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AQA chemistry A level notes Diana Tut > acids & bases > Flashcards

acids & bases Flashcards

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1
Q

define a Bronsted-Lowry acid

A

H+ donor (not pH dependent so the pH doesn’t need to be <7)

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2
Q

define a Bronsted-Lowry base

A

H+ acceptor (not pH dependent so the pH doesnt need to be >7)

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3
Q

define lewis acid

A

e- pair acceptor

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4
Q

define lewis base

A

e- pair donor

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5
Q

what happens in a Bronsted-Lowry acid-base reaction?

A

involves the transfer of H+ from B-L acid to B-L base

using this definitions, not all acids and bases react via a neutralisation reaction (i.e the products are not always a salt + water)

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6
Q

identify the Bronsted-Lowry acid and base in the following reaction:

KOH + HCOOH → HCOOK + H2O

A
  • HCOOH becomes HCOOK so has lost a H+
  • therefore HCOOH is the B-L acid
  • KOH dissociates into OH- and K+ , the OH- accept H+ to become H2O
  • therefore OH- is the B-L base
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7
Q

define conjugate acid-base pair

A

2 species (one a B-L base and the other a B-L acid) which differ by a H+

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8
Q

identify the conjugate acid-base pair in this reaction and identify which species within the pair is a B-L acid/ base:

NH3 + HCl → NH4Cl

A

NH4Cl is made from NH4+ and Cl- ions

NH3 & NH4 + are conjugate acid-base pair, where the NH3 is the base and NH4 + is the acid

HCl & Cl- are the second conjugate acid-base pair, where HCl is the acid and Cl- is the base

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9
Q

define amphoteric

A

a substance which can act both as a B-L base / acid

e.g H2O can accept H+ to form H3O+ (hydronium or oxonium ions) or donate H+ to form OH- ions (hydroxide ions)

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10
Q

define monoprotic acid

A

a strong acid which produces 1 mole of H+ for every mole of acid

only one H atom in the formula

e.g HCl and HNO3

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11
Q

define diprotic acid

A

a strong acid which produces 2 moles of H+ for every 1 mole of the acid

contains 2 H atoms in its formula

e.g H2SO4

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12
Q

give the formula used to calculate the pH of strong acids

A

pH = -log [H+]

always given to 2 d.p.

pH of very strong acids may be negative because y= -logx < 0 when [H+] > 1

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13
Q

give the formula used to find the H+ conc of strong acids

A

[H+] = 10-pH

give to 3 s.f.

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14
Q

calculate the new pH when 100cm3 of water is added to 50 cm3 of 0.1 mol dm-3 HNO3

A
  • first determine the original [H+]
    • original [H+] = [HNO3] (bc its a monoprotic acid) = 0.1 mol dm-3
  • then determine the new [H+]
    • new [H+] = original [H+] x old volume/ new volume
    • new [H+] = 0.1 x 50/150 = 0.1 x 1/3 = 0.0333 mol dm-3
  • find the new pH using the new [H+]
    • pH = -log 0.0333 = 1.48
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15
Q

explain how to convert between g dm3 to mol dm-3

A

mol dm-3 x Mr → g dm3

g dm3 divided by Mr → mol dm-3

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16
Q

define a neutral solution

A

when [OH-] = [H+] (hydroxide ion conc is equal to hydrogen ion conc)

pH does not have to be 7 exactly

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17
Q

explain what is meant by Kw (the ionic product of water) and the factor that affects it

A

as water molecules can act as both B-L acids and B-L bases, in pure water the 2 following equilibirum reactions occur:

2H2O ⇌ H3O+ + OH-

H2O ⇌ H+ + OH-

Kw = [H+] [OH-]

pure water only dissociates very slightly

Kw is only affected by temperature

fowards reaction endo so backwards reaction exo

increasing temp shifts position of eqm to the right to oppose change by lowering temperature

both [H+] and [OH-] increase so Kw increases also

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18
Q

how do you calculate the hydrogen ion conc of neutral solutions?

A

[H+] = √ Kw

19
Q

how to calculate pH of bases using pKw

A

pKw = pH + pOH

which means:

-log Kw = -log [H+] + -log [OH-]

20
Q

calculate the pH of a 2.00 gdm-3 NaOH solution given that pKw = 14.00 at 25º

A
  • 2.00 gdm-3 NaOH = 2/40 = 0.05 mol dm-3 NaOH
  • [OH-] = 0.05 mol dm-3 (1 mol NaOH forms 1 mole OH- )
  • -log [OH-] = -log 0.05 = 1.30
  • pH = pKw - pOH = 14.00 - 1.30 = 12.70
21
Q

explain how to calculate the pH of strong bases

A
  • determine the [OH-] from the conc of the base
  • use [H+] = Kw / [OH-] to find the [H+]
  • calculate the pH using
    • pH = -log [H+]
22
Q

explain how to calculate the pH of a new solution when water is added to a strong base

A
  • determine the original [OH-]
  • determine the new [OH-] using
    • new [OH-] = original [OH-] x original volume/ new total volume
  • determine the [H+] using Kw and the new [OH-]
    • [H+] = Kw / [OH-]
  • calculate the pH using
    • ​pH = -log [H+]
23
Q

how do you convert between degrees celius and kelvin?

