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Chemistry > 21. Further Aspects of Equilibria > Flashcards

21. Further Aspects of Equilibria Flashcards

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1
Q

What is the ionic product of water (Kw)?

A

Kw = [H+][OH-] 1.0 × 10 ^ -14 (298K)

  • [H2O] constant (water ionises very little)
  • water is covalent but has small conducting ability
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2
Q

What is pH?

A

pH = -log[H+]

- negative logarithm to the base 10 of [H+].

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3
Q

How do you calculate [H+] from pH?

A

10 ^ -pH

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4
Q

What is a monobasic acid?

A

An acid which only contains one replaceable H atom per molecule.

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5
Q

What is a strong acid?

A

An acid that is completely ionised in solution. [H+] in solution roughly = [acid], assuming that [H+] from the ionisation of water is negligible.

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6
Q

How do you calculate the pH of a strong base?

A

1) Rearrange the Kw equation: [H+] = Kw / [OH-]
2) Substitute Kw and [OH-] in to find [H+]
3) Calculate pH

OR: 14 - (-log[OH-])

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7
Q

What is Ka?

A

Ka = [H+][A-] / [HA] (ethanoic acid = 1.74 x 10 ^ -5)

  • (weak) acid dissociation constant
  • [H+] = [A-] so the numerator is [H+] ^ 2
  • assumes that EQM [acid] =original
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8
Q

What do different values of Ka indicate?

A

High Ka = EQM to the right, ionised more, more acidic.

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9
Q

What is pKa?

A

pKa = -log Ka

  • negative logarithm to the base 10 of Ka
  • easier way of writing Ka values for comparisons
  • high Ka = less acidic
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10
Q

How do you calculate Ka for a weak acid?

A

1) Convert pH to [H+]
2) Write the equilibrium expression
3) Enter [H+] and [acid] values.

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11
Q

How do you calculate the pH of a weak acid?

A

1) Rearrange the equilibrium expression ([H+] the subject)
2) Enter Ka and [acid] values
3) Calculate pH using [H+].

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12
Q

What is Kb?

A

Kb = [B+][OH-] / [BOH]

- base dissociation constant

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13
Q

What is an acid-base indicator?

A

Dye / mixture of dyes that changes colour over a specific pH range.
Many can be considered weak acids, where the acid (HIn) and the CB (In-) have different colours.
Adding acid shifts EQM to the right (A), adding alkali shifts it to the left (CB).

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14
Q

What are the pH range and colour range of bromothymol blue?

A
  • Yellow in acid -> grey-green endpoint -> blue in alkaline.

- pH range from 6 - 7.6, endpoint at pH 7

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15
Q

What are the pH range and colour range of phenolphthalein?

A
  • Colourless -> pink-violet

- pH range from 8.2 - 10, endpoint at 9.3

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16
Q

What are the pH range and colour range of methyl orange?

A
  • Red -> yellow

- pH range from 3.2 - 4.4, endpoint at 3.7

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17
Q

What are the pH range and colour range of methyl red?

A
  • Red -> yellow

- pH range from 4.2 - 6.5, endpoint at 5.1

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18
Q

What is the equivalence point?

A

Point of colour change, where the solutions have been mixed in exactly equation proportions.

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19
Q

Describe the pH curve for the SA and SB titration.

A
  • Slope falls/rises slightly initially
  • Steep vertical region from 3.5 - 10.5
  • Endpoint at 7
  • Bromothymol blue, phenolphthalein
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20
Q

Describe the pH curve for the SA and WB titration.

A
  • Rapid fall, then slows down (buffer solution set up)
  • Above is for acid into alkali
  • Steep vertical region from 3.5 - 7.5
  • Endpoint at 5
  • Methyl red
21
Q

Describe the pH curve for the WA and SB titration.

