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Chapter 20 Ionic Equilibria
20
...
2 pH & pOH
pH of strong acids :

pH = - log10 [H+]
∴ [H+] = 10–pH

pH of strong bases : Input [OH-] into the ionic product of water equation to obtain the [H+]
20
...


pKa = - log10 Ka
Two assumptions are being made
...
The
II
...


ionisation of weak acid is so small that the concentration of HA is very similar
before and after dissociation
...
4 Indicator & Acid-base titrations


End-point: the point at which the colour changes



Equivilence point: Amt of acid = Amt of base (neutralisation)
...
5 Buffer solution


A buffer solution is a solution whose pH does not change significantly (remains
almost unchanged) when a little acid or base is added to it
...




Substitute into the formula

Calculate pH when acid / base is added into a buffer


Write equation, build I
...
E equilibrium table filling only no of MOLES
...


Uses of buffer


Electroplating, Manufacture of dyes, Treatment of leather, pH meters



Controlling pH in blood by few systems:

a)

Hydrogencarbonate ions HCO3CO2 (aq) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)
Add H+, shift LHS, add OH-, shift RHS

b)

Dihydrogenphosphate (H2PO4-) & hydrogenphosphate (HPO42-)

c)

Haemoglobin and plasma proteins

20
...


Solubility product, Ksp - is the product of the concentrations of each ion in a saturated
solution of a sparingly soluble salt at 298 K, raised to the power of their relative
concentrations
...


Calculate solubility product Ksp from its solubility or

ii
...


Predicting precipitation - predict whether the formation of precipitate will occur when
2 solutions are mixed
...




If Q < Ksp, no precipitate will be formed
...


Common Ion effect - reduction in the solubility of a dissolved salt achieved by adding a
solution of a compound which has an ion in common with the dissolved salt
...

Step 1: Find the ion concentration (solubility) of the limiting ion from its given Ksp
Step 2: Assume the common ion concentration comes from the added solution
Step 3: Find the NEW ion concentration of the limiting ion from the Ksp and the common ion
concentration
...

20
...

Kpc

=

[I2 organic layer]
[I2 in aqueous layer]

20
...

Hydroxonium ions

H2O(l) +H2O(l) ⇌ H3O+(aq) + OH- (aq)
acid

Simplified ver:

H2O (l) ⇌ H+ (aq)

base

+ OH- (aq)



Hydroxonium ions are strong acids & hydroxide ions are strong base, they
react quickly to produce water again
...
2 M
-x

+x

+x

0
...
2 pH & pOH


pH is a measure of hydrogen ion concentration in an
aqueous solution
...
00 x 10-7 mol dm-3



Due to the small magnitude of the [H+], they are often
represented in logarithmic form
...
00 X 10-4 M
[H+] = 1
...
00 X 10-8 M
[H+] = 5
...
80 X 10-10 M

Calculate the concentration of hydrogen ions in solutions
having the following pH values:
pH 2
...
70,
pH 11
...
4,
pH 12
...
3 x 10-14
...

pH = ………
So is the water acidic, neutral or basic?






pH of Strong Acids and Strong Bases


Strong acids/Strong bases undergoes complete dissociation in
water
...

 Therefore, the pH of a 0
...

HCl → H+ + Cl0
...
01 + 0
...
00056 mol dm-3 of HNO3
...
0000334 mol dm-3 of H2SO4
...
35
...
003 mol dm-3 of NaOH?



Calculating pH of strong bases
...
05 M
...
00123 mol dm-3 of KOH
...
2 g of NaOH per dm3
...
26
...
45
...
57
...

[H+][OH-] = Kw = 1×10-14 mol2 dm–6
∴ -log [H+] +(-log [OH-]) = -log (1×10-14M)
pH + pOH = 14
...
3 Weak acid, Ka


Weak acids dissociate partially in water
...


α=

amount of molecules ionised in equilibrium
amount of molecules present initially



For strong acids/bases, α = 1



For weak acids/bases
...
Usually the values are
for 298 K
...




Due to the small magnitude of these figures, they are often
represented in logarithmic form
pKa = -log10[Ka]
pH = -log10[H+]



If Ka is very high, it means that the position of equilibrium lies
to the right
...




If Ka is very low, it means that the position of equilibrium lies
to the left
...


