Oceans - Acid/Base Equilibria Flashcards

1
Q

what is an acid?

A

a substance that donates a hydrogen ion in solution

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2
Q

what is a base?

A

a substance that can accept a H+ in solution

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3
Q

what is a conjugate base?

A

an acid that has donated a proton but then can re-accept a proton back

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4
Q

what is a conjugate acid?

A

a base that has accepted a proton but could donate it back into solution

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5
Q

how can water act as an acid?

A

H2O -> OH- + H+

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6
Q

how can water act as a base?

A

H2O + H+ -> H3O+

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7
Q

what is an alkali?

A

a base dissolved in water

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8
Q

what is an oxonium ion?

A

H3O+

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9
Q

what are examples of acids that don’t dissolve in water?

A

benzoic acid

carboxylic acids

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10
Q

what is the equation for the dissociation of a strong acid?

A

HA -> H+ + A-

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11
Q

what is pH

A

the representation of protons in solution

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12
Q

what is the equation for calculating pH?

A

pH = -log10[H+]

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13
Q

what is a strong acid?

A

an acid that dissociates fully in solution

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14
Q

what are examples of strong acids?

A

hydrochloric acid

sulfuric acid

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15
Q

what is the relationship between concentration of the acid and pH in strong acids?

A

1 x 10-x molar solution = pH x

eg

0.001 molar solution = pH 3

because -log10[0.001] = 3

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16
Q

what is the equation for calculating the concentration of hydrogen ions from pH?

A

[H+] = 10-pH

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17
Q

what is a weak acid?

A

an acid that doesn’t dissociate fully in solution

18
Q

what is the equation for the dissociation of a weak acid?

A

HA ⇔ H+ + A-

19
Q

what is an example of a weak acid?

A

carboxylic acids

20
Q

what is Ka?

A

the acid dissociation constant

similar to Kc

21
Q

what assumptions can be made when calculating the Ka for weak acids?

A
  1. conc of conjugate base formed = conc of hydrogen ions present
  2. initial conc of acid = conc of acid at equilibria
22
Q

when is the assumption that the conc of the conjugate base is equal to the conc of the hydrogen ions in weak acids less valid?

A

in dilute acids - more dissociation of water into hydroxide + hydrogen

23
Q

when is the assumption that the initial conc of a weak acid is equal to the conc of the acid at equilibrium less valid?

A

when the acid is stronger / when Ka is bigger

24
Q

how can Ka be calculated using the example of the dissociation of ethanoic acid?

A

[CH3COO-] [H+]

[CH3COOH]

25
Q

what is the dissociation of a strong base?

A

dissociates completely

26
Q

what is Kw?

A

ionisation product of water

27
Q

what is the ionisation product of water?

A

the product of the conc of H+ ions and the conc of hydroxide ions from the dissociation of water

28
Q

how can Kw be calculated?

A

Kw = [H+] [OH-]

also given on datasheet

29
Q

what is the expression for the Ka of water?

A
30
Q
A
31
Q

what is a buffer solution?

A

a solution that resists change in pH for small additions of acid or alkali

32
Q

how are simple buffer solutions made?

A

weak acid mixed with salt of a strong base and weak acid

33
Q

if ethanoic acid is used in a buffer, how will it resist hydroxide ions being added to the solution?

A

ethanoic acid deprotonated by reacting with hydroxide ions to form ethanoate ions and water

34
Q

if ethanoic acid is used in a buffer solution, how will it resist the change in pH from hydrogen ions being added?

A

ethanoate ions from the salt will react with the hydrogen ions, producing more ethanoic acid

35
Q

what assumptions can be made about simple buffers, using example HA ⇔ H+ + A-

A

all A- ions come from the salt because the weak acid has low levels of dissociation

almost all HA molecules put into the buffer remain unchanged

36
Q

what is the equation for calculating the Ka of a buffer solution?

A
38
Q

what is the solubility product / Ksp?

A

represents conditions for equilibrium between sparingly soluble solid and its saturated solution

39
Q

using example:

CaCO3(s) ⇔ Ca2+(aq) + CO32-(aq)

give the expression for the Ksp of calcium carbonate

A

Ksp (CaCO3(s)) = [Ca2+(aq)] [CO32-(aq)]

40
Q

what will happen when solids are added to equilibria involving solubility products?

A

the equilibrium will not shift because the solution is saturated - causing the equilibrium to be unaffected by any solids present

41
Q

what can Ksp be used to predict?

A

whether a precipitate will be formed from a solution

42
Q

using example:

CaCO3(s) ⇔ Ca2+(aq) + CO32-(aq)

explain when Ksp can be used to predict a formation of a precipitate

A

if [Ca2+(aq)] x [CO32-(aq)] gives a value in excess of Ksp, then CaCO3 will precipitate out of the solution

43
Q

using example:

CaCO3(s) ⇔ Ca2+(aq) + CO32-(aq)

explain how Ksp can be used to predict that a precipitate will not be formed

A

if [Ca2+(aq)] x [CO32-(aq)] is < or = Ksp, a calcium carbonate precipitate will not be formed - ions will stay in solution