A
  • Rapid rise, then slows down (alkali into acid)
  • Steep vertical region from 7.5 - 11
  • Endpoint at 9
  • Phenolphthalein
  • Buffer solution past EP
22
Q

Describe the pH curve for the WA and WB titration.

A
  • No sharp fall

- Endpoint at 7 (point of inflection)

23
Q

What is a buffer solution?

A

A solution which resists changes in pH when small volumes of acid/alkali are added.

24
Q

What are two types of buffer solution?

A
  • Acidic buffer - weak acid + salt (eg. ethanoic acid + sodium ethanoate).
  • Alkaline buffer - weak base + salt (eg. ammonia + ammonium chloride).
25
Q

How do acidic buffer solutions work, in terms of reserve supplies?

A

ethanoic acid ⇌ H+ + ethanoate
- high acid concentration
sodium ethanoate -> Na+ + ethanoate
- high ethanoate concentration

Solution has high ethanoic acid and ethanoate concentrations (reserve supplies of acid and base). As long as the ratio of these concentrations is constant, so is pH.

26
Q

How do acidic buffer solutions work, in terms of equilibria?

A

ethanoic acid ⇌ H+ + ethanoate
ethanoic acid + OH- ⇌ ethanoate + H2O

  • Increasing [H+] shifts EQM to the left
  • Increasing [OH-] shifts EQM to the right
  • High ethanoate and ethanoic acid reserve supplies ensure that pH does not change.
  • If H+ and OH- combine, the EQM shifts to replace the H+ lost.
27
Q

How do alkaline buffer solutions work, in terms of reserve supplies?

A

ammonia + water ⇌ ammonium + OH-
- high ammonia concentration
ammonium chloride -> ammonium + Cl-
- high ammonium concentration

  • High reserve supplies of unreacted ammonia and ammonium ions.
28
Q

How do alkaline buffer solutions work, in terms of equilibria?

A

ammonia + H+ ⇌ ammonium
ammonia + water ⇌ ammonium + OH-

  • Increasing [H+] shifts EQM to the right
  • Increasing [OH-] shifts EQM to the left
  • If OH- and H+ combine, the EQM shifts to replace the OH- lost.
29
Q

How do you calculate the pH of an acidic buffer solution?

A

1) Write EQM expression, making [H+] the subject
2) Enter Ka and concentration values
3) Calculate pH
- Assumes that [acid] has not changed, and that [ethanoate] = [salt].

OR pH = pKa + log ([salt] / [acid])

30
Q

How can you tell the buffer pH of a pH curve?

A

pH will = pKa, because [acid] and [salt] will be equal. Half of the (reacting) titrating solution will have been added.

31
Q

How do you calculate the pH of an alkaline buffer solution?

A

ammonium ⇌ ammonia + H+

1) Write EQM expression, making {H+] the subject
2) Enter Ka (ammonium) and concentration values
3) Calculate pH

32
Q

Give an environmental example of excess addition to buffer solutions.

A

Unpolluted regions - rainwater pH 5.7 (dilute H2CO3 solution).
H2CO3 ⇌ H+ + HCO3- (buffer for rainwater)
Polluted regions - reserve supplies of acid and CB are not enough to counter the large amounts of acidic pollution.

33
Q

Name some uses of buffer solutions.

A

Electroplating, manufacture of dues, treating leather, checking pH meter readings (make up a buffer solution of certain pH and adjust the pH meter to read that value).

34
Q

Describe the use of buffer solutions in maintaining blood pH.

A

pH range 7.35 - 7.45, using HCO3-, haemoglobin and plasma proteins, H2PO4- and HPO4 2-.

carbon dioxide + water ⇌ H+ + HCO3-

  • catalysed by carbonic anhydrase
  • reverse when blood is in lung capillaries
  • overproduction of H+ can lead to acidosis -> coma
35
Q

What is the solubility product, Ksp?

