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A level- Chemistry > Acids and pH > Flashcards

Acids and pH Flashcards

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1
Q

Define an acid

A

A species that releases H+ ions in aqueous solution

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2
Q

Define a base

A

A compound that neutralises an acid to form a salt- proton acceptor

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3
Q

Define an alkali

A

A water soluble base that releases OH- ions in solution

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4
Q

Define salt

A

The product of a reaction in which the H+ ions from the acid are replaced by metal or ammonium ions

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5
Q

What observations would you expect to see if solid magnesium carbonate is affed to excess hydorchloric acid

A
  1. White solid is added to colourless solution
  2. Effervescence can be seen and the solid dissolves
  3. Colourless solution remains
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6
Q

Why can an H+ ion not exist on its own and much immediately attatch to another material

A

Strong electrostatic force of attraction attract a lone pair of electrons from another molecule

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7
Q

Explain whether 100% sulfuric acid is acidic

A

No- requires water to release H+ ions

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8
Q

Describe what a monobasic, dibasic and tribasic acid is

A
  1. Mono- transfer of one H+ ion- HNO3
  2. Di- transfer of two H+ ions- H2SO4
  3. Tri- transfer of 3H+ ions- H3PO4
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9
Q

Define conjugate acid

A

A species formed when a proton is added to a base

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10
Q

Define conjugate base

A

A species frmed when a proton is removed from an acid

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11
Q

For the following reactions state the acid, base, conjugate acid and conjugate base:

  1. HNO3 + H2O ↔ H3O+ + NO3-
  2. H2SO4 + HCl ↔ H2Cl+ + HSO4-
A
  1. Acid- HNO3, Base- H2O, Conjugate acid-H3O+, Conjugate base- NO3-
  2. Acid- H2SO4, Base- HCl, conjugate acid- H2Cl+ conjugate base- HSO4-
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12
Q

Define conjugate base pairs

A

A conjugate acid-base pair contains two species that can be inter converted by the transfer of a proton.

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13
Q

What is the conjugate base of these acids:

HCl,NH3, H2PO4-, H3O+

A
  1. HCl= Cl-
  2. NH3= NH2-
  3. H2PO4- = HPO4 2-
  4. H3O+ = H2O
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14
Q

Write an equation to show how each of these bases form conjugate acids
HSO3-, HPO4 2-, CO3 2-, H2PO4-

A
  1. H+ + HSO3 ⇌ H2SO4
  2. H+ + HPO4 2–⇌ H2PO4–
  3. H+ + CO3 2– ⇌ HCO3–
  4. H+ + H2PO4– ⇌ H3PO4
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15
Q

Define strong acid

A

Acid that completely dissociates in solution

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16
Q

Define weak acid

A

Acid that partially dissociates in solution

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17
Q

Give two examples of weak acids and 2 examples of strong acids

A

Weak- CH3COOH, NH4

Strong- HCl, HNO3

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18
Q

Define pH

A
  1. -log( [H+ (aq)] )

2. - sign shows it is an inverse relationship, pH goes down as strength goes up

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19
Q

What does a change of one pH number represent

A

It is equal to 10 times difference in [H+] concentration because it is a log scale.

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20
Q

How would you dilute a solution from pH 1 to pH 4

A

Requires dilution 10 x 10 x 10= 1000 times

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21
Q

How can you find the pH using the H+ concentration and the reverse

A
  1. -log ([H+])

2. 10 ^ -pH = [H+]

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22
Q

What is the relationship between the concentration of a strong monobasic acid and [H+]

A

They are equal, so the pH of a strong acid can be calculated directly from the concentration of the acid.

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23
Q

Define the Acid dissociation constant Ka

A 
Ka= ( [H+ (aq)] * [A- (aq)] ) / [HA (aq)]
Units= mol dm-3
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24
Q

What is the link between Ka value and acid strength

A

The larger the Ka value the greater the dissociation and the greater the acid strength.