Strong
acid

Weak
acid

The ↑Ka, the  pKa, the stronger the acid
...



Calculation of Ka of weak acids
HA (aq) ⇌ H+ (aq) + A- (aq)
[H+] = [A-]
Ka = [A-][H+]
[HA]
Ka = [H+]2
[HA]


If the concentration of the weak acid and Ka of the weak acid
is known, the pH of the solution could be calculated
...


I
...


Eg: NH4+(aq) → H+ + NH3, H2O(l) ⇌ H+(aq) + OH-(aq)
Total [H+] is not [H+] + [H+], but rather [H+]
...


The ionisation of weak acid is so small that the concentration of
HA is very similar before and after dissociation
...
2 M
-x

+x

+x

0
...
2 M

+x

+x

Calculating pH of weak acids
...
Requires the concentration of weak acid [HA]
2
...
The acid dissociation constant, Ka, for methanoic acid is 1
...
In a 0
...
24 X 10-3 mol/dm3
b) the pH; 2
...
2
...
0 x
10-4 mol dm-3
...
01 mol dm-3 solution of the acid, what is the
concentration of H+ (aq) ions and the pH?
 1
...


3
...
0 mol dm-3 solution of a weak monobasic
acid is 4
...
0 x 10-2 mol dm-3
b) 1
...
0 x 10-7 mol dm-3
d) 1
...


For benzoic acid, pKa = 4
...
What concentration of the
acid, in mol dm-3, would have a pH = 5
...


Calculate the Ka value of a 0
...
88
...


Calculate the value of Ka and pKa for the following acids:
i
...
02 mol/dm3 2-aminobenzoic acid, which has a pH of
4
...
0
...
10
iii
...
10 mol/dm3 2-nitrophenol, which has a pH of 4
...
4 Indicator & Acid-base titrations


When performing a titration, indicator is use to detect endpoint, and to show that acid and base are neutralised by each
other, and are mixed in exactly the right proportions
...




An acid-base indicator is a substance which changes colour
over a specific pH range
...




The chemical equilibrium of the indicator when it ionise is as
follow:
HIn ⇌ H+ + In–



Example: Methyl orange indicator



In the methyl orange case, the half-way stage where the mixture
of red and yellow produces an orange colour happens at pH
3
...




At ↓pH (↑[H+]): system shift to left; backward reaction favours;
colour of HIn predominate



Example: Phenolphthalein



The half-way stage happens at pH 9
...
Since a mixture of pink
and colourless is simply a paler pink, this is difficult to detect
with any accuracy!



Indicators don't change colour sharply at one particular pH
(given by their pKind)
...




Indicators must be carefully chosen so that their colour
changes take place within the region



Red cabbage juice contains a mixture of substances whose
color depends on the pH
...
0 (far left) to pH = 11
...
At pH
= 7
...




However, the term neutral point not applicable to all
neutralization process when involving different strength
of acid and base
◦ Strong acid-weak base: product slightly acidic
◦ Strong base-weak acid: product slightly base




Burette – Acid
Conical flask - Alkali

Choosing suitable indicators
I
...


Bromothymol Blue

II
...
(Cation hydrolysis)
 NH4+ + H2O → H3O+ + NH3 , resulting in an acidic solution
 Sharp fall between pH 3
...



Bromothymol Blue

III
...
(Anion hydrolysis)
C6H5COO- + H2O → C6H5COOH + OHSharp fall between pH 7-11
...
No acid-base indicator is suitable
to determine the end point of this reaction
...


Bromothymol Blue

Mid point = [H+] = [OH-]
Acid

Base

pH
range

Mid
Methyl orange
point 4
...
3

Bromothymol
Blue 6
...
6

Phenolphthalein
8
...
5-3
...
5-3
...
5

9

Weak

Weak

Can

Can

Diprotic acid curves


The reaction between NaOH and HOOCCOOH (ethanedioic
acid/oxalic acid) takes place in 2 stages because one of the
hydrogens is easier to remove than the other
...
5 Buffer solution


In many experiments, particularly in biochemistry, need to be
carried out in solutions of constant pH
...




It has to contain substance which will remove any hydrogen
ions or hydroxide ions otherwise the pH will change
...
CH3COOH/ CH3COONa (pH 4
...
25)

Acid buffer solution
Acid buffer consist of weak acid & its conjugate salts
...

