AS 91392 Chemistry 36 5 external credits Equilibrium Systems N 2 g 3H 2 g 2NH 3 g H What is the formula for the equilibrium constant Kc What can be said about the concentrations if Kc is very small ID: 755919
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Slide1
Demonstrate understanding of equilibrium principles in aqueous Systems
A.S. 91392
Chemistry 3.6
5
external creditsSlide2
Equilibrium Systems
N
2
(g) + 3H
2
(g) 2NH
3(g) - HWhat is the formula for the equilibrium constant Kc?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 leftWhat 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
Don’t include solids or the solvent in KcSlide3
Le Chatelier
Principle
H
2(g)
+ I2(g) 2HI(g)Kc = [HI]2 [H2][
I2]At equilibrium the concentration of species doesn’t changeExplain the changes [H2
] is increased so forward reaction is favoured to use up H2 and re-establish Kc, therefore the [I2] decreases and [HI] increasesWhat would the graph look like if the pressure was increased?What if a catalyst was added?
ESA pg216 Q5Slide4
Solubility Definitions
Solvent – a liquid that dissolves something
Solute – gets dissolved (by the solvent)
Saturated – no more solute can dissolveSolubility – how much solute will dissolve in a volume (molL
-1 or gL-1)Slide5
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+ present2H2O H3O+ +
OH- If the solution is acidic then [H3O+]>[
OH-]Kw = [H3O+ ][OH-] = 10-14 therefore low level of dissociationIn neutral solutions [H3O+ ] = [OH-] = 10-7These 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, ethanolSlide6
Calculating Ks
At the point of saturation an equilibrium is formed.
Adding Ag
2
CrO4 to water will create an equilibrium. Write the equation.Ag2CrO4(s) 2Ag+(aq) + CrO
42-(aq)Ag2CrO4 has a solubility of 6.5 x 10
-5 molL-1What is the concentration of CrO42-?What is the concentration of Ag+?What is the value of Ks?Ks = [Ag+]2[CrO42-] thereforeKs = (1.3 x 10-4)2 x (6.5 x10-5)
Ks = 1.1 x 10-12This is a special type of equilibrium constant
The bigger the value of Ks, the more likely the substance will dissolve
6.5 x 10
-5
molL
-1
1.3
x
10
-4
molL
-1Slide7
Special Equilibrium Constants
K
s
= [A] x [B] for AB salt (
NaCl
), giving
s =
K
s
= [A]
2
[B] for A
2
B salt (Na
2
S), giving
s =
K
s
=
[A][B]
2
for AB
2
salt (MgCl
2
)
s = solubility (molL
-1
)
Memorise these!
Plus Kw,
Ka
and Kb
Use these to calculate solubility if you are given KsSlide8
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-1B) Zn(OH)2 where Ks = 2.0 x 10 -171.71 x 10 -6 molL
-1Don’t forget the equilibrium arrowDon’t include solids or solvents as their concentration doesn’t varySlide9
Does a Precipitate form?
Known as the
Ionic Product
Use the Ks expressionIf IP>Ks a precipitate will form
Ks(AgCl) = 1.6 x 10 -10If 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 19ESlide10
Ionic Product calculations
Will a precipitate form if we mix 50ml of 0.01molL
-1
AgNO
3(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 expressionKs = [Ag+][Cl-]Calculate I.P. by substituting values for the concentrations of ionsIf you increase the volume you will decrease the concentrationUse this formula to work out the new concentrations[new] = original vol x [original] Total
volE.g. for Ag+ new concentration is:(50/100) x 0.01 = 0.005molL-1Comment 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 AgNO
3 were 10ml and NaCl were 40ml?
