Revision notes on chemical equilibrium - concentration equilibrium expression/constant (Kc) calculations for Advanced Level Theoretical-Physical Chemistry

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Doc Brown's Chemistry Advanced A Level Notes - Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 2

2b. Kc chemical equilibrium expression, Kc equilibrium constant and Kc equilibrium calculations

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How do we write out Kc chemical equilibrium expressions. What is the Kc equilibrium constant? How do you do Kc equilibrium calculations using concentrations? All is explained with examples including the units for concentrations (mol dm–3) and how to solve equilibrium problems using concentration units.

The equilibrium constant Kc is deduced from the equation for a reversible reaction, NOT experimental data as for rate expressions in kinetics. The concentration, in mol dm-3, of a species X involved in the expression for Kc is represented by in square brackets i.e. [X] The value of the equilibrium constant is not affected either by changes in concentration or addition of a catalyst. You need to be able to construct an expression for Kc for a homogeneous system in equilibrium, calculate a value for Kc from the equilibrium concentrations for a homogeneous system at constant temperature, perform calculations involving Kc and predict the qualitative effects of changes of temperature on the value of Kc.

How to write equilibrium expressions and work out the units of K, the equilibrium constant

• For heterogeneous equilibria, K expressions do not normally include values for solid phases, since their chemical potential cannot change since the concentration of a solid cannot change.
• Please remember, only temperature changes K, because only changing temperature can change the energy of the molecules.

• First consider the simple equilibrium: aA + bB cC + dD   {(i) ΔH -ve, (ii) ΔH +ve}

• Using the convention described above, writing out the concentration equilibrium expression gives ...

•  Kc = [C]c [D]d ––––––––– [A]a [B]b
• The equilibrium constant, Kc, is governed by temperature, which is the only factor that can alter the internal potential energy of the reactants or products.

• The 'rule' for the trend in Kc value change is provided by Le Chatelier's Principle.

• (i) If the forward reaction is endothermic, then Kc increases with increase in temperature (so Kc decreases if temperature decreased).

• (ii) If the forwards reaction is exothermic, then Kc decreases with increase in temperature. (so Kc increases if temperature decreased).

Some 'VERY rough rules of thumb' for an equilibrium K value and the 'position' of the equilibrium in terms of LHS (e.g. original reactants or products of backward reaction) and the RHS (products of the forward reaction):

• For: LHS RHS

• (for A + B C + D the rules below work ok BUT once the ratios of reactants or products are not 1:1, things are not so simple)

• If Kc is >> 1 the equilibrium is mainly on the RHS, maybe virtually 100% completion of the forward reaction i.e. a very large RHS yield i.e. and likely to be very thermodynamically feasible.

• If Kc is approx. 1 the equilibrium is more evenly distributed between the RHS and LHS.

• If Kc is << 1 the equilibrium is mainly on the LHS, maybe virtually 0% of products of the forward reaction i.e. a very low RHS yield i.e. likely to be less thermodynamically feasible.

• BUT remember, K changes with temperature considerably changing the position of an equilibrium, AND, at constant temperature, and therefore constant K, the position of an equilibrium can change significantly depending on relative concentrations/pressures of 'reactants' and 'products'.

• Finally a catalyst may speed up getting to the equilibrium but a catalyst cannot affect the position of the equilibrium constant or the value of the equilibrium constant K (Kc or Kp).

• Equilibrium expression example 2b.1 The formation of hydrogen iodide

• H2(g) + I2(g) 2HI(g)

