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Oceans
Explain the factors determining the relative solubility of a solute in aqueous and non-aqueous solvents and
explain hydration of ions

Before an ionic solute can dissolve in a solvent, the intermolecular bonds between solute
molecules must be broken – this is done by taking in energy to overcome charges between
electrostatically charged ions (an endothermic process)
...

(ΔHLE) Lattice enthalpy is the enthalpy change where 1 mole of solid is formed from its
constituent ions
...

Therefore -ΔHLE is the opposite of this; where 1 mole of solid is broken up into its
constituent ions
...

The lattice enthalpy is dependent on two features of the ions it contains:



Ionic charge
Ionic radii

The greater the ionic charge, the greater the attraction between ions in the lattice – and
therefore more energy needed to overcome these bonds
...
This makes them less likely to dissolve, as more energy is required to
break apart the lattice
...

In an aqueous solvent of water, the difference in electro-negativities and bent shape of the
molecule makes water molecules polar
...
When a solute first touches the water, ions are pulled
from the lattice into solution by the water molecules
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These ions are therefore hydrated
...


The number of water molecules which are able to form ion-dipole bonds is again,
dependent on ionic charge and radii
...

However it is also important to take into account attractions between water molecules
...
This process, like the breaking of the
intermolecular bonds in a lattice, also requires energy input (is endothermic)
...

This is where the ions have been separated from the lattice, and are forming ion-dipole
bonds with water molecules
...

If the aqueous solvent is not water, the value is known as enthalpy of solvation rather than
hydration - ΔHsolv
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The arrows for lattice enthalpy and hydration will be pointing downwards, because they
are exothermic
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The gaseous ions always have the highest energy, so are written at the top
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Describe acid as proton donators and bases as proton acceptors, using the terms conjugate base/acids

Acids are able to dissociate in solution to donate a proton, whilst bases accept a proton
...
g
...

Some substances can act as either an acid or a base depending on the environment – these
are known as amphoteric
...

It makes sense that a strong acid has a weak conjugate base, as it is far more stable after
donating a proton than receiving one
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Explain and use the terms weak acid, acidity constant Ka, and pKa

A weak acid will only partially dissociate in water, as it has a lower tendency to donate
protons
...
This is known as the acidity constant Ka:
[𝑯+ ] × [𝑨− ]
𝑲𝒂 =
[𝑯𝑨]
The greater the Ka value, the stronger the acid (as the position of equilibrium is shifted to
the right, meaning the acid has dissociated more)
...
Similar to the pH scale, the lower the pKa value to stronger the acid
...

pH = -log (H+ concentration)

pH is simply a logarithmic scale, where changing H+ by a factor of 10 changes pH by a factor
of 1
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Working out pH of a strong acid
One of the assumptions of strong acids is that the acid fully dissociates into its H + and Aions, therefore the concentration of H+ is the same as that of HA
...

𝐾𝑎 =

[𝐻 + ] × [𝐴− ]
[𝐻𝐴]

We may know the Ka value, and can make the assumption that the concentration of HA at
equilibrium is the same as that put in, giving us two values in the equation:
1
...
1]

We can use the second assumption, that A- and H+ are the same concentration to convert
the top of the equation into:
1
...
1]

We then rearrange to isolate [H+]2:
1
...
1 = [𝐻 + ]2 = 1
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7 × 1𝑂−6 = 1
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3 × 10−3 = 𝒑𝑯 𝟐
...
A reversible
reaction can therefore produce an equilibrium constant:
𝐾𝑎 =

[𝐻 + ] × [𝑂𝐻 − ]
𝐻2 𝑂

Water is present in excess, so is removed from the equation to produce:

Kw = [H+] [OH-]
Kw is the ionic product of water - the extent to which water ionises
...


As breaking bonds (ionisation of water) is endothermic, increasing the temperature shifts
the position of equilibrium to the right, increasing the concentration of H+ and therefore
decreasing the pH
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Dissolving a strong base in water will cause
it to completely dissociate to give the same concentration of OH- ions
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1 moldm-3, then OH- concentration will be 0
...
1 (𝑂𝐻 − )
As pH is a measure of H+ ions, we need to rearrange to make H+ the subject so we can work
this out:
𝐻+ =

1 × 10−14
= 1 × 10−13
0
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A buffer solution can be made up of either:



A weak acid (donates H+) and its salt (accepts H+)
A weak base (accepts H+) and its salt (provides a proton donating substance)

We can use the equation for a weak acid to explain buffer action:

𝑯𝑨 ⇄ 𝑯+ + 𝑨−
HA is the weak acid, which acts as a reserve of H+ whilst A- is the salt, acting as an H+ sink
...
All the A- ions come the salt – HA is a weak acid, therefore the amount it dissociates
to produce H+ and A- will be negligible
2
...