I.P. = 5.0 x 10-6Slide11
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 -10So the [Ag+] = 1.56 x 10 -9 molL-1which will be equivalent to the concentration of AgCladding more Cl- to the RHS shifts the equilibrium to the left
To maintain Ks [Ag
+] must decrease
A common ion will
reduce
the solubility of a substanceSlide12
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-, SCN-The number of ligands is twice the charge on the
cation. E.g. Cu2+ forms [Cu(NH3)4]2+Zn(OH)2 Zn2+ + 2OH-According 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…Zn(OH)2 + 2OH
- [Zn(OH)4]2- When excess OH- is added the zincate ion forms which
removes Zn2+ from the solution thus shifting the equilibrium forwards
to produce more Zn2+ and maintain Ks, this causes more Zn(OH)
2
to dissolve
.
Can you predict what might happen if the reaction took place in acidic conditions?
ESA
pg
227 activity 19C and D
Formation of a complex ion
increases
the solubility of a substanceSlide13
Solubility in acidic conditions
If an excess of
H
3O+
Zn(OH)2 Zn2+ + 2OH-2H2
O 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.Slide14
Solubility in basic conditions
Calculate the solubility of Fe(OH)
2
in pH10 solution. Ks(Fe(OH)2) = 8.0 x 10
-16Fe(OH)2(s) Fe2+ + 2OH-Ks=[Fe2+
][OH-]2We need to work out[OH-][
H3O+] = 10-pH = 1 x10-10molL-1Kw =1x10-14=[H3O+][OH-]So [OH-] =1x10-4molL-1Substitute this into the Ks expression
8.0x10-16 =[Fe2+](1x10-4)2 [Fe2+
] = 8x10-8molL-1This is equal to the solubility of Fe(OH)2
due to stoichiometry.Slide15
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
-12Mg(OH)2 Mg2+
+ 2OH-Ks= [Mg2+ ][OH-] 27.1x10-12 > [Mg2+ ][OH-]27.1x10-12 > 0.555 x [OH-] 2[OH-] > 3.58x10-6molL-1
pOH = -log[OH-]pOH = 5.45Minimum pH = 14 – 5.45 = 8.55Slide16
Acids and Bases
You should know:
pH = -log[H
3O+]
Kw = [H3O+] x [OH-] = 10 -14[H3O
+] = 10 -pHpOH + pH = 14pOH = -log[OH
-][OH-] = 10 -pOHStrong 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+]) Slide17
Weak Acids
CH
3
COOH + H
2O CH3COO- + H
3O+Work out the equation for the equilibrium constant for the dissociation of weak acids (Ka)Ka is the
acid dissociation constant or acidity constantKa =[product] [reactant][water] remains very high and can be thought of as constant therefore… Ka = [H3
O+
][A-
]
[HA]
So if
Ka
is small the level of dissociation is low, the [H
3
O
+
] is low so the
pKa
and pH are high
K
a
= 10
-pK
a
pK
a
= - log
K
a
[H
3
O
+
] =
Also as [
H
3O+] = 10-pH
then…
Can you rearrange this to find [H3O+
] for a weak acid?Clue: use stoichiometry
MEMORISESlide18
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 pHHOCl + H2O H3
O+ + OCl-Ka =[H3O+][OCl-] [HOCl]Ka=10-pKaKa = 2.95 x 10-8Ka
is very small, not much dissociation so assume [HOCl]=0.1molL-1 [H3O+
] =√(Ka x[HOCl] )
[H3O+] = 5.43 x 10
-5
pH =-log[
H
3
O
+
]
pH = 4.27 therefore
HOCl
is a weak acid
Note:
pKa
is high,
Ka
is small, partial dissociation means low [
H
3
O
+
] and so high pH
Slide19
Weak Bases
NH
3
+ H
2O 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 waterNH4+ + H2O NH3 + H3O+Therefore Ka = [NH3][
H3O+] [NH4+ ]
But due to stoichiometry [NH4+ ] = [OH-] and [OH-
]=kw/[H3O+]So
Ka
=[
NH
3
][
H
3
O
+
]
Kw/[
H
3
O
+
]
Which rearranges to
Ka
= [
NH
3
][
H
3
O
+
]
2
