The Oceans

Question 1

Carbon dioxide dissolves in water giving the overall equation

Answer 1

CO2(aq) + H2O(l) 2H+(aq) + CO3^2- (aq)

Product is carbonic acid

Question 2

Methods to reduce CO2 in atmosphere

Answer 2

More economical use of fossil fuels
Alternatives to fossil fuels
Capture and storage of CO2
Increased levels of photosynthesis

Question 3

Entropy is a measure of

Answer 3

Order

Question 4

Entropy change for a chemical system is

Answer 4

The difference between the entropies of the reactants and products. P-R

Question 5

Entropy change of the surroundings

Answer 5

Depends on the transfer of heat to and from the surroundings I.e enthalpy change.

Delta Ssurr = delta H (enthalpy change) / T

T is in kelvin. 0 –> +273

Question 6

Total entropy change is…

Answer 6

Sum of delta system add delta surroundings.

Question 7

Total entropy predictions:

Answer 7

Delta total is positive then a reaction will occur spontaneously
Delta total = 0 then reaction is in equilibrium

Question 8

Ionic compounds in water

Answer 8

Dissolve in water, ionic lattice is broken up and the ions seperate and become hydrated. Energy is used to break up lattice but energy is given out when ions are hydrated, ion- dipole bonds are formed.

Question 9

Lattice (formation) enthalpy definition :

Answer 9

Enthalpy change when one mole of a solid is formed from its separate gaseous ions.
Magnitude of delta LE is an indication for bond strength in the lattice
ALWAYS NEGATIVE - bonds forming

Question 10

Bigger charges and or smaller ions…._____ ionic bonds_______ lattice enthalpy

Answer 10

Stronger

Larger (or more negative )

Question 11

Ionic radius of an element depends on

Answer 11

Nuclear charge (atomic number) - the bigger the nuclear charge the smaller the ion ( for atoms with the same number of filled energy levels) 
The number of full energy levels - the more levels, the bigger the ion

Question 12

Enthalpy change of hydration definition

Answer 12

Enthalpy change when an aqueous solution is formed from one mole of gaseous ions.
Ion-dipole bonds form - ALWAYS NEGATIVE.

Question 13

The greater the enthalpy change of hydration the…

Answer 13

Stronger the ion-dipole attractions and the greater the number of water molecules surrounding the ion.

Question 14

Strength of ion - dipole attraction depends on the charge and size of the ions:

Answer 14

Bigger the charge&/or small the size of ions the stronger the ion -dipole bonds - larger the enthalpy of hydration.

Question 15

Enthalpy change of solution definition

Answer 15

Is the enthalpy change when one mole of a solute dissolves to form a dilute solution. Can be thought of as a two step process - breaking down the lattice and then hydrating the gaseous ions produced (water is solvent)

Question 16

Enthalpy change of solution =

Answer 16

Delta Hhyd (cation) + delta Hhyd(anion) - deltaHle

Question 17

Enthalpy change of solvation

Answer 17

Enthalpy change when a solution is formed from 1 mole of gaseous ions using a solvent OTHER than water.

Question 18

If delta H solution is negative or slightly positive

Answer 18

Then solid will dissolve as entropy change will be favorable

Question 19

If delta H solution is large and positive then

Answer 19

Solid will not dissolve even tho enthalpy change is favorable because too much energy is required.

Question 20

Ionic solutes in non polar solvents

Answer 20

-Delta H solvation is tiny because there is little attraction between the ions and solvent.

Question 21

Strong acids in water

Answer 21

Almost donate all their protons - said to have undergone complete dissociation
Eg HCL, sulfuric acid and nitric acid (HNO3)

Question 22

Weak acids in water ..

Answer 22

Only a small proportion of the acid molecules donate their protons. Said to have undergone incomplete dissociation
Eg Carboxylic acids and carbonic acid

Question 23

Ka

Answer 23

Acidity constant or acid dissociation constant. Greater the value of Ka the Stronger the acid.

Question 24

Equation for Ka

Answer 24

= (H+) (acid-) / (acid)

All aqueous

Question 25

For very weak acids we can use something instead of Ka for ease

Answer 25

pKa = -log (Ka)

Question 26

Calculation to work out pH

Answer 26

pH= -log(H+ (aq))

Question 27

Monoprotic acids

Answer 27

Have one acidic hydrogen - they release one proton per molecule

Question 28

Strong acid assumptions

Answer 28

Because they fully dissociate we can assume that concentration of acid is the same as concentration of hydrogen ions in that solution if acid is Monoprotic

Question 29

Weak acids assumptions

Answer 29

Because it hardly dissociates we say that acid concentration remains same.
Also for every one mole of H+ there’s one mole of acid-. conc =

Question 30

What’s needed to calculate weak acid

Answer 30

Concentration of the acid and its dissociation constant

Question 31

Strong bases in water

Answer 31

Gives (OH-) is fully dissociated

Question 32

Ionic product of water:

Answer 32

Used to calc pH of strong bases

Kw = (H+)(OH-) = 1X10-14 mol2dm-6 at 298K

Question 33

Calculating pH of strong base

Answer 33

Fully dissociates so conc of base = conc of OH-
Kw =(H+)(OH-)
So (H+) = Kw/(OH-)
Take -log(H+)

Question 34

Buffer solution definition

Answer 34

Solutions that have almost constant pH, despite dilution of small additions of acid or alkali

Question 35

Buffer solutions contain

Answer 35

Either a weak acid and on of its salts - e.g ethanoic acid and sodium ethanoate
Or a weak base and one of its salts - ammonia and ammonium chloride

Question 36

All buffers contain large amounts of…

Answer 36

Proton donor - weak acid or conjugate acid
And large amounts of proton acceptor - weak base or conjugate base.
Any additions react with these large amounts and maintain the stable pH

Question 37

Two assumptions made in buffer calculations

Answer 37

All the anions (A-) Have come from the salt, so contribution from the acid is negligible
Concentration of the acid in solution (HA(aq)) is the same as conc of acid put into solution.

Question 38

Calculations with buffers

Answer 38

Ka= (H+)(A-) / (HA)

Using assumptions can be shown that Ka=(H+) X ((salt)/(acid))

Question 39

Finding pH of a buffer solution

Answer 39

If Ka of the weak acid is known and conc of both salt and weak acid then (H+) can be calculated and hence the pH can be also.

Whenever solutions are mixed, the individual concentrations are reduced in proportion to their volumes in the mixture. Equal amounts = halved

Question 40

Making buffer solution of a specific pH

Answer 40

Use Ka= (H+) X (salt)/acid

Choose right weak acid with correct Ka or pKa and then calculate ratio of acid : salt

Question 41

Buffers in action

Answer 41

CO2 dissolved in oceans and limestone rocks (Calcium carbonate)

Carbonic acid produced acts as weak acid
Limestone acts as as anion sink (CO3^2-)

Means as CO2 atmospheric levels increase and dissolve to form carbonic acid the pH can be maintained

Question 42

Unique properties of water

Answer 42

High bp for a compound with low relative molecular mass
Has a high specific heat capacity ( takes a lot of energy to raise temperature) large masses of water can carry a lot of energy in the oceans and distribute it across the worlds oceans.
Has a high enthalpy change of vaporization
Density of ice is lower than liquid water, result of bond angles - ice has a very open structure with large spaces.

ALL DUE TO H BONDING, WATER CAN FORM 2 HBONDS PER MOLECULE.