•  Kc = [HI(g)]2 ––––––––––– (no units) [H2(g)] [I2(g)]
• units = (mol dm–3)2/[(mol dm–3) x (mol dm–3)], all cancel out, no units
• Kc has no units as all the concentration units cancel out.
• An example of the quantitative connection between kinetics (rates of reaction) and equilibrium expressions.
• This, historically, has been one of the most studied reactions in terms of kinetics and equilibrium and is a good example to study for comparing and amalgamating two important conceptual frameworks in chemistry. (If you haven't studied kinetics – rate expressions etc. then just miss out this paragraph.)
• The concentrations of reactants and products have been followed quantitatively by starting with either hydrogen iodide or a hydrogen iodine gas mixture at temperatures of 250–450oC.
• The graph below show in principle what happens ...
• • Whatever mixture you start with, eventually all the concentrations level out.
• Both the forward (f) and backward (b) reactions occur via a simple one step mechanism i.e. via a single bimolecular collision and this simple reaction mechanism leads to simple and verifiable second order kinetics rate expressions.
• ratef = kf [H2(g)] [I2(g)] and rateb = kb [HI(g)]2
• Now at the point of dynamic equilibrium, with no net change in concentrations, the rate of the forward reaction = rate of the backward reaction, so
• ratef = kf [H2(g)] [I2(g)] = rateb = kb [HI(g)]2
• therefore [H2(g)] [I2(g)] = ratef / kf  and  [HI(g)]2 = rateb / kb
• and substituting into the equilibrium expression, with the 'rates' cancelling out at equilibrium, gives
•  Kc = rateb x kf        kf –––––––– = –––– kb x ratef         kb
• So the equilibrium constant is equal to the ratio of the two rate constants of the forward and backward reaction.
• Also see calculation example 2.2a.2
• Equilibrium expression example 2b.2 The formation of the ester ethyl ethanoate

• CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

•  Kc = [CH3COOCH2CH3(l)] [H2O(l)] ––––––––––––––––––––––––––– (no units) [CH3COOH(l)] [CH3CH2OH(l)]
• units = [(mol dm–3) x (mol dm–3)]/[(mol dm–3) x (mol dm–3)], all cancel out, no units
• -
• Equilibrium expression example 2b.3 The formation of phosphorus(V) chloride (gaseous phase)

• PCl3(g) + Cl2(g) PCl5(g)

•  Kc = [PCl5(g)] ––––––––––––––––– (units mol–1 dm3) [PCl3(g)] [Cl2(g)]
• units = (mol dm–3)/[(mol dm–3) x (mol dm–3)] = mol–1 dm3
• Equilibrium expression example 2b.4 The synthesis of ammonia

• N2(g) + 3H2(g) 2NH3(g)

•  Kc = [NH3(g)]2 ––––––––––––––– (units mol–2 dm6) [N2(g)] [H2(g)]3
• units = [(mol dm–3)2]/[(mol dm–3) x (mol dm–3)3] = mol–2 dm6
• Equilibrium expression example 2b.5 The synthesis of phosphorus(V) fluoride

• P4(g) + 10F2(g) 4PF5(g)

•  Kc = [PF5(g)]4 ––––––––––––– (units mol–7 dm21) [P4(g)] [F2(g)]10
• units = ](mol dm–3)4}/[(mol dm–3) x (mol dm–3)10] = mol–7 dm21
• Equilibrium expression example 2b.6 A cobalt(II) complex ion ligand exchange reaction

• [Co(H2O)6]2+(aq) + 4Cl(aq) [CoCl4]2–(aq) + 6H2O(l)

•  Kc = [[CoCl4]2–(aq)] ––––––––––––––––––––––––– (units mol–4 dm12) [[Co(H2O)6]2+(aq)] [Cl–(aq)]4
• units = (mol dm–3)/[(mol dm–3) x (mol dm–3)4] = mol–4 dm12
• Note that the concentration of water is effectively constant in an aqueous solution and omitted from the equilibrium expression, but is effectively subsumed into the Kc value which in complex ion chemistry is called the stability constant denoted by Kstab.

• Equilibrium expression example 2b.7 The acid behaviour of high oxidation state hexa–aqua ion

• The hexaaquairon(III) ion is quite acidic in aqueous solution due to proton transfer giving the oxonium ion.

• [Fe(H2O)6]3+(aq) + H2O(l) [Fe(H2O)5OH]2+(aq) + H3O+(aq)

 Kc = [[Fe(H2O)5OH]2+(aq)] [H3O+(aq)] –––––––––––––––––––––––––– (units mol dm–3) [[Fe(H2O)6]3+(aq)]
• units = [(mol dm–3) x (mol dm–3)]/(mol dm–3) = mol dm–3
• Again [H2O(l)] incorporated into Kc.