A

degrees celsius + 273 = kelvin

24
Q

define weak acid

A

an acid partly dissociates in aqueous solution

25
Q

what is Ka and how is it used

A
  • weak acids partly dissociate according to the following equilibrium:
    • HA ⇌ H+ + A-
  • Ka is the same as the Kc expression for the eqm
    • Ka = Kc = [H+][A-] / [HA]
    • Ka is the acid dissociation constant - it gives us a measure of the dissociation of the weak acid
    • its units are mol dm-3
  • for a weak acid at eqm (when the weak acid has been added to water alone with nothing else)
    • [H+] = [A-] (bc of molar ratio)
    • [HA] = orgininal conc of the weak acid
    • so Ka can be written as Ka = [H+]2 / [HA]initial
  • this equation can be rewritten to find the [H+] and hence the pH of a weak acid
26
Q

what is pKa ?

A

p means -log

so pKa = -log Ka

therefore Ka = 10-pKa

27
Q

explain how Ka /pKa relate to weak acid strength and pH?

A

As long as [HA] remains constant:

  • in a very weak acid, position of eqm lies very far to the left as very few weak acid molecules dissociate
  • [HA] is very high so Ka is small (because Ka = [H+] [A-] / [HA])
  • [H+] is very low so the pH is high (because pH = -log[H+])
  • this means the larger the value of Ka, the stronger the weak acid and the lower the pH

the bigger the pKa value however, the higher the pH

28
Q

explain why methanoic acid is a stronger acid than ethanoic acid

methanoic acid = CHCOOH

ethanoic acid = CH3COOH

[6]

A
  • weak acids partially dissociate in water in the following reaction
    • HA ⇌ H+ + A-
  • strength of an acid depends on its ability to dontate protons
    • ​therefore the further the position of eqm is to the right, the stronger the weak acid (because more protons are donated)
  • in methanoic acid, COOH group is bonded to a H atom
  • in ethanoic acid, the COOH group is bonded to a CH3 group
  • this methyl group has a positive inductive effect meaning that it donates electron towards to O-H bond of the carboxyl group
  • higher e-density around the O-H bond of the carboxyl group than that in the methanoic acid
  • the H atom bonded to the carboxyl group in methanoic acid has no positive inductive effect = density around the OH bond of the carboxyl group is unaffected
  • OH bond in methanoic acid is weaker than in ethanoic acid so less energy is required to break the OH bond in methanoic acid
  • methanoic acid donates more H+ ions in water than ethanoic acid
29
Q

explain how to calculate the pH of a mixture of strong acid and and base

A
  1. calculate the moles of H+ from the acid
  2. calulate the moles of OH- from the base
  3. calculate the moles of excess H+ or OH-
  4. calculate excess [H+] or excess [OH-] using:
    • excess [H+] = excess mols of H+ / total volume in dm3
    • excess [OH-] = excess moles of OH- / total volume in dm3
  5. calculate the [H+] using Kw/[OH-] if you have excess OH-
  6. calculate pH using -log[H+]
30
Q

explain how to find the pH of a solution when a weak acid reacts with a strong base

A
  • you can tell the question involves a weak acid if Ka or pKa is given
  • for every mole of OH- added, one mole of HA is used up and one mole of A- is formed
    • therefore moles of OH- = moles of A-

method:

  1. calculate moles of HA (because it is a weak acid it will mostly remain undissociated)
    • moles HA = volume x concentration
  2. calculate the moles of OH- formed by the base (check if it forms 2 moles of OH- for every mole of base)
  3. calculate the moles of excess HA or OH-
  4. calculate the moles of A- formed
    • moles A- = moles OH-
  5. calculate excess [HA] and [A-] formed using total volume in dm3
    • excess [HA] = excess moles of HA/ total vol
    • [A-] formed = moles A- / total vol
  6. use Ka or Kw to find the [H+]
    • [H+] = Ka [HA] / [A-]
    • [H+] = Kw / [OH-]
  7. find the pH using -log[H+]
31
Q

what are the uses of buffers?

A
  • keep pH at optimal level for enzymes
  • keep blood pH constant
    • CO2 produced by respiration lower the pH by forming H+ ions in aqueous solution
    • HCO3- react with the H+ ions thus removing then
    • H+ + HCO3- ⇌ H2CO3 (aq) (carbonic acid)
  • shampoo - counteract the alkilinity of soap and prevent irritation
  • baby lotion - maintain pH at a level that prevent bacteria from multiplying
32
Q

define buffer solution

A

maintains the pH within narrow limits when a small volume of acid or base is added

33
Q

define an acidic/basic buffer

A

maintains the pH of a solution approximately constant at a value below/above 7.00

