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A Level Chemistry > Topic 12: Acid Base Equilibria > Flashcards

Topic 12: Acid Base Equilibria Flashcards

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1
Q

What are Bronsted-Lowry acids?

A
  • acids are proton donors
  • they release H+ ions when they’re mixed with water
  • H+ ions are always combined with H2O to form hydroxonium ions, H3O+
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2
Q

What is the reaction to form hydroxonium ions?

A
  • HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
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3
Q

What are Bronsted-Lowry bases?

A
  • bases are proton acceptors
  • when they’re in solution, they take hydrogen ions from water molecules
  • B(aq) + H2O(l) → BH+(aq) + OH-(aq)
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4
Q

What are strong acids and strong bases?

A
  • strong acids dissociate almost completely in water (nearly all the H+ ions will be released
  • e.g. HCl is a strong acid
  • strong bases dissociate almost completely too
  • e.g. NaOH
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5
Q

What are weak acids?

A
  • weak acids dissociate only very slightly in water
  • only small numbers of H+ ions are formed
  • an equilibrium is set up, which lies well over to the left
  • e.g. ethanoic acid
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6
Q

What are weak bases?

A
  • weak bases only slightly protonate in water
  • equilibrium lies on the left
  • e.g. ammonia
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7
Q

What are conjugate pairs?

A
  • conjugate pairs are species that are linked by the transfer of a proton
  • the species that has lost a proton is the conjugate base
  • the species that has gained a proton conjugate acid
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8
Q

What the definition of the standard enthalpy change of neutralisation?

A
  • the standard enthalpy change of neutralisation is the enthalpy change when the acid and alkali neutralise each other under standard conditions to form one mole of water
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9
Q

How do you calculate pH?

A
  • pH = -log10 [H+]
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10
Q

How do you work out [H+]?

A
  • 10-pH
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11
Q

What is the H+ concentration of strong monoprotic acids?

A
  • the H+ concentration is the same as the acid concentration
  • they dissociate fully
  • monoprotic: each mole of acid produces one mole of hydrogen ions
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12
Q

How many protons are released per molecule of diprotic acid?

A
  • two protons
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13
Q

How do you find the pH of a weak acid?

A
  • use Ka (the acid dissociation constant)
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14
Q

What are the assumptions you have to make to find Ka?

A
  • [HA(aq)](start) ≈ [HA(aq)](equilibrium)
  • [H+(aq)] ≈ [A-(aq)]
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15
Q

What is the ionic product of water?

A
  • KW = [H+] [OH-]
  • units are always mol2dm-6
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16
Q

How do you use KW to find the pH of a strong base?

A
17
Q

What is KW at 25°C (298K)?

A
  • Kw = 1.0 x 10-14 mol2dm-6
18
Q

What is pKw at standard conditions?

A
  • always 14.00
19
Q

How do you calculate pKa and Ka from pKa?

A
  • pKa = -log10 Ka
  • Ka = 10-pKa
20
Q

How do you calibrate a pH meter?

A
  • place the bulb of the pH meter into deionised water and allow reading to settle
  • adjust reading so that it read 7.0
  • do the same with a standard solution of pH4 and pH 10
  • rinse the probe with deionised water between each reading
21
Q

How does the pH of a strong acid change when diluted?

A
  • diluting the acid by a factor of 10 increases the pH by 1
  • you can see by using pH = -log10[acid]
22
Q

How does the pH of a weak acid change when diluted?

A
  • diluting a weak acid by a factor of 10 increases the pH by 0.5
23
Q

Describe these titration curves

A
  1. strong acid/strong base
  2. strong acid/weak base
  3. weak acid/strong base
  4. weak acid/weak base
24
Q

How do you explain the shapes of titration curves?

A
  • the initial pH depends on the strength of the acid
  • a strong acid titration will start at a much lower pH than a weak acid
  • to start with, the addition of small amounts of the base have little impact on the pH of the solution
  • final pH depends on the strength of the base
  • the stronger the base, the higher the final pH
25
Q

What is the equivalence point?

A
  • the centre of the equivalence line
  • [H+] ≈ [OH-]
  • the acid is just neutralised
  • a tiny amount of base causes a sudden, big change in pH
26
Q

How do you find the pKa of a weak acid using titration curves?

A
  • find the pH at the half-equivalence point
  • half of the acid has been neutralised
  • at the half-equivalence point, [HA] = [A-]
  • therefore, Ka = [H+] and pKa = [pH] (see image)
27
Q

What is a buffer?

A
  • a buffer is a solution that minimises changes in pH when small amounts of acid or base are added
28
Q

What are the two ways of making acidic buffers?

A
  1. Mix a weak acid with the salt of its conjugate base:
    e. g. ethanoic acid and sodium ethanoate
  2. Mix an excess of weak acid with a strong base
    e. g. ethanoic acid and sodium hydroxide
29
Q

What is the equilibrium set up for acidic buffers?

A
30
Q

How do acidic buffers work?

A
  • the conjugate base accepts any extra H+ and conjugate acid release H+ if there is too much base
  • if you add a small amount of acid, the H+ concentration increases
  • most of the extra H+ ions combine with CH3COO- ions to form ethanoic acid
  • this shifts equilibrium to the left, reducing the H+ concentration close to original value
  • if a small amount of base is added, the OH- concentration increases
  • most of the extra OH- ions reaction with H+ ions to form water
  • this causes more CH3COOH to dissociate to form H+ ions, shifting equilibrium to the right
31
Q

What are alkaline buffers made of?

A
  • a weak base and one of its salts
  • e.g. a mixture of ammonia solution (a base) and ammonium chloride (a salt of ammonia)
32
Q

How do alkaline buffers work?

A
33
Q

How can buffer solutions been seen on a titration curve?

A
34
Q

Why are buffer solutions important in blood?

A
  • blood needs to be kept at pH 7.4
  • this is controlled by carbonic acid-hydrogencarbonate buffer system
  • carbonic acid dissociates into H+ ions and HCO3- ions

H2CO3(aq) ⇔ H+(aq) + HCO3-(aq)

  • if the concentration of H+ rises in the blood, then the HCO3- ions will react with the excess H+ ions
  • the equilibrium will shift to the left, reducing the H+ concentration to its original value
  • vice versa
  • H2CO3 concentration is controlled by respiration

H2CO3(aq) ⇔ H2O(l) + CO2(aq)

  • HCO3- levels are controlled by kidneys
35
Q

What assumptions need to be made to calculate the pH of a buffer solution?

A
  • the salt of the conjugate base is fully dissociated, so assume that the equilibrium concentration of A- is the same as the initial concentration of the salt
  • HA is only slightly dissociated, so assume that its equilibrium concentration is the same as its initial concentration
36
Q

How do you calculate the concentration of salt and acid or base to create a buffer with a specific pH?

A

A Level Chemistry (19 decks)

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