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2b Exemplar calculations and concept Questions

These exemplar questions involve both numerical calculations and application of Le Chatelier's Principle.

2.2a Kc and concentration calculations

• Kc Example Q 2b.1 Esterification

• Given the esterification reaction: ethanoic acid + ethanol  ethyl ethanoate + water

• CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

• A mixture of 1.0 mol of ethanoic acid and 1.0 mol of ethanol was left to reach equilibrium at 25oC.

• On analysis of the equilibrium mixture, it was found that by titration with standard sodium hydroxide solution, 0.333 mol of the ethanoic acid was left unreacted.

• Calculate the value of the equilibrium constant Kc and give its units.

• The reactant–product mole ratios are 1:1 ==> 1:1
• If 0.333 mol ethanoic acid was left unreacted, then 0.667 mol of the acid had reacted.
• Therefore 0.667 mol of ethanol must also have reacted, leaving 0.333 unreacted.
• If 0.667 of the acid/ethanol reacted, 0.667 mol of  ethyl ethanoate and 0.667 mol of water must be formed.
• The concentrations = mol / volume, but the volume terms cancel each other out, so substitution in the equilibrium expression can be done in terms of final numbers of moles reactants and products
•  Kc = [CH3COOCH2CH3(l)] [H2O(l)] ––––––––––––––––––––––––– [CH3COOH(l)] [CH3CH2OH(l)]
•  Kc = 0.667 x 0.667 ––––––––––––––– = 4.01 (no units) 0.333 x 0.333
• -
• Kc Example Q 2b.2 Formation of hydrogen iodide or the decomposition of hydrogen iodide

• In these examples, don't forget that all the V's cancel, so you can work your logic in moles when solving these equilibrium problems. In the first example, concentrations are given, BUT after that its all moles, logic and maybe algebra!

• Example (a) For the reaction: H2(g) + I2(g) 2HI(g)

• An equimolar mixture of hydrogen and iodine was heated in a sealed flask at 491oC and the concentration of iodine was found to be 2.5 x 10–2 mol dm–3 and that of hydrogen iodide 1.71 x 10–1 mol dm–3.

• (a) Calculate the value of the equilibrium constant Kc.

• The concentration of the remaining hydrogen and iodine must be the same since they started at a 1:1 ratio and react in a 1:1 ratio.

• Therefore substituting in the equilibrium expression:

•  Kc = [HI(g)]2 ––––––––––– [H2(g)] [I2(g)]
•  Kc = (1.71 x 10–1)2 ––––––––––––––––––––– = 46.8 (no units) (2.5 x 10–2) x (2.5 x 10–2)
• -
• Example (b) More logic–mathematical solutions to the hydrogen, iodine and hydrogen iodide equilibrium

• PLEASE NOTE - for UK A level chemistry courses you do NOT need to solve quadratic equations!

• HI/H2/I2 moles logic to the fore!

• (i) For the equilibrium reaction: 2HI(g) H2(g) + I2(g)

• Suppose we start with 1 mole of hydrogen iodide and fraction x of it decomposes into hydrogen and iodine (x x 100 = % decomposition).

• For every mole of HI that decomposes, you will form 0.5 moles of hydrogen and 0.5 moles of iodine.

• This arises from the stoichiometry of the equation i.e. the 2 : 1 : 1 mole ratio, so we can tabulate this logic as follows ...

•  HI H2 I2 initial moles 1.0 0.0 0.0 moles at equilibrium 1–x 0.5x 0.5x
• Noting that all the V's cancel, so you can write the equilibrium expression in terms of the relative moles of hydrogen iodide, hydrogen and iodine, giving .....
•  Kc = [H2(g)] [I2(g)] –––––––––––– [HI(g)]2
•  Kc = (0.5x)(0.5x)       0.25x2 –––––––––– = –––––––– = ..... (no units) (1–x)2             (1–x)2
• So, if for example, the hydrogen iodide was 20% decomposed, then obviously the proportion decomposed x = 0.20, and simple substitution into the equation enables you to calculate the equilibrium constant K without any difficulty.
• Incidentally, since K is dimensionless, Kc = Kp even if you were working in partial pressures.
• However, if you were given K instead, and had to solve the equation for x, the proportion decomposed, this requires some carefully worked our algebra and the formula for solving quadratic equations (shown on the right).
• Rearranging the equilibrium expression gives
• K(1–x)2 = 0.25x2
• K(x2 – 2x + 1) = 0.25x2
• Kx2 – 2Kx + K = 0.25x2
• (K – 0.25)x2 – 2Kx + K = 0
• So, in the quadratic equation formula
• a = K – 0.25, b = –2K and c = K, to solve for x.
• bon voyage and watch the sign!
• (ii) For the equilibrium reaction: H2(g) + I2(g) 2HI(g)

