-pKa
pKa = - log Ka
H3O+
So if Ka is small
the level of
dissociation is
low, the [H3O+] is
low so the pKa
and pH are high
MEMORISE
Can you rearrange this
to find [H3O+] for a
weak acid?
Clue: use stoichiometry
[H3O+] = K [HA ]
a
Weak Acid Calculations - Ka and pKa
• Ka is dependent on concentration
and temperature
• If a 0.1molL-1 solution of HOCl has a
pKa of 7.53 calculate its
Ka,[H3O+]and hence pH
• HOCl + H2O
H3O+ + OCl• Ka =[H3O+][OCl-]
[HOCl]
• Ka=10-pKa
• Ka = 2.95 x 10-8
• Ka is very small, not much
dissociation so assume
[HOCl]=0.1molL-1
• [H3O+] =√(Ka x[HOCl] )
• [H3O+] = 5.43 x 10-5
• pH =-log[H3O+]
• pH = 4.27 therefore HOCl is a weak
acid
Note: pKa is high, Ka is small,
partial dissociation means
low [H3O+] and so high pH
Weak Bases
• NH3 + H2O
NH4+ + OH-
• Ka can be used to calculate the pH of a weak
base e.g 0.1molL-1 NH3 (Ka = 5.75x10-10)
• This is because the conjugate acid NH4+
further reacts with water
• NH4+ + H2O
NH3 + H3O+
• Therefore Ka = [NH3][H3O+]
[NH4+ ]
• But due to stoichiometry [NH4+ ] = [OH-] and
[OH-]=kw/[H3O+]
• So Ka =[NH3][H3O+]
Kw/[H3O+]
• Which rearranges to Ka = [NH3][H3O+]2
Kw
• [H3O+]=√(Ka xKw)/[NH3]
• The very low value for Ka means
we can assume [NH3] = original
concentration
• Can you prove the pH of the
solution is 11.1?
• An alternative method is to use the
base dissociation constant (Kb) and
pOH
• Kb = Kw/Ka
• [OH-] = √(Kb x [NH3])
Species in Solution
•
•
•
•
•
HCl(g) and CH3COOH(l) are both soluble in water
Write separate equations to demonstrate this
HCl + H2O
H3O+ + ClCH3COOH + H2O
CH3COO- + H3O+
What are the relative concentrations of ALL the
species present in each equation?
• H2O >> H3O+ >Cl->OH• H2O >>CH3COOH> H3O+ > CH3COO- >OH• Which solution would have the greater
conductivity and why?
• The partial
dissociation of water
adds a small amount
of H3O+ and OH- to
any aqueous system
Full dissociation means a high
concentration of ions and
greater conductivity in the HCl
solution
Conductivity
• Electrical conductivity is due to the
movement of charged particles.
• A solution which fully dissociates will
have a greater concentration of ions
and will therefore conduct better
than one which only partially
dissociate.
• HCl(g) + H2O(l)
H3O+ + Cl• NH3(g) + H2O (l)
NH4+ + OH-
Exceptions
• CH3CH2OH dissolves but no
ions form
• CH3COOH and NH3 only
partially dissociate
• Ca(OH)2 limited solubility
means weak electrolyte
Species in solution
• Consider the following compounds dissolving in water and
determine the relative concentration of species and their pH
• 5.85g of NaCl, 1 litre of water (MNa=23, MCl=35.5)
• 0.1molL-1 CH3COONa, Ka (CH3COOH)= 1.74 x10-5
•
•
•
•
NaCl
Na+ + Cln=m/M so [NaCl]=[Na+]=[Cl-]=0.1molL-1
[Na+]=[Cl-]>[H3O+]=[OH-]
[H3O+] = 10-7 and pH =7 so solution is neutral
•
•
•
•
•
•
CH3COONa H2O CH3COO- + Na+
CH3COO- + H2O
CH3COOH + OH[H2O]>>[Na+]>[CH3COO-]>[OH-]>[CH3COOH]>[H3O+]
[H3O+] < 10-7 and pH >7 so solution is basic
Kb = Kw/Ka and [OH-] = √(Kb x [CH3COO-])
[OH-] = 5.71 x 10-11 therefore pH = 8.88
• pH measures the [H3O+]
• Kw= 10-14=[H3O+][OH-] so
if [OH-] increases then
[H3O+] decreases
Therefore salts of weak acids are basic.