 The high conc of both CH3COOH & CH3COONa keeps
them from dissociating to a large extent, ensuring their
concentration in solution does not change significantly
...


CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

At eqm:

0
...
1 M

2
...
1 M
0
...
1 M

At eqm:



The high conc of both CH3COOH & CH3COONa keeps
them from dissociating to a large extent, ensuring their
concentration in solution does not change significantly
...


CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

At eqm:

0
...
1 M

2
...
1 M
0
...
1 M

At eqm:

When more acid (H+) is added,
 the equilibrium shifts to the left and more ethanoic acid is
produced
...





1
...

CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

At eqm:

0
...
1 M

2
...
1 M
0
...
1 M

At eqm:

1
...

Because most of the new OH- are removed, the pH doesn't
increase very much
...


The high conc of both CH3COOH & CH3COONa keeps
them from dissociating to a large extent, ensuring their
concentration in solution does not change significantly
...
1 M – x ≈ 0
...


CH3COONa (aq) → Na+ (aq) + CH3COO- (aq)
0
...
1 M
0
...


When more base (OH-) is added, (2 ways to remove)
Reacts with H+ from the ionisation of the ethanoic

acid to form H2O
...

 For example [HA] & [A-] of equal molar
...



At eqm:


HA (aq)
⇌ H+ (aq) +
0
...
1 M

A- (aq)
0
...
Hence max buffering
capacity is when pH = pKa
Ka = [H+] [A-]
[HA]
Ka = [H+]
-log10 Ka = -log10 [H+]
pKa = pH

Alkaline buffer solution
Alkaline buffer consist of weak base & its conjugate acids
...

 The high conc of both NH3 and NH4+ keeps them from
dissociating to a large extent, ensuring their concentration in
solution does not change significantly
...
1 M – x ≈ 0
...

NH4Cl (aq)
→ Cl– (aq) + NH4+
At eqm:
0
...
1 M
0
...




The high conc of both NH3 and NH4+ keeps them from
dissociating to a large extent, ensuring their concentration in
solution does not change significantly
...
Reacts with NH3
 NH3 (aq) + H+ ⇌ NH4+ (aq)
 Because most of the new H+ are removed, the pH doesn't
decrease very much
...


Reacts with OH- from the ionisation of NH3 to form H2O
...




When more base (OH-) is added,
the equilibrium shifts to the left and more NH3 is
produced
...


At eqm:
2
...
1 M – x ≈ 0
...
1 M
0
...
1 M

1st equation used to maintain the original [OH–] therefore
remaining the pH
...
10 mol dm3 of ethanoic acid and 0
...

Ka for ethanoic acid is 1
...


*Assume the ethanoate ions comes solely from sodium
ethanoate
...


A buffer is prepared by dissolving 0
...
60 mol dm-3 aqueous CH3COOH
...
85 x 10-5 mol dm-3
Calculate the pH of the buffer
...


How many moles of CH3COONa must be added to 2
...
200 mol dm-3 ethanoic acid to produce a buffer
solution of pH 4
...
85 x 10-5 mol dm-3

Calculate the pH of 250 cm3 solution containing 3
...
06 g HCOONa
...
8 x 10-4 mol dm-3
3
...


Calculate the proportion of ethanoic acid and sodium
ethanoate to get a buffer solution with a pH of 4
...
Ka for
ethanoic acid is 1
...


...


What is the mass of NaOH which must be added to 1
...
20 mol/dm3 H3BO3 to make a buffer solution of pH 9
...
3 X 10-10]

Calculate pH when acid/base is
added into a buffer
1
...
10 mol/dm3 KOH(s) is added
...
8 X 10-4]
a) Solution mixture of 0
...
20 mol/dm3
HCOOK
b) Solution mixture of 1
...
00 mol/dm3
HCOOK

Step 1: Write equation,
 Step 2: build I
...
E equilibrium table filling only no of MOLES
...



Calculate the pH of a mixture containing 0
...
25 mol dm-3 sodium ethanoate
...
8 X 10-5 mol dm-3)



Longman pre-U text STPM chemistry pg167 Ex 11
...


Calculate the change in pH when 10 cm3 of 1
...




Calculate the change in pH when 10 cm3 of 1
...