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25
Q

Show the general form for the dissociation of any weak acid HA

A

HA (aq) ↔ H+ (aq) + A- (aq)

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26
Q

Describe how to convert from Ka to pKa and back

A
  1. pKa= -log(Ka)

2. Ka= 10^-pKa

27
Q

What is the relationship between pKa and acid strength

A

A pKa increases the acid strength will decrease

28
Q

As you add chlorine atoms to ethanoic acid what happens to the strength of the acid and why

A
  1. The strength increases

2. Electronegative chlorine atoms draw electron density away from the carboxylic acid group weakening the O-H bond

29
Q

What approximations/ assumptions are made when calculating the pH of weak acids

A
  1. Dissociation of water is negligible[H+]=[A-]
    a) Problems- The approximation breaks down for very weak acids or very dilute solutions.
  2. For a weak acid very little dissociation occurs: [HA]start = [HA]eqm
    a) Problems- Approximation is not justified for stronger weak acids with Ka>10^-2 and for very dilute solutions.
30
Q

Using assumptions write the Ka expression for a weak acid

A

Ka= [H+]^2/ [HA]

31
Q

Write the steps to calculate pH using concentration of acid and Ka value

A
  1. Multiply Ka value by concentration of acid to find [H+]^2
  2. Root this value
  3. -log([H+])
32
Q

A 0.01 mol dm-3 solution of bromic acid HBrO has pH 5.35 show that is is a weak acid

A
  1. If strong acid [H+]= 0.01
  2. So pH= -log(0.01)= 2
  3. But pH is 5.35 so therefore not a strong acid
33
Q

What is likely to happen to the validility of the approximation of pH of a weak acid as the Ka value increases

A
  1. Less valid
  2. As proportion of acid molecules which dissociates increases
  3. Meaning greater difference between initial and equilibrium concentration of undissociated acid.
  4. Assumptions are not as valid.
34
Q

Describe how the Ka value for a weak acid can be determined experimentally

A
  1. Prepare a standard solution of the weak acid of known concentration
  2. Measure the pH of the standard solution using a pH meter
  3. Find [H+]= 10^-pH
  4. Calculate Ka from [H+] and [HA]
35
Q

Define ionic product of water Kw

A

Kw= [H+ (aq)] * [OH- (aq)]

Units- mol2 dm-6

36
Q

What is the value of Kw at 298K

A

1.00 x10^-14 mol2 dm-6

37
Q

What is the importance of Kw

A
  1. It is an equilibrium constant that controls the concentrations of H+ and OH- ions in aqueous solutions
    2.
38
Q

When are solutions acidic, neutral and alkaline

A
  1. Acidic [H+] > [OH-]
  2. Alkaline [H+] < [OH-]
  3. Neutral [H+] = [OH-]
39
Q

How can you find the [H+] and [OH-] in a solution with a pH of 3.25

A
  1. Calculate [H+]= 10^-pH

2. Calculate [OH-] from Kw and [H+]

40
Q

How can you calculate the pH of a solution of a strong base e.g NaOH

A
  1. Convert [NaOH] into [OH-], NaOH is a strong monobasic alkali [OH-]= [NaOH]
  2. Use Kw and [OH-] to find [H+]
  3. Then find the pH using the [H+]
41
Q

Define buffer solution

A

A system that minimises pH changes on addition of small amounts of an acid or a base

42
Q

What do buffer solutions contain

A
  1. A weak acid- removes added alkali

2. Its conjugate base- removes added acid

43
Q

Describe how to prepare a buffer solution from a weak acid and its salt for ethanoic acid

A
  1. Mixing solution of ethanoic acid and one of its salts e.g. sodium ethanoate
  2. When ethanoic acid is added to water, the acid partially dissociates and the amount of ethanoate ions is very small- ethanoic acid is the source of the weak acid component of the buffer solution.
    CH3COOH ↔ H+ + CH3COO-
  3. Salts of weak acids are ionic compounds and provide a source of the conjugate base. When added to water the salt completely dissolves- dissociates into its ions so the salt is the source of the conjugate base component.
    CH3COONa ↔ CH3COO- + Na+
  4. When CH3COO- ions are added to CH3COOH, the equilibrium position for ethanoic acid shifts further to teh left reducing the small concentration of H+ ions leaving a solution mainly containing the two components.
    CH3COOH ↔ H+ + CH3COO-
44
Q

