Chemistry: Oceans

    a Solubility:

      Test yourself
    • When a solute dissolves, the solute and solvent bonds break and new ones are formed
    • Usually, substances won't dissolve if the broken bonds are stronger than the new ones
    • Solvents can be:
      - Polar (e.g. water)
      - Non-polar (e.g. hexane)

    Ionic substances:

    • These only usually dissolve in polar solvents, creating an aqueous solution
    • This is because:
      - The H atoms in water have a positive partial charge and the O atoms have a partial negative charge
      - The ions bond to these dipoles to form ion-dipole bonds
      - This causes the ions to leave the ionic lattice, and they become randomly distributed
      - A dissolved ion can be drawn like this. If you're drawing this in an exam, make sure to draw at least three water molecules around an ion
    • Solvation is when ions are being surrounded by a solvent. It's called hydration if the solvent is water
    • All lattices which can be pulled apart by the ion-dipole bonds are soluble

    Covalent substances:

    • Usually, the intermolecular bonds between covalent substances are weak
    • Similarly, non-polar solvents also have weak instantaneous dipole—induced dipole bonds
    • Because the forces are similar, most covalent substances will dissolve in non-polar solvents

    • In polar solvents, the hydrogen bonds between water molecules are much stronger than the bonds that would form if the substance dissolved. So most covalent substances are not soluble in polar solvents
    • The only exceptions are when the molecule can hydrogen bond with water

    b Ionic enthalpy measurements:

      Test yourself
    • Standard lattice enthalpy (ΔLEHo) is the enthalpy change when 1 mole of an ionic lattice is formed from its gaseous ions under standard conditions
      - Covalent bonds are broken, so it's always negative (exothermic)
    • The standard enthalpy change of hydration of ions (ΔhydHo) is the enthalpy change when 1 mole of aqueous ions are formed from gaseous ions (dissolving in water)
      - This is also always negative (exothermic)
    • Standard enthalpy change of solution (ΔsolHo) is the enthalpy change when 1 mole of an ionic substance dissolves in enough solvent to form an infinite dilution
      - An infinite dilution is where there is so much solvent that adding solvent would not affect the concentration
         - In reality, this is impossible, but we can experimentally determine the value by extrapolating from known concentrations

    Using an enthalpy cycle (Hess cycle):

    • The below cycle relates the above three enthalpy measurements:
    • So, for example, to calculate the enthalpy change of solution for NaCl from the lattice enthalpy (-787 kJ mol-1) and hydration enthalpies (-406 kJ mol-1 and -378 kJ mol-1), you would use the equation
      enthalpy change of solution = -(-787) + (-406 + -378) = +3 kJ mol-1

    Enthalpy level diagrams:

    • Enthalpy level diagrams can also be used to visualise and calculate enthalpy values for ionic substances
    • Start by drawing a short horizontal line and writing the solid ionic substance above it. From here, draw two more lines and write on them the ions in gaseous and aqueous form. Use arrows to connect them with the enthalpies:
    • As with Hess cycles, you can use enthalpy of route 1 = enthalpy of route 2 to calculate the unknown value. So route 1 would be -(-787) + -784 = 3 kJ mol-1 and route 2 would be the enthalpy of solution. Therefore, the enthalpy of solution is 3 kJ mol-1

    Measuring enthalpy change of solution:

    • You can directly measure enthalpy change of solution experimentally - you just need to use a thermometer and insulated polystyrene beaker with a lid
    • Measure the number of moles of the salt, the change in temperature (in K) and the mass of liquid and solid (in g). You would usually use about 50 cm3 of water, which has a mass of about 50 g
    • Use the equation q = mcΔ T to calculate the energy change in J. Divide this by 1000 to convert it to kJ, and then divide it by the number of moles of salt. This will get you its enthalpy change of solution
    • This method isn't very accurate mainly due to energy loss. It will usually have an error between 20% and 50%. It is also impossible to do this under standard conditions in a school lab since temperature would need to be kept at 298 K

    c Factors affecting lattice and hydration enthalpy:

      Test yourself
    • The greater the charge density of the ions:
      - the greater the electrostatic attraction and the more exothermic the lattice enthalpy
      - water molecules are more attracted to the solute and the more exothermic the hydration enthalpy
    • Charge density increases with charge and decreases with size
      - So, for example, Mg2+ has a higher charge density than Na+ because it has twice the charge and a very similar atomic radius
      - But K+ has a lower charge density than Na+ because it has an additional electron shell but the same charge

    d Entropy:

      Test yourself
    • Entropy, S, measures the number of combinations that particles can be arranged in, and how many ways quanta can be distributed amongst the particles (quanta are fixed packets of energy)
    • To increase stability, the overall entropy of a system will increase over time

    e Factors affecting entropy:

      Test yourself
    • Gases and liquids have higher entropy than solids because there is lots of disorder (more possible arrangements and more randomness)
    • Substances with more energy have higher entropy because entropy also depends on the number of ways quanta can be distributed
    • Increasing the amount of a substance will increase the entropy, because this will mean that there are more possible arrangements

    f, g Entropy calculations:

      Test yourself
    • Entropy is measured in J K-1 mol-1
    • Total entropy change, , is the sum of the entropy changes of the system and surroundings. This is written as
    • These variables are calculated with:
      and
      where Δ H is the enthalpy change in J mol-1 and T is the (initial) temperature in K

    Reaction feasibility:

    • If is:
      - negative: the reaction won't be feasible
      - equal to 0: the reaction will happen as an equilibrium reaction
      - positive: the reaction will be feasible
    • Many reactions are only feasible at high temperatures. This is because increases as T increases

    h Solubility product:

      Test yourself
    • A sparingly soluble salt is one which only slightly dissolves
    • A saturated solution is a solution which has the maximum number of particles dissolved in it - any additional solid added will not dissolve
    • When a sparingly soluble ionic solid is dissolved in water, there will be an equilibrium
      - The lattice will dissolve into its individual ions with their standard charges
      - For example, NaCl(s) ⇌ Na+(aq) + Cl-(aq)
      - Or MgF2(s) ⇌ Mg2+(aq) + 2F-(aq)
      - This can be written as an equilibrium constant, e.g. [Mg2+(aq)][F-(aq)]2
         - Remember that solids are not included in expressions, so they have a value of 1
         - Therefore, in the above expression, I was dividing the products by 1, which does nothing
      - Also, remember to include state symbols inside the square brackets
      - This equilibrium constant is called the solubility product and is written as

    Determining a solubility product experimentally:

    • First, make a saturated solution:
      - Warm distilled water in a conical flask and add the salt whilst shaking
      - Once no more salt will dissolve, leave the mixture to cool
      - If there is no salt at the bottom, warm again and add more salt
      - Otherwise, filter through filter paper
    • Now calculate the concentration
      - First, measure the temperature since solubility product is temperature dependent
      - Use a titration (e.g. with HCl or NaOH) or colorimetry if the solution is coloured
      - Once you know the concentration of one of the ions, you can calculate the solubility product

    i The Brønsted-Lowry theory of acids and bases:

      Test yourself
    • Acids are proton donors, releasing H+ ions in water in the form of H3O+ (oxonium ions)
    • All acids contain a hydrogen atom, so the general formula is HA
    • Bases are proton acceptors. The general formula is B
    • When an acid is added to water, the water acts as a base because it accepts a proton to form H3O+
    • When a base is added to water, the water acts as an acid because it donates a proton and becomes OH-

    Conjugate pairs:

    • When an acid or base dissociates in water, a reversible reaction is set up. Each side of the ⇌ has an acid and a base
    • The acid on the left and the base on the right form a conjugate pair
    • Similarly, the base on the left and the acid on the right form a conjugate pair
    • For example, the conjugate base of HCl is Cl- because it has donated a proton (H+). The conjugate base of H2SO4 is HSO4- for the same reason

    j Strong acids:

      Test yourself
    • Strong acids dissociate completely or almost completely in water, i.e. nearly all H+ ions are released
    • Although dissociation is a reversible reaction, for strong acids a → is usually used because the equilibrium lies very far to the right
    • For example, HCl is a strong acid: HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

    Strong bases:

    • The same applies for strong bases; they nearly completely dissociate in water to form OH-(aq) ions

    General dissociation equations for strong acids/bases:

    • HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
    • B(aq) + H2O(l) → BH+(aq) + OH-(aq)

    k Weak acids:

      Test yourself
    • Only a small number of weak acid molecules dissociate in water, forming very few H+ ions. Therefore, these reactions are usually written with a ⇌ symbol
    • For example, CH3COOH is a weak acid: CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)

    Calculating (for weak acid calculations):

    • Weak acids only slightly dissociate in water, so at equilibrium, the H+ concentration will be lower than HA concentration
    • Acid dissociation has the equilibrium HA(aq) ⇌ H+(aq) + A-(aq)
    • Only a small amount of HA dissociates, so we assume that its concentration remains constant
    • We can create an expression
    • Because the acid still dissociates more than the water, we can also assume that all of the H+ ions come from the acid, so the H+ concentration is the same as the A- concentration
    • Therefore the above expression simplifies to
      - The units of are mol dm-3
      - Remember, this only works for weak acids due to the above assumptions
      - It is temperature dependent

    Calculating :

    • is a way of showing how strong or weak an acid is
    • The smaller the , the stronger the acid

    Measuring pH:

    • You can use an acid-base titration to find the concentration and then convert this to pH using the method in the section below
    • Alternatively, use universal indicator or a pH probe
    • pH is temperature dependent. This means that you need to calibrate a pH probe for the solution temperature
      - First, wash the electrode with distilled water
      - Put the electrode into a pH 7.00 buffer solution, wait for the reading to stabilise, and then calibrate to 7.00
      - Now repeat the above two steps with a pH 4.00 solution if you're measuring the pH of an acid
      - Or use a pH 10.00 solution if you're measuring the pH of an alkaline solution

    Finding of a dilute acid experimentally:

    • Use a burette to add NaOH(aq) a few cm3 at a time, measuring the pH after each addition with a pH probe
    • After one addition, you'll get a significant change in pH. Repeat the titration around this point but with smaller volume increments and more readings to increase accuracy
    • Plot a graph of pH against the volume of sodium hydroxide solution and draw a curve to connect the points. You should get a graph similar to the one below (which is for a weak acid)
    • The volume of alkali with the gradient (in the centre of this graph) is called the equivalence point. Here, all of the HA molecules have dissociated (because it dissociates as H+ ions get used up)
    • Halve the volume at the equivalence point. At this point (the half-equivalence point), [HA] = [A-]
    • Read off the value for pH at the half-equivalence point. Use this to calculate the value for [H+]
    • =
      So

    l pH:

      Test yourself
    • pH is a measure of H+ concentration. Because there is a huge variation, it is measured on a log scale
    • Most universal indicators will measure pH 0-14, but pH can sometimes be outside of this range
    • A solution is neutral (pH 7.00) if [H+] = [OH-]

    Calculating the pH of a strong acid:

    • Use the concentration of H+ ions to calculate pH: pH
    • To convert pH to H+ concentration, use

    Using to calculate the pH of a weak acid:

    • Use the expression in the above section (k) to calculate [H+]2
    • Now square root it and take the negative log to get the pH

    Calculating :

    • Water can behave as an acid or base - so there'll always be both oxonium ions (H3O+) and hydroxide ions (OH-)
    • This creates the equilibrium H2O(l) ⇌ H+(aq) + OH-(aq)
    • We can use this equation to write an expression for the equilibrium constant:
    • Because water is present in much larger amounts than the ions, [H2O] is considered to have a constant value
    • If we multiply by the water concentration, we get [H+][OH-]. We call this , the ionic product
      - So [H+][OH-]
      - The units of are
      - It varies depending on temperature only
      - The value is given in the data sheet at 298 K:

    Using to calculate the pH of a base:

    • Use the equation to calculate the H+ concentration
    • Now use this to calculate the pH

    m Buffers:

      Test yourself
    • A buffer is a solution that resists changes in pH (nearly keeping it constant) if a small amount of an acid/base is added
    • There are multiple types of buffer; we only need to know about acidic buffers

    Acidic buffers:

    • Acidic buffers are acidic solutions, made by mixing a weak acid with one of its salts
    • The weak acid will dissociate by a small amount
    • The salt will fully dissociate and dissolve
    • We'll use ethanoic acid (CH3COOH) and sodium ethanoate (CH3COO-Na+) as an example
    • This gives an overall equilibrium of CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)
      - If H+ is added, the equilibrium moves left, reducing the H+ concentration back to normal
      - If OH- is added, it reacts with the H+, causing the CH3COOH to dissociate more

    Calculating the pH of a buffer solution:

    • First, write the expression for (the numerator should be [H+][salt], and the denominator should be [acid])
    • Now rearrange to make [H+] the subject use this to calculate pH

    n The greenhouse effect:

      Test yourself
    • Most of the solar energy reaching the Earth is visible and UV.
      The above point is almost directly quoting the OCR specification; I'm not sure why IR is not also listed since ~45% of the Sun's radiation is IR. I assume it's because they're trying to simplify the process, but please email me/tell me if you know a more correct reason
    • Some visible light is reflected by clouds, but most is absorbed and re-emitted as infrared radiation
    • Some of this radiation escapes into space. This is because some IR frequencies are not absorbed by atmospheric gases (the IR window)
    • Some is absorbed by gases in the troposphere (between the ground and 10-20 above the surface) and re-emitted in all directions. Therefore, some will be directed back towards the surface. This partial retention of IR is the greenhouse effect
    • Although the main greenhouse gases (methane, carbon dioxide and water) have an IR window, some human-created gases fill this window, contributing to global warming
    • Because humans are increasing the concentrations of greenhouse gases in the atmosphere, the effects of the IR retention are increasing. This is called the enhanced greenhouse effect

    Energy transfer:

    • Greenhouse gases absorb some IR frequencies. This causes their bonds to vibrate more
    • This additional energy is transferred to other gas molecules in collisions, raising the temperature