What about salts of weak bases?
Prove, using Ka that the
pH is 8.88
Buffer Solutions
Buffers are solutions of [weak acid] ≈[conjugate base]
Or
• [weak base]≈[conjugate acid]
• E.g. CH3COOH and CH3COO- or NH3 and NH4+
• Buffers resist a change in pH by neutralising the
addition of H3O+ or OH• As dissociation of weak acid / bases is low additional
conjugate is added to ensure approximately
equivalent concentrations
• The buffer zone is the area where the pH does not
change significantly
Common Buffers
• Methanoic acid/ methanoate
• Ammonium ion/ ammonia
• Ethanoic acid/ ethanoate
• Hydrogen carbonate/ carbonate
Adding acid (the conjugate base reacts)
• NH3 + H3O+
NH4+ + H2O
• A- + H3O+
HA + H2O
Adding base(the conjugate acid reacts)
• NH4+ + OHNH3 + H2O
• HA + OHA- + H2O
Buffer calculations
• HA + H2O
A- + H3O+
• HA = acidic species, A- = basic species
• So using the Ka expression we can say
[HA]
Ka x
[A-]
•
• Or
-]
[A
• pH = pKa + log
[HA]
[H3O+]=
• If [HA] = [A-]then [H3O+]= Ka and pH =pKa
(half way to equivalence)
Diluting a buffer will not change its pH but will
reduce its buffering capacity.
Calculate the pH of a buffer made from
0.100mol of NH4Cl dissolved in 100ml of
0.100molL-1 NH3(aq) Ka(NH4+) = 5.75 x 10-10
• [NH4+]= 0.1 x 10 = 1molL-1
• [H3O+]= 5.75 x 10-10 x 1/0.1 = 5.75 x 10-11
• pH = -log(5.75 x 10-11) = 10.2
Buffer calculations
• You will need to find the concentration of
acidic and basic species or the pH
• Write the equilibrium
• Calculate the pH of a buffer made from
adding 2.23g NaF to 150ml of 0.05molL-1
HF(weak acid). pKa(HF) = 3.17, M NaF = 42
• NaF
Na+ + F• HF +H2O
H3O+ + F• nNaF =m/M = 2.23/42 = 0.0531
• 1:1 mol ratio therefore 0.0531 mol F- added
to 0.15L (n/v =c) so [F-] = 0.354molL-1
• HF partially dissociates so assume [HF] =
0.05molL-1
-]
[A
• pH = pKa + log
[HA]
• pH = 3.17 + log(0.354/0.05)
• pH =4.02
Buffer calculations
• Calculate the pH of a buffer
made from 50ml of 1molL-1
sodium methanoate added to
50ml of 0.2molL-1 methanoic
acid (pKa HCOOH = 3.74)
[A-]
• pH = pKa + log [HA]
• [A-]= methanoate
• [HA]= methanoic acid
• HCOONa
HCOO- + Na+
• 1:1 mol ratio therefore
[HCOONa]=[HCOO-]
• n=cv= 1 x0.05= 0.05 mols HCOO• New solution will be 100ml so
0.05/0.1 = 0.5molL-1
• Dilution of [HA] makes the new
concentration= (50/100) x 0.2 = 0.1
molL-1
[A-]
• pH = pKa + log
[HA]
• pH = 4.44
-]
[A
• Now use [H3O+]= Ka x
to get
[HA]
the same value
Preparing a buffer
• If pH = pKa then add twice the number of
moles of weak acid to the base. Eg
• NaOH + HCOOH
HCOONa + H3O+
• X mol 2X mol (initially)
•
X mol
X mol (finally)
• If pH ≠ pKa
• Make a buffer of pH5 by adding CH3COONa
to 1L of 0.1 molL-1 ethanoic acid. Assume no
change in volume, what mass needs to be
added?