Finding pH from buffers

As we have a weak acid in the solution of a buffer, we can use the acidity constant equation:
[𝐻 + ] × [𝐴− ]
𝐾𝑎 =
[𝐻𝐴]
If we think about our assumptions, A- will come purely from the salt, so can be represented
by the salt concentration
...
This creates the equation:
𝐾𝑎 = [𝐻 + ] ×

[𝑠𝑎𝑙𝑡]
[𝑎𝑐𝑖𝑑]

The value of H+ is affected by the value of Ka, and the ratio of salt : acid
...

With a Ka value of 1
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1dm-3 ethanoic acid with 0
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7 × 10−5 = [𝐻 + ] ×

0
...
1

The equation is rearranged to make H+ the subject by switching salt and acid around, and
switching H+ and Ka:
𝐻 + = 1
...
1
= 8
...
2

We then input the H+ concentration into the pH equation:
− log (8
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𝟎𝟕
Explain hydrogen bonding in water and the unusual properties that arise from it

There are 3 conditions which must be met for hydrogen bonding to occur:




A hydrogen must be covalently bonded to a highly electronegative element (O, N, F)
to produce a large dipole
The small hydrogen atom must be close to an O, N or F on another nearby molecule
A lone pair of electrons on the O, N or F for hydrogen to line up with

When H is bonded to a highly electronegative substance, the H develops a strong partial
positive charge whilst the other atom develops a strong partial negative charge
...
Similarly, the partially negative substance is
attracted to the partial positive charge on the hydrogen
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Of the group 6 halides, HF
has a substantially higher boiling point; this is because whilst the other halides can form DP
-DP bonds, HF can form hydrogen bonds
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Water, on the other hand, has an even higher boiling point
...

It is important to remember that hydrogen bonding is the strongest
form of intermolecular bond, but is weaker than atomic bonding e
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covalent
...

This is relatively similar to boiling point, which is
shown on the graph
...

The boiling point demonstrated in group 4,
starting with methane, is as expected –as the size
of the molecule increases so does the strength of the ID-ID bonds, increasing the boiling
point
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This can be explained with hydrogen bonding
...
The individual discrepancies can also be
explained:





Nitrogen is the least electronegative of the 3 hydrides and only forms one hydrogen
bond (having one lone pair)
Whilst fluorine also only forms one hydrogen bond, it is far more electronegative
than nitrogen
Oxygen is able to form 2 hydrogen bonds, as it has 2 lone pairs, and is more
electronegative than nitrogen

Specific heat capacity
Specific heat capacity is the energy taken to raise 1g of a substance through 1 K
...
This is a result of the hydrogen bonding in water
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Density of water and ice
As most liquids reach their freezing point and become solid their
density increases
...

This is also due to hydrogen bonding
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These open spaces
are not present in the liquid; therefore ice has a lower density than
water
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Explain entropy in a qualitative manner

Entropy is a measure of the number of ways molecules can be arranged
...

A collection of molecules has greater entropy when spread out – therefore the more
spread out molecules are across energy levels, the more ways there are of arranging them
and thus the greater the entropy
...
This in turn increases entropy
...
As more quanta = greater
entropy, we know that the closer the energy levels, the greater the quanta of energy and
therefore the higher the entropy
...
Furthermore, substances with a larger number of atoms increases the number of
energy levels, making them closer together and also increasing entropy
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Calculate entropy change of a reaction, given entropies of reactants or products

Calculating entropy change of the system
ΔSᶿ (entropy change) = ΣS (products) – ΣS (reactants)
Calculate the entropy change of the formation of ammonia
...
For a reaction to occur entropy must increase – therefore
entropy must be increasing somewhere
...
Whilst this energy loss
decreases entropy for the system, increasing the energy of the surroundings increases
entropy there
...
The equation to give us the
entropy change of the surroundings is:
𝚫𝐒 𝐬𝐮𝐫𝐫 =

−𝚫𝐇(𝐞𝐧𝐭𝐡𝐚𝐥𝐩𝐲 𝐜𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐉𝐎𝐔𝐋𝐄𝐒)
𝐓 (𝐢𝐧 𝐊)

It is important to note the equation is – 𝚫𝐇, meaning an exothermic reaction will have a
double negative and therefore a positive value
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Calculating total entropy change
𝐶𝑎𝐶𝑂3 → 𝐶𝑎𝑂 + 𝐶𝑂2

𝛥𝑆 = +159𝐽𝐾 −1 𝑚𝑜𝑙 −1

𝛥𝐻 = +179𝑘𝐽𝑚𝑜𝑙 −1

What is the entropy change at 298K, and is this reaction spontaneous?
ΔSsys = +159
ΔSsurr =

− (+179,000)
= −601 𝑘𝐽𝑚𝑜𝑙 −1
298

𝛥𝑆 𝑡𝑜𝑡 = (+159) + (−601) = −𝟒𝟒𝟐 𝑱𝑲−𝟏 𝒎𝒐𝒍−𝟏
As the overall entropy change is negative, the reaction will not occur spontaneously
...
This means that increasing the temperature
reduces the significance of ΔSsurr, increasing the possibility that the reaction can occur
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