• Suppose we start with 1 mole of hydrogen and 1 mole of iodine and no hydrogen iodide.

• For every x moles of hydrogen or iodine that react, 2x moles of hydrogen iodide will be formed, so we can tabulate this argument as follows ...

 H2 I2 HI initial moles 1.0 1.0 0.0 moles at equilibrium 1–x 1–x 2x
 Kc = [HI(g)]2 ––––––––––––– [H2(g)] [I2(g)]
 Kc = (2x)(2x)             4x2 –––––––– = –––––––––  .... (no units) (1–x)2             (1–x)2
• Again, if you know the amount of HI formed OR the amount of iodine or hydrogen reacted, you can readily calculate K via the molar logic.
• BUT if you are given K, to solve for x, this requires solving a quadratic equation.
• Rearranging the equilibrium expression
• K(1–x)2 = 4x2
• K(x2 –2x + 1) = 4x2
• Kx2 – 2Kx + K = 4x2
• (K – 4)x2 – 2Kx + K = 0
• So, in the quadratic equation formula
• a = K–4, b = –2K and c = K, to solve for x.
• watch the signs and enjoy!
• (iii) For the equilibrium reaction: H2(g) + I2(g) 2HI(g)

• A general solution for ANY combination of hydrogen and iodine forming hydrogen iodide.

• Suppose we start with A moles of hydrogen and B moles of iodine and no hydrogen iodide.

• For every x moles of hydrogen OR iodine that react, 2x moles of hydrogen iodide will be formed, so we can tabulate this argument as follows ...

 H2 I2 HI initial moles A B 0.0 moles at equilibrium A–x B–x 2x
 Kc = [HI(g)]2 ––––––––––– [H2(g)] [I2(g)]
 Kc = (2x)2 –––––––––––– = .... (no units) (A–x) (B–x)
• Again, if you know the amount of HI formed (2x) OR the amount of iodine/hydrogen reacted (x), you can readily calculate K.
• Note that x is NOT a fraction here!
• Its the actual moles of hydrogen or iodine that have reacted to form hydrogen iodide.
• BUT if you are given K, to solve for x, this requires solving a quadratic equation.
• Rearranging the equilibrium expression
• K(A–x)(B–x) = 4x2
• K(x2 –Ax – Bx + AB) = 4x2
• Kx2 –AKx –BKx + ABK = 4x2
• (K–4)x2 –(A + B)Kx + ABK = 0
• So, in the quadratic equation formula
• a = K–4, b = –(A+B)K and c = ABK, to solve for x.
• This methodology is not designed for late night working! and watch the signs!
• (iv) For the equilibrium reaction: 2HI(g) H2(g) + I2(g)

• A general solution for the decomposition of hydrogen iodide.

• Suppose we start with A moles of hydrogen iodide and x moles of it decomposes into 0.5x moles of hydrogen and 0.5x moles of iodine.

• Remember, for every mole of HI that decomposes, you will form 0.5 moles of hydrogen and 0.5 moles of iodine.

• This arises from the stoichiometry of the equation i.e. the 2 : 1 : 1 mole ratio, so we can tabulate this logic as follows ...