pKa CH3COOH = 4.76, M CH3COONa = 82
14.2g must be dissolved in
ethanoic acid and made up to 1L
CH3COOH +H2O
CH3COO- + H3O+
Ka =[CH3COO-][H3O+]
[CH3COOH]
[H3O+] =10-pH and Ka =10-pKa
So 10-pKa=[CH3COO-] x 10-pH
0.1
10-pKa x 0.1 =[CH3COO-]
10-pH
0.174molL-1 =[CH3COO-]
m=n x M = 0.174 x 82= 14.2g
Acid –Base Titrations
• When an unknown [NaOH] is
titrated against 50ml of 0.1M HCl,
the following graph can be plotted.
• What was the initial pH?
• Calculate the [NaOH]
• HCl + NaOH
NaCl + H2O
• n=CV
HCl
NaOH
Mol ratio
1
1
C
0.1
V
0.05
0.1
0.05
n
0.005
0.005
With a strong acid and base
titration, both will fully
dissociate and the equivalence
point (half way up the steepest
part of the graph) will be pH 7
Titration curves
•
•
•
•
•
•
•
•
Shape of graph indicates:
SASB, SAWB, WASB
initial and final pH
Equivalence point (ep) is when acid
and base are in equal concentrations
(not necessarily neutral)
½ volume at ep, pH=pKa, Ka = [H3O+]
Buffer region is 1 pH above or below
pKa
Range of a suitable indicator
Note the ep is neutral, lack of buffer,
long vertical section
Strong Acid/ Strong Base Titration
• Calculate the pH of the solution at the
end of the titration to 3S.F.
• HCl + NaOH
NaCl + H2O
• After 50ml of base has been added the
solution is neutral so this will dilute the
concentration of additional base.
• C=n/V
Volume after
• [OH-] = 0.1 M x 0.05
equivalence
(0.1 + 0.05)
[OH-] = 0.033
pOH = 1.47
Total volume
pH = 12.5
Titrating Weak Acid / Strong Base
• The initial pH is low so [H3O+] = √Ka x [HA]
• CH3COOH + H2O
CH3COO- + H3O+
• Equivalence point – when ALL the weak acid
has reacted with the base
• CH3COOH + NaOH
CH3COONa + H2O
At equivalence the [CH3COO-] is half of the
original [CH3COOH] due to the dilution effect
At half way to equivalence
[CH3COO-]=[CH3COOH]
• Ka = [CH3COO-][H3O+]
[CH3COOH]
After equivalence pH is dependent on the
dilution effect on the [base]
If 0.01molL-1 NaOH is added to 50ml of 0.01molL-1
CH3COOH calculate:
(Ka CH3COOH = 1.74 x 10-5)
3.38
• the initial pH
• pH half way to equivalence 4.76
• pH at equivalence
8.23
• pH at end
11.3
Answers Initial pH
• It is a weak acid so [H3O+] = √Ka x [HA]
• [H3O+] = √ 1.74 x 10-5 x 0.01
• [H3O+] = 4.17 x 10-4
• pH = -log[H3O+]
• pH = -log[4.17 x 10-4]
• pH = 3.38
Think: does this look right, it is a
weak acid and pH is inversely
proportional to [H3O+]
Answers half way to equivalence pH
• Ka = [H3O+] = 1.74 x 10-5
• pH = -log[H3O+]
• pH = -log[1.74 x 10-5]
• pH = 4.76
Answer pH at Equivalence
• All the [CH3COOH] has been neutralised so the number of
mols of CH3COO- = the initial mols of CH3COOH BUT it is in
twice the volume therefore the [CH3COO-] is half of the
original ie 0.005 molL-1
• CH3COOH + NaOH
CH3COONa + H2O
• [H3O+]=√(Ka x Kw/[CH3COO-])
• [H3O+]=√(1.74 x 10-5 x 10-14/0.005)
• [H3O+]= √1.74 x 10-19/0.005