34
Q

explain how acidic buffers are made

A
  • weak acid + salt of that weak acid (the conjugate base)
  • includes a conjugate acid base pair
    • B-L acid = weak acid
    • B-L base = negative RCOO- ion from the salt
  • conc of acid and salt together is much greater than the H+ conc
  • the pH value of the acidic buffer = ratio of [acid]:[salt]
35
Q

explain how basic buffers are made

A
  • weak base + salt of weak base
  • contains a conjugate acid base pair
    • B-L acid = positive ion from salt
    • B-L base = weak base
  • conc of base and salt is much greater than the OH- conc
  • pH of basic buffer = ratio [base] : [salt]
36
Q

explain how an acidic buffer works

A
  • weak acid dissociate via this eqm in aqueous solution
    • HA ⇌ H+ + A-
  • acidic buffers are made from HA and A- (from the salt of the weak acid)
  • [HA] is high and [A-] is higher
  • when small amounts of acid or alkali are added, the ration [HA] : [A-] remains approximately constant so pH hardly changes (because [H+] doesn’t change by much and pH = -log [H+] )
37
Q

explain how you would calculate the pH of an acidic buffer made from excess weak acid and a strong base solution

100 cm3 of 0.0575 mol dm-3 HNO3 and 50 cm3 of 0.0428 mol dm-3 NaOH (aq)

Ka of HNO3 = 7.20 x 10-4 mol dm-3

A
  • first find the moles of HA, OH- and so the excess moles of HA
    • moles HA = 5.75 x 10-3
    • moles OH- = 2.14 x 10-3
    • excess moles HA = 3.61 x 10-3
  • find the excess [HA] and [A-]
    • [HA] = 0.02407
    • [A-] = 0.01427
  • find the [H+] using Ka
    • (7.2x10-4x0.02407)/0.01427 = 1.21 x 10-3
    • pH = -log 1.21x10-3 = 2.92
38
Q

explain how you would calculate the pH of acidic buffers made from weak acid and aqueous salt of weak acid

0.1 mol dm-3 ethanoic acid (aq) and 0.4 mol dm-3 sodium ethanoate (aq)

Ka of ethanoic acid = 1.78 x 10-5 mol dm-3

A
39
Q

explain how you would find the pH of an acidic buffer made from weak acid and solid salt of weak acid

2.05g of sodium ethanoate and 0.5 dm3 of 0.01 mol dm-3 ethanoic acid

Ka of ethanoic acid = 1.74 x 10-5 mol dm-3

A
  • first find the moles of A- present (same as moles of salt)
    • moles of A- = m/Mr =2.05/82 = 0.025
  • find [A-] using the total volume
    • [A-] = 0.025/0.5 = 0.05
  • find [H+] using Ka
    • [H+] = (1.74 x 10-5 x 0.01) / 0.05 = 3.48 x 10-6
  • find pH
    • pH = -log (3.48 x 10-6) = 5.46
40
Q

explain how you would find the new pH of a buffer when a small amount of acid is added

A
  • new moles HA = initial moles HA + moles H+
    • because position of eqm shifts to the left to remove added H+
      • HA ⇌ H+ + A-
  • new moles A- = inital moles A- - moles H+ added
  • calculate new [H+] using
    • ​[H+] = Ka x new moles HA / new moles A-
    • no need to find concentrations as the new total volume stays the same
  • find pH
41
Q

explain how you would find the new pH of a buffer solution when a small amount of base is added

A
  • new moles HA = initial moles HA - moles OH- added
    • because OH- added react with H+ so the position of eqm shifts to the right to replace lost H+
  • new moles A- = initial moles A- + moles OH- added
  • calculate new [H+] using:
    • [H+] = Ka x new moles HA/ new moles A-
  • find pH
42
Q

explain why the pH of an acidic buffer solution remains almost constant when you add a small amount of base?

A
  • HA⇌H+ + A-
  • OH- from the base react with the H+ in the solution to form H2O
  • position of eqm shifts to the right to replace lost H+
  • added OH- are removed and lost H+ are reformed

further detail for questions with more marks

  • [HA] decreases slightly , [A-] increases slightly
  • ratio [HA] : [A-] remains roughly constant bc :
    • [HA] + [A-] in buffer is much greater than the [H+]
  • therefore [H+] remains roughly constant bc
    • [H+] = Ka x [HA] / [A-]
43
Q

explain why the pH of a buffer solution remains almost constant when you add a small amount of acid

A
  • HA ⇌ H+ + A-
  • position of eqm shifts to the left to remove added H+
  • added H+ also reacts with the A- to reform HA
  • added H+ therefore removed

further detail for qs with more marks

  • [A-] decreases slightly , [HA] increases slightly
  • ratio [HA] : [A-] remains constant bc:
    • [HA] + [A-] much greater than [H+]
  • so [H+] in expression remains roughly constant
    • [H+] = Ka x [HA] / [A-]
44
Q

find pH of a mixture of 20.0cm3 of 0.1 moldm-3 butanoic acid and 40.0cm3 of 0.025moldm-3 NaOH

pKa for butanoic acid = 4.82

A

moles HA = 20 x 10-3 x 0.1 = 0.002

moles OH- = 40 x 10-3 x 0.025 = 0.001

moles OH- = 1/2 moles HA

therefore butanoic acid is at half equivalence

pH at half equivalence = pKa = 4.82

AQA chemistry A level notes Diana Tut (15 decks)

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