Kw[H3O+]=√(Ka xKw)/[NH3]The very low value for Ka means we can assume [NH3] = original concentrationCan you prove the pH of the solution is 11.1?An alternative method is to use the base dissociation constant (Kb) and pOHKb = Kw/Ka[OH-] = √(Kb x [NH3])Slide20
Species in Solution
HCl
(g) and CH
3COOH(l) are both soluble in waterWrite separate equations to demonstrate this
HCl + H2O H3O+ + Cl-
CH3COOH + H2O CH3COO
- + H3O+ What are the relative concentrations of ALL the species present in each equation?H2O >> H3O+ >Cl->OH- H2O >>CH3COOH> H3O+ > CH3
COO- >OH-Which solution would have the greater conductivity and why?The partial dissociation of water adds a small amount of H
3O+ and OH- to any aqueous system
Full dissociation means a high concentration of ions and greater conductivity in the
HCl
solutionSlide21
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) + H
2O (l) NH4+ + OH-
ExceptionsCH3CH2OH dissolves but no ions formCH3COOH and NH3 only partially dissociateCa(OH)2 limited solubility means weak electrolyteSlide22
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+ + Cl-n=m/M so [NaCl]=[Na+]=[Cl-]=0.1molL-1[Na+]=[Cl-]>[H3O+]=[OH-][H
3O+] = 10-7 and pH =7 so solution is neutralCH3COONa CH3COO
- + Na+CH3COO- + H
2O CH3COOH + OH-[H
2
O
]>>[
Na
+
]>[
CH
3
COO
-
]>[
OH
-
]>[CH
3
COOH]>[
H
3
O
+
]
[H
3
O
+
]
<
10
-7
and pH
>7 so solution is basicKb = Kw/Ka and [OH-] = √(Kb x [CH3COO-])[OH-] = 7.58 x 10-6 therefore pH = 8.88pH measures the [H3O+]Kw= 10-14=[H3O+][OH-] so if [OH-] increases then [H3O+] decreasesH2OProve, using Ka that the pH is 8.88Therefore salts of weak acids are basic.What about salts of weak bases?Slide23
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-HA + OH- H2O + A-A- + H3O+ H2O + HAAs dissociation of weak acid / bases is low additional conjugate is added to ensure approximately equivalent concentrationsThe buffer zone is the area where the pH does not change significantlySlide24
Common Buffers
Methanoic
acid/
methanoateAmmonium ion/ ammonia
Ethanoic acid/ ethanoateHydrogen carbonate/ carbonateAdding acid (the conjugate base reacts) NH3 + H3O
+ NH4+ + H2OA
- + H3O+ HA + H2OAdding base(the conjugate acid reacts) NH4+ + OH- NH3 + H2O HA + OH- A- +
H2O Slide25
Buffer calculations
HA + H
2
O A-
+ H3O+HA = acidic species, A- = basic speciesSo using the Ka expression we can say
[H3O+
]= Ka x OrpH = pKa + logIf [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.
[HA]
[A-]
[A
-
]
[HA]
Calculate the pH of a buffer made from 0.100mol of NH
4
Cl dissolved in 100ml of 0.100molL
-1
NH
3(
aq
)
Ka
(NH4+)
= 5.75 x
10
-10
[NH
4
+
]= 0.1
x
10 = 1molL
-1
[H
3
O
+
]=
5.75 x 10
-10 x 1/0.1 = 5.75 x 10-9pH = -log(5.75
x 10-9) = 8.24Slide26
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 +H2
O H3O+ + F-nNaF =m/M = 2.23/42 = 0.05311:1 mol ratio therefore 0.0531 mol F- added to 0.15L (n/v =c) so [F-] = 0.354molL-1HF partially dissociates so assume [HF] = 0.05molL-1 pH =
pKa + logpH = 3.17 + log(0.354/0.05)pH =4.02
[A-][HA]Slide27
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)pH =
pKa + log[A-]= methanoate[HA]=
methanoic acidHCOONa 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-1Dilution of [HA] makes the new concentration= (50/100) x 0.2 = 0.1 molL-1pH = pKa + logpH = 4.44
Now use [H3O+]= Ka x to get the same value
[A-][HA]
[A
-
]
[HA]
[HA]
[A-]Slide28
Preparing a buffer
If pH =
pKa
then add twice the number of moles of weak acid to the base.