3 a) What is meant by the term "buffer solution"?
b) Calculate the pH of a buffer solution which contains the weak monoprotic
acid, propanoic acid (CH3CH2COOH), in concentration 0
...
05 moldm-3
...
26 x 10-5 moldm-3
...

d) Calculate the pH of the solution after 0
...

AQA

e) Calculate the pH of the solution after 0
...

f) Calculate the pH after 0
...

g) Comment on your answers to (d) and (f)
...
35-7
...
4, it could be fatal
...
35 – acidosis
If the pH of the blood is pH > 7
...

 CO2 is produced as a result of aerobic respiration, which
then dissolves in water to give HCO3-
...



Carbonic anhydrase
CO2 (aq) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)
Addition of acid (H+) will shift eqm to LHS; whereas addition
of base will remove H+ and and shift eqm to RHS
...



b) Phosphate buffer system
H2PO4- ⇌ HPO42- +
H+
Weak acid
Conjugate base / weak base


Addition of acid (H+) will shift eqm to LHS



Addition of base will be neutralised by H2PO4-
...


Exercise
I
...
The
concentration of the acid is 0
...
What must
be the concentration of the sodium salt, so that the
pH of the solution is equal to the pKa of the acid HA?

II
...
20
...
2 mol
dm-3 HA, producing a solution with a pH = 3
...
How
many moles of NaA were added to the solution?

20
...
The eqm lies far to the left
...




It can be expressed in terms of mol dm-3 or g dm-3
...


Solubility product, Ksp




Solubility product, Ksp is the product of the concentrations of
each ion in a saturated solution of a sparingly soluble salt at
298 K, raised to the power of their relative concentrations
...
Calculate solubility product Ksp from its solubility or
ii
...





Example 1:
The solubility of calcium sulfate, CaSO4 is found
experimentally to be 4
...
Calculate the value
of Ksp for calcium sulfate
...
0168 g dm-3
CaF2
...
(Ksp for Cu(OH)2 = 2
...
(Ksp for Cu(OH)2 = 2
...



Calculate the solubility of silver carbonate Ag2CO3
...
3 X 10-12 mol3 dm-9
...


A
...


C
...


The solubility of cadmium phosphate, Cd3(PO4)2 is x mol dm-3 at
25 °C
...


The solubility of lead chromate, PbCrO4 is 1
...

Calculate the value of Ksp for lead chromate
...


The solubility of manganese carbonate, MnCO3 is 4
...
Calculate the value of Ksp for manganese carbonate
...


The solubility of M2X3 (molar mass 288g/mol) is 3
...
Calculate the
value of Ksp for M2X3
...


Calculate the solubility (in mol dm–3 ) for the following ionic compound:

a)

Silver(I) chloride, AgCl; Ksp = 1
...
3×10–9

c)

Cobalt(II) sulfide, CoS; Ksp = 4
...
0×10–14 ;

6
...
8
...


1
...


1
...


1
...


1
...




The formation of a precipitate is highly dependent on the
concentration of the ions
...
No precipitate would be formed
...




When aqueous NaCl is added to a solution of AgCl, will a
precipitate of AgCl be formed?



To predict whether a precipitate would be formed or not, we
have to calculate the solubility quotient, Q of the mixture,
and compare it with the given Ksp
...

 If Q < Ksp, no precipitate will be formed
...





Example:
25
...
0×10–3 mol dm–3 potassium chromate are mixed
with 75
...
25×10–4 mol dm–3 lead(II) nitrate
...
8×10–14)
Step 1: write ionic equation and derived the expression
Step 2: Calculate the NEW ion concentration (total new volume)
Step 3: Calculate Q
Step 4: Compare Q with Ksp
K2CrO4(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbCrO4(s)
Pb2+(aq) + CrO42– (aq) → PbCrO4(s)
PbCrO4(s) ⇌ Pb2+(aq) + CrO42– (aq)
Ksp = [Pb2+][CrO42–]





Will a precipitate form if equal volumes of solutions of 1
...
00 X 10-5 mol dm-3 BaCl2 are
mixed? Given Ksp BaCO3 = 5
...
0 cm3 of 4
...
0 cm3 of 8
...
Will a precipitate of barium sulfate form? (Ksp of barium
sulfate is 1
...
0 cm3 of 0
...
0 dm3 of
0
...
0×10–6)

3)

Will precipitation occur if equal volumes of solutions of 1
...
25 mol dm–3 AgNO3 are mixed? (Ksp of
Ag2CrO4 is 1
...
(Ksp of Zn(OH)2 is 1
...
01 mol dm–3 ZnCl2 + 0
...
01 mol dm–3 KOH
b) 200 cm3 0
...
10 cm3 0
...