• [H3O+]= √3.48 x 10-17 Think: does this look
right, it is almost neutral
• [H3O+]= 5.89 x 10-9
• pH =
• pH = -log[5.89 x 10-9]
• pH = 8.23
-log[H3O+]
but weak acid and strong
base results in a pH>7 at
equivalence because
CH3COO- acts as a weak
base
Weak conjugate
base
Answer pH at end
• [OH-] = [NaOH] x volume beyond equivalence
total volume
• [OH-] = 0.01 x 25
125
• [OH-] = 0.002
Alternatively
• pOH = -log[OH-]
use Kw = [OH-] [H3O+]
• pOH = 2.69897
• pH =14 - pOH
Think: does this look right, it is a
strong base, diluted
• pH = 11.3
Although 75ml has been
added the first 50ml has
reacted with the acid
50ml of acid and 75ml of
base results in a dilution
Titrating Weak Base Vs Strong Acid
• NH4+ + H2O
NH3 + H3O+
Initial
• [H3O+]=√(Ka x Kw/[NH3])
• [H3O+]=√(5.75 x 10-10 x 10-14/0.1)
• [H3O+]= √5.75 x 10-24/0.1
• [H3O+]= √5.75 x 10-23
• [H3O+]= 7.58 x 10-12
• pH = -log[H3O+]
• pH = -log[7.58 x 10-12]
• pH = 11.1
Half way to equivalence
• Ka = [H3O+] = 5.75 x 10-10
If 0.1molL-1 HCl is added to 40ml of 0.1molL-1 NH3 calculate: (Ka NH4+ = 5.75 x 10-10)
• pH = -log[H3O+]
11.1
• the initial pH
• pH = -log[5.75 x 10-10]
• pH half way to equivalence
9.24
• pH = 9.24
• pH at equivalence
• pH at end
Titrating Weak Base Vs Strong Acid
Equivalence
• NH3 + HCl
NH4+ + Cl• All the [NH3] has been neutralised so the
number of mols of NH4+ = the initial mols
of NH3 BUT it is in twice the volume
therefore the [NH4+] is half of the original
ie 0.05 molL-1
• NH4+ + H2O
NH3
+ H3O+
• NH4+ is a weak acid so [H3O+] = √Ka x [HA]
• [H3O+] = √ 5.75 x 10-10 x 0.05
• [H3O+] = 5.36 x 10-6
• pH = -log[H3O+]
-1 HCl is added to 40ml of 0.1molL-1 NH calculate: (Ka
-10)
If
0.1molL
=
5.75
x
10
3
NH4+
• pH = -log[5.36 x 10-6] • the initial pH
11.1
• pH = 5.28
• pH half way to equivalence
9.24
• pH at equivalence
• pH at end
5.28
Titrating Weak Base Vs Strong Acid
At End
• [H3O+] = [HCl] x volume beyond equivalence
total volume
• [H3O+] = 0.1 x 40
120
• [H3O+] = 0.033
• pH = -log[H3O+]
• pH = 1.48
If 0.1molL-1 HCl is added to 40ml of 0.1molL-1 NH3 calculate: (Ka NH4+ = 5.75 x 10-10)
11.1
• the initial pH
• pH half way to equivalence
9.24
• pH at equivalence
5.28
• pH at end
1.48
Indicators
• Indicators are weak acids, their conjugate base is a
different colour.
• Let HMe represent Methyl red
• HMe + H2O
H3O+ + Me• In acidic conditions the equilibrium shifts left so the
indicator is red.
• In basic conditions H3O+ is neutralised so the
equilibrium shifts right and the indicator is yellow.
• Colour change represents the titration endpoint so an
indicator is chosen which changes colour on the steep
part.
• Pka of indicator ± 1pH of the pH at equivalence
• Strong acid Vs weak base results in a slightly acidic salt
(NH4Cl ), so equivalence point is <7 HMe has a pKa of
5.1 which means it works as an indicator
phenolpthalein (pKa 9.3) does not