•  HI H2 I2 initial moles A 0.0 0.0 moles at equilibrium A–x 0.5x 0.5x
• Noting that all the V's cancel, so you can write the equilibrium expression in terms of the relative moles of hydrogen iodide, hydrogen and iodine, giving .....
•  Kc = [H2(g)] [I2(g)] ––––––––––– [HI(g)]2
•  Kc = (0.5x)(0.5x)            0.25x2 –––––––––– = ––––––––– = ..... (no units) (A–x)2                 (A–x)2
• So, if you know how much HI reacted–decomposed, OR, the amount of iodine formed, then simple substitution into the equilibrium equation enables you to calculate the equilibrium constant K without any difficulty.
• Note that x is NOT a fraction here!
• Its the actual moles of hydrogen iodide that have decomposed to form hydrogen and iodine.
• However, if you were given K, to solve the equation for x, the amount decomposed, this requires some carefully worked our algebra and the formula for solving quadratic equations (shown on the right).
• Rearranging the equilibrium expression
• K(A–x)2 = 0.25x2
• K(x2 –2Ax + A2) = 0.25x2
• Kx2 –2KAx + KA2 = 0.25x2
• (K – 0.25)x2 –2KAx + KA2 = 0
• So, in the quadratic equation formula
• a = K–0.25, b = –2KA and c = KA2, to solve for x.
• keep a clear head and watch the signs!
• Kc Example Q 2b.3 Formation of complex ion

• Aqueous iron(II) ions can complex with chloride ions to form the tetrachloroferrate(III) ion.

• For the equilibrium: Fe3+(aq) + 4Cl(aq) FeCl4(aq)

• Kc = 8.0 x 10–2 mol–1 dm3 at 298K.

• (a) If the concentration of the free chloride ion is 0.80 mol dm–3 and that of the free iron(III) ion 0.20 mol dm–3, calculate the concentration of the tetrachloroferrate(III) complex ion.

•  Kc = [FeCl4–(aq)] –––––––––––––––––– [Fe3+(aq)] [Cl–(aq)]4
• rearranging the equilibrium expression gives ...
• [FeCl4(aq)] = Kc x [Fe3+(aq)] x [Cl(aq)]4
• [FeCl4(aq)]  = 8.0 x 10–2 x 0.20 x (0.80)4 = 6.55 x 10–3 mol dm–3
• (b) Equal volumes of 5 molar sodium chloride solution and 0.02 molar iron(III) nitrate Fe(NO3)3 solution were mixed together.
• (i) Assuming the chloride ion concentration changes very little and c represents the concentration of the tetrachloroferrate(III) ion, show how the concentration of the complex ion can be approximately calculated.
• The nitrate ion is a spectator ion and can be ignored.
• Let [FeCl4(aq)]equilib = c, and
• since 1 mole of Fe3+ forms 1 mole of FeCl4 complex
• then [Fe3+(aq)]equilib = [Fe3+(aq)]init – c, and assuming
• [Cl(aq)]equilib ~ [Cl(aq)]init since [Cl(aq)]  >> [Fe3+(aq)] + [FeCl4(aq)]
• (ii) Calculate the concentration of the [FeCl4(aq)] ion.
• The dilution factor on mixing equal volumes is 2, therefore
• [Fe3+(aq)]init = 0.02/2 = 0.01 mol dm–3 and
• Cl(aq)]equilib ~ [Cl(aq)]init = 5.0/2 = 2.5 mol dm–3
• from (a) [FeCl4(aq)] = Kc x [Fe3+(aq)] x [Cl(aq)]4
• c = 8.0 x 10–2 x (0.01 – c) x (2.5)4
• c = 3.125 x (0.01 – c) = 0.03125 – 3.125c
• 4.125c = 0.03125
• [FeCl4(aq)] = c =  0.03125/4.125 = 0.7575 = 7.58 x 10–3 mol dm–3
• -
• (iii) What percentage of the original Fe3+ ion is converted into the complex?
• If all of the Fe3+ ion had been converted to the chloro complex, the concentration of the complex would be 0.01 mol dm–3
• Therefore the % conversion = 7.58 x 10–3 x 100/0.01 = 75.8%
• -

• Kc Example Q 2b.4 Iodine–iodide equilibrium in aqueous solution

• Iodine is much more soluble in potassium iodide solution than pure water because of the equilibrium:

• I(aq) + I2(aq) I3(aq)    for which Kc = 7.10 x 102 mol–1 dm3 at 298K.