The common ion effect is the reduction in the solubility of a
dissolved salt achieved by adding a solution of a compound
which has an ion in common with the dissolved salt
...


In a saturated solution of AgCl,
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq);
Ksp = [Ag+] [Cl-]
 What if you add NaCl?
 The equilibrium shifts to the left and AgCl will precipitate
...
75×10–
5
...
1 mol/dm3 NaCl
...
Therefore the Pb2+ ion
is the limiting factor
...


Step 1: Find the ion concentration (solubility) of the limiting ion from its given
Ksp
Step 2: Assume the common ion concentration comes from the added solution
Step 3: Find the NEW ion concentration of the limiting ion from the Ksp and
the common ion concentration
...

http://www
...
c
o
...
html

Step 1: Find the ion concentration (solubility) of the limiting ion from its given
Ksp
Step 2: Assume the common ion concentration comes from the added solution
Step 3: Find the NEW ion concentration of the limiting ion from the Ksp and
the common ion concentration
...


http://www
...
c
o
...
html

Step 1: Find the ion concentration (solubility) of the limiting ion from its given Ksp
Step 2: Assume the common ion concentration comes from the added solution
Step 3: Find the NEW ion concentration of the limiting ion from the Ksp and the
common ion concentration
...








Exercise 1
Calculate the solubility of silver chloride in a 6
...
(Ksp of AgCl is 1
...

We have to assume the Ag+ is entirely from AgNO3 = 6
...
2 x 10-7
mol2 dm-6
...

b) Work out the concentration of the dissolved strontium ions in mol
dm-3
...

d) Suppose the concentration of the sulphate ions in the mixture was
0
...


Exercise 3
 Calculate the mass of PbBr2 precipitated when 200 cm3 of
1
...
0 mol dm-3
NaBr
...
0 X 10-5 mol3 dm-9
...
7 Partition coefficient, Kpc


Separation of components of sample mixture occurs because
of partition
...


Example:
 Solute: Iodine
 Solvent: water and CCl4 mixture




Does iodine dissolve well in water or in hexane?



iodine will dissolve in both solvent



equilibrium is achieve when the mixture is left to settle; at
this point iodine molecules travel from CCl4 layer to water
layer at the same rate as molecules travel from water layer
to the CCl4 layer
I2(aq) ⇌ I2(CCl4)

I2(aq) ⇌ I2(CCl4)


The value for this equilibrium constant at a specific
temperature can be calculated and it is call partition
coefficient, Kpc
...




Kpc is the ratio of the concentrations of a solute in two
immiscible or slightly miscible liquids when it is in
equilibrium
...

◦ Ion-ion attraction
◦ Ion-dipole interaction
◦ Hydrogen bonding
◦ Van der Waals



Generally, the interaction of the substances is based on their
polarity
...

◦ Non-polar solvents are more likely to be able to dissolve those
solutes whose molecules are only attracted to each other by van
der Waals’ forces
...
00

g of iodine solids are shaken with a mixture of 100 cm3 of
water and 40 cm3 of hexane and left in a separating funnel for
equilibrium to be established
...
15 mol dm–3 of sodium
thiosulfate
...
05 cm3 of sodium
thiosulfate
...

b)Write the balance equation for this reaction
...


Question 2
When 100 cm3 of an aqueous solution containing 2
...
6 g of the dye had been extracted into the
hexane
...

b) Calculate the minimum volume of hexane needed to reduce
the amount of X in the aqueous layer to less than 0
...



Question 3
500 cm3 of water contains 6
...
The solution is shaken with 100 cm3 of
trichloromethane to extract the caffeine
...

[Partition coefficient for caffeine distributed between
trichloromethane and water = 10]


Question 4


1
...
Calculate how much mass of X was extracted into the
ether layer
...




1
...

Calculate how much mass of X was extracted into the ether
layer
...




How much of X would be extracted in total by using the 2
times of 5 cm3 ether for each extraction instead of 1
extraction by using 10 cm3 ether?



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