EgNaOH + HCOOH HCOONa + H3O+ X mol
2X mol (initially) X mol X mol
(finally)If pH ≠ pKaMake 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 = 82CH3COOH +H2O CH3COO- + H3O+
Ka =[CH3COO-][H3O+] [CH3COOH]
[H3O+] =10-pH and Ka
=10-pKaSo 10-pKa=[CH3COO-
] x 10
-pH
0.1
10
-pKa
x 0.1 =[
CH
3
COO-]
10
-pH
0.174molL
-1
=[
CH
3
COO-]
m
=n x M = 0.174 x 82= 14.2g
14.2g must be dissolved in ethanoic acid and made up to 1LSlide29
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 + H2
On=CV
HClNaOHMol ratio11C
0.1V0.05
n
0.0050.005
0.05
0.1
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 7Slide30
Titration curves
Shape
of graph
indicates: SASB, SAWB, WASB
initial and final pHEquivalence point (ep) is when acid and base are in equal concentrations (not necessarily neutral)½ volume at ep, pH=pKa, Ka = [H3
O+]Buffer region is 1 pH above or below pKaRange of a suitable indicatorNote the ep is neutral, lack of buffer, long vertical sectionSlide31
Strong Acid/ Strong Base Titration
Calculate the pH of the solution at the end of the titration to 3S.F.
HCl
+
NaOH NaCl + H2OAfter 50ml of base has been added the solution is neutral so this will dilute the concentration of additional base.C=n/V[OH-] = 0.1 M x 0.05
(0.1 + 0.05)[OH-] = 0.033pOH = 1.47pH = 12.5
Volume after equivalence
Total volumeSlide32
Percentage Purity
0
.5g of impure
LiOH was added to water and the solution made upto 100ml. On average 17.2ml
of 0.25molL-1 HCl was needed to neutralise the 25ml samples. Calculate the percentage purity of the LiOH. MLiOH=23.9gmol
-1 LiOH + HCl LiCl + H2O
HClC0.25V0.0172n
0.0043
1:1 mole ratio, so at neutralisation 25ml LiOH had 0.0043
molsTherefore the original 100ml had 4x 0.0043= 0.0172 mol
n x M = m
0.0172 x 23.9 = 0.411g
Percentage composition
0.411/0.5 x100 = 82%Slide33
Apple Juice
20ml apple juice was titrated against 0.1molL
-1
NaOH and the average was 10.36ml. Assuming all apple juice is citric acid and that 1mole reacts with 3 moles of NaOH. Calculate the concentration of citric acid in gL-1. Apple juice must contain between 0.3 and 0.8g per 100ml. Is this considered apple juice or apple drink?