• If the concentration of the I ion is 0.122 mol dm–3, and that of the I3 ion is 0.153 mol dm–3, calculate the concentration of free aqueous iodine.

•  Kc = [I3–(aq)] ––––––––––––– [I–(aq)] [I2(aq)]
• Rearranging gives ...

•  [I2(aq)] = [I3–(aq)] ––––––––––– Kc x [I–(aq)]
• [I2(aq)] = 0.153 / (7.10 x 102 x 0.122) = 1.77 x 10–3 mol dm–3

• Kc Example Q 2b.5 Ester equilibrium – titrimetric analysis

• A titration method for determining the equilibrium constant for an esterification reaction.

• 12.0g of pure ethanoic acid was mixed with 11.5g of pure ethanol and left to stand for a week at room temperature (298K/25oC). The mixture was then mixed with deionised water and made up to 250 cm3 in a calibrated volumetric flask. When a 25.00 cm3 aliquot of the mixture was titrated with 0.50 mol dm–3 sodium hydroxide solution using phenolphthalein indicator (colourless ==> 1st permanent pink end–point), 10.60 cm3 of the alkali was needed for complete neutralisation.

• (i) CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) (esterification reaction)

• (ii) CH3COOH(l) + NaOH(aq) ==> CH3COONa+(aq) + H2O(l) (neutralisation titration reaction)

• (a) Calculate the moles of ethanoic acid unreacted in the original mixture.

• 1 mole CH3COOH : 1 mole NaOH from equation (i) above.

• moles = molarity x vol(dm3), so moles acid = 0.50 x 10.6/1000 = 0.0053 mol

• since the 25.00 cm3 aliquot titrated is equal to 1/10th of the total mixture,

• the total moles of unreacted acid = 10 x 0.0053 = 0.053 mol CH3COOH left

• -

• (b) Calculate the moles of ethanoic acid and ethanol in the starting mixture.

• Mr(CH3COOH) = 60, mol ethanoic acid = 12/60 = 0.20

• Mr(CH3CH2OH) = 46, mol ethanol = 11.5/46 = 0.25

• -

• (c) Calculate the moles of ethanol left unreacted and the moles of ethyl ethanoate ester and water formed.

• This requires a little bit of logic, best appreciated with a little table showing the 'logical thinking' and the final mol numbers.

•  equation CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l) moles at start 0.20                      0.25                               0                             0 mol at equilibrium 0.20 – x                0.25 – x                         x                             x mol at equilibrium 0.053                   0.103                             0.147              0.147
• x = moles of ester or water formed,

• so x mol of acid or alcohol must be used to reach equilibrium,

• since mol ratios are 1:1 ==> 1:1 in the reaction equation.

• Therefore 0.20 – x = 0.053 from calculation (a), therefore x = 0.20 – 0.053 = 0.147

• so all the molar quantities can be logically deduced and are shown in the final line of the table.

• TIP In an exam, doing the working of this part of the Q under the equation is not a bad idea.

• -

• (d) Calculate the equilibrium constant Kc for this esterification.

•  Kc = [CH3COOCH2CH3(l)] [H2O(l)] ––––––––––––––––––––––––––– [CH3COOH(l)] [CH3CH2OH(l)]
•  Kc = (0.147/V) x (0.147/V) ––––––––––––––––––––––– = 3.96 (no units) (0.053/V) x (0.103/V)
• Note that all the volume terms cancel out, so you can work purely in moles to calculate Kc.
• -
• (e) Suggest several reasons why it may take a week for the equilibrium point to be reached and suggest ways of speeding up the reaction.

• (i) The reaction will be slow at room temperature, refluxing the mixture will speed things up!

• (ii) The reaction is catalysed by hydrogen ions (H+) and since ethanoic acid is a weak acid, the hydrogen ion concentration will be very low. Adding a strong mineral acid e.g. conc. sulfuric acid.

• (iii) Not surprisingly, esters are often prepared by refluxing the acid and alcohol with a few drops of conc. sulfuric acid added to the mixture.

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WHAT NEXT?

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product Ksp, common ion effect, ion–exchange systems * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces watch out for sub-indexes to multiple sections or pages

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