A

Ksp = [C+ (aq)]a [A-(aq)]b

  • Product of concentrations of each ion in a saturated solution of a sparingly soluble salt at RTP, each raised to the power of their relative concentrations.
  • [solid] remains constant
36
Q

How is solubility usually quoted?

A

moldm-3 or g per 100g.

37
Q

How is the Ksp EQM established?

A

Undissolved ionic compound is in contact with a saturated solution of its ions. Rate of movement of solid -> solution = ions from solution -> solid.

38
Q

What is the relationship between magnitude of Ksp and solubility?

A

Larger Ksp = more soluble

39
Q

How can you use Ksp values to predict precipitation?

A

If the ionic concentrations multiply to:

  • < Ksp : no precipitate forms (unsaturated)
  • > Ksp: precipitate forms (supersaturated)
40
Q

How do you calculate solubility from Ksp?

A

1) Write EQM equation
2) Write EQM expression in terms of one ion only (s)
3) Substitute in value of Ksp
4) Calculate concentration

OR: s = √Ksp (AB)
s = 3√Ksp/4 (A2B or AB2)

41
Q

How do you calculate Ksp from solubility?

A
  • AB: s ^ 2 (because A and B are of equal concentrations)
  • A2B: 4s ^3 (because A is double the concentration of B)
  • AB2: 4s ^3 (because B is double the concentration of A)
42
Q

What is the common ion effect?

A

Reduction in solubility of a dissolved salt, achieved by adding a solution of a compound containing an ion in common with the dissolved salt. Often results in precipitation.
The solubility of an ionic compound in an aq solution containing its common ion is lower than that in water.

43
Q

Give an example of the common ion effect with AgCL (s).

A

AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)

  • adding NaCl: Cl- common ion, shifts EQM to the left
  • AgCl solubility has reduced as Ksp has been exceeded.
44
Q

How do you calculate the common ion effect in AB type salts?

A

eg. BaSO4 and Na2SO4 (0.1moldm-3)
1) Write the EQM equation and Ksp expression
2) [Ba2+] will = s, but [SO4 2-] will = s + 0.2
3) Ksp will therefore = s (s + 0.2), but s is negligible
4) Therefore Ksp = s x 0.2
5) Rearrange to solve for s.

45
Q

How do you calculate the common ion effect in AB2 type salts?

A

eg. Mg(OH)2 and NaOH (0.5moldm-3)
1) Write the EQM equation and Ksp expression
2) [Mg2+] will = s, but [OH-] will = 2s + 0.5
3) Ksp will therefore = s (2s + 0.5) ^2, but 2s is negligible
4) Therefore Ksp = s x 0.5 ^2
5) Rearrange to solve for s.

46
Q

How do you convert between moldm-3 and gdm-3?

A
  • mol/dm3 to g/dm3: multiply by Mr

- g/dm3 to mol/dm3: divide by Mr

47
Q

What is a partition coefficient, Kpc?

A

Kpc = [X (solvent A)] / [X (solvent B)]

  • EQM constant relating conc. of solute partitioned between two immiscible solvents at a particular temperature.
  • eg. ammonia (aq) ⇌ ammonia (organic), exchange of solute occurs via the interface.
48
Q

How do you calculate partition coefficients in titrations?

A

1) Write necessary equations
2) For a titration, find the concentration of the solution used in the titration
3) Use this value to calculate the mol in the original volume, after shaking
4) Subtract this from the initial mol of the unmixed solution
5) Use these mole values to find the concentrations
6) Enter these values into the Kpc expression.

49
Q

How do you calculate partition coefficients in successive extractions?

A

1) Calculate [X (org)], using (mass of X - m) as the mass extracted
2) Calculate [X (aq)], using m as the mass remaining
3) Enter Kpc value and rearrange to solve for m
4) Subtract m from the original mass of X
5) Repeat this for the next extraction, this time using m as the original mass of X, and a new m in (mass of X - m) for mass extracted.

Chemistry (12 decks)

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