M(citric acid)=192gmol-1
Moles NaOH = 1.036x10-3
molMoles citric = 3x less
ie 3.453x10
-4
Conc
citric = n/v
= 0.01727molL
-1
so 0.01727
x 192 = 3.315g/L therefore 0.3315g per 100ml it is juice JUST
!Slide34
Titrating Weak Acid / Strong Base
The initial pH is low so [H
3
O
+] = √Ka x [HA]CH3COOH + H2O CH3COO
- + H3O+Equivalence point – when ALL the weak acid has reacted with the base
CH3COOH + NaOH CH3COONa + H2O At half way to equivalence [CH3COO-]=[CH3COOH] Ka = [CH3COO-][H3O+]
[CH3COOH]At equivalence the [CH3COO-] is half of the original [CH3COOH] due to the dilution effect
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
CH
3
COOH
calculate: (
Ka
CH3COOH
=
1.74 x 10
-5
)
the initial pH
pH half way to equivalence
pH at equivalence
pH at end
3.38
4.76
8.23
11.3Slide35
Answers Initial pH
It is a weak acid so
[H
3O+] = √Ka x [HA]
[H3O+] = √ 1.74 x 10-5 x 0.01[H3
O+] = 4.17 x 10-4pH = -log[H3O+]
pH = -log[4.17 x 10-4] pH = 3.38Think: does this look right, it is a weak acid and pH is inversely proportional to [H3O+] Slide36
Answers half way to equivalence pH
Ka
=
[H3O+] =
1.74 x 10-5pH = -log[H3O+] pH = -log[1.74 x 10-5
] pH = 4.76Slide37
Answer pH at Equivalence
All the [
CH
3COOH]
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
-1CH3COOH + NaOH CH3COONa + H2O[H3O+]=√(Ka x Kw/[CH3
COO-])[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
[H
3
O
+
]=
5.89 x 10
-9
pH = -log[H
3
O
+
]
pH =
-log[5.89
x
10
-9
]
pH =
8.23
Think: does this look right, it is almost neutral but weak acid and strong base results in a pH>7 at equivalence because CH
3
COO
-
acts as a weak base
Weak conjugate baseSlide38
Answer pH at end
[OH-] = [
NaOH
] x volume beyond equivalence total volume[OH-] = 0.01 x 25
125[OH-] = 0.002pOH = -log[OH-]pOH = 2.69897pH =14 - pOH
pH = 11.3Although 75ml has been added the first 50ml has reacted with the acid
50ml of acid and 75ml of base results in a dilution
Alternatively use Kw = [OH-]
[H3O+]
Think: does this look right, it is a strong base, diluted Slide39
Titrating Weak Base Vs Strong Acid
NH
3
+ HCl
NH4+ + Cl -NH4+ + H
2O 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[H
3O+]= √5.75 x 10-23
[H3O+]= 7.58 x 10
-12
pH = -log[H
3
O
+
]
pH = -
log[7.58
x
10
-12
]
pH =
11.1
Half way to equivalence
Ka
= [H
3
O
+
] =
5.75
x
10
-10
pH = -log[H
3
O
+
]
pH = -log[5.75 x 10-10] pH = 9.24If 0.1molL-1 HCl is added to 40ml of 0.1molL-1 NH3 calculate: (Ka NH4+ = 5.75 x 10-10)the initial pHpH half way to equivalencepH at equivalence pH at end11.19.24Slide40
Titrating Weak Base Vs Strong Acid
Equivalence
NH
3
+ 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 [NH
4+] is half of the original ie 0.05 molL-1NH4+ + H2O NH3 + H3O+NH4+ is a weak acid so [H3O
+] = √Ka x [HA][H3O+] = √ 5.75
x 10-10 x 0.05[H3
O+] = 5.36 x 10-6
pH = -log[H
3
O
+
]
pH = -
log[
5.36 x
10
-6
]
pH =
5.28
If 0.1molL
-1
HCl
is added to 40ml of 0.1molL
-1
NH
3
calculate: (
Ka
NH4+
=
5.75 x 10
-10
)
the initial pHpH half way to equivalencepH at equivalence pH at end11.19.245.28Slide41
Titrating Weak Base Vs Strong Acid
At End
[
H3
O+] = [HCl] x volume beyond equivalence total volume[H3O+
] = 0.1 x 40 120[H
3O+] = 0.033pH = -log[H3O+] pH = 1.48If 0.1molL-1 HCl is added to 40ml of 0.1molL-1 NH3 calculate: (
Ka NH4+ = 5.75 x 10-10)the initial pH
pH half way to equivalencepH at equivalence
pH at end11.1
9.24
5.28
1.48Slide42
Indicators
Indicators are weak acids, their conjugate base is a different colour.
Let
HMe
represent Methyl redHMe + 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 equivalenceStrong 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