Describe how to prepare a buffer solution by half neutralisation for ethanoic acid

A
  1. Buffer solution can be prepared by adding an aqueous solution of an alkali e.g. NaOH to an excess of weak acid.
  2. The weak acid is partially neutralised forming the conjugate base.
  3. Some of the weak acid is left over unreacted.
  4. The resulting solution contains a mixture of the salt of the weak acid and any unreacted weak acid.
45
Q

How does the buffer solution work when acid is added

A
  1. [H+] increases
  2. H+ ions react with the conjugate base
  3. The equilibrium position shifts to the left removing most of the H+ ions
    HA ↔ H+ + A-
46
Q

How does the buffer solution work when alkali is added

A
  1. [OH-] increases
  2. The small concentration of H+ ions reacts with the OH- ions: H+ + OH- →H2O
  3. HA dissociates shifting the equilibrium position to the right to restore most of the H+ ions.
47
Q

When is a buffer solution most effective

A
  1. When there are equal concentrations of weak acid and its conjugate base [HA] = [A-]
48
Q

Write the 2 equations to calculate the pH of a buffer solution

A
  1. [H+]= Ka * ([HA]/[A-])

2. pH = pKa + log ([A-]/[HA])

49
Q

What values does the pH of blood plasma need to be between and what are the buffers used.

A
  1. 7.35-7.45

2. (H2CO3/ HCO3-)

50
Q

What happens if the pH of blood is too high or low

A
  1. If below 7.35 people can develop a condition called acidosis, which can cause fatigue, shortness of breath and in extreme cases shock or death.
  2. If above 7.45- condition called alkalosis which can cause muscle spasms, light-headedness and nausea.
51
Q

Explain how the buffer solution in blood works to minimise the changes in pH upon addition of small amounts of acid or alkali.

A

H2CO3 ↔ H+ + HCO3-
1. If acid is added, then the position of the above equilibrium will move to the left, to produce more H2CO3 and use up most of the added H+
H+(aq) + HCO3- (aq) ↔ H2CO3(aq)
2. If alkali is added, then OH- ions will combine with H+ to produce water
H+(aq) + OH- (aq) → H2O(l)
3. More H2CO3 will dissociate to produce more H+ to replace most of that which has formed water.
H2CO3(aq) ↔ H+(aq) + HCO3-(aq)

52
Q

How do you calculate the concentration ratio of A-/HA using the Ka and pH

A
  1. Write our the expression for Ka
  2. Divide the Ka value by [H+] : Ka/[H+] = [A-]/ [HA]
  3. Convert pH to H+
  4. Then calculate by dividing the value of Ka by the value of [H+]
53
Q

How do you calculate the pH of an alkaline buffer solution e.g. ammonia and ammonium ion

A
  1. Ammonium ion is the acid form and ammonia is the base form
  2. Write Ka and use this to calculate [H+]
  3. Calculate pH from this.
    Note- for alkalis the positive ion is the HA
54
Q

You are supplied with sodium hydroxide and ethanoic acid each solution has concentration of 0.1 mol dm-3. State how you would prepare a buffer solution with pH=pKa of ethanoic acid

A
  1. NaOH and CH3COOH, we want pH= pKa so +log([A-]/[HA])= 0
  2. This would happen if [A-]=[HA] because both NaOH and CH3COOH have a concentration of 0.1 mol dm-3, you would need the same volume of both.
55
Q

When does pKa= pH

A
  1. At t=1/2 at the end point

2. And when [A-]=[HA]

56
Q

How would you use a pH meter

A
  1. Using a pippette add a measured volume of acid to a conical flask
  2. Place electrode of teh pH meter in the flask
  3. Add the aqueous base 1cm^3 at a time (from burette), after each addition swirl the contents and record the pH and total volume added
  4. Repeat until the pH starts to change more rapidly then add dropwise for each reading until the pH changes less rapidly.
  5. Then add the base 1 cm3 at a time until excess has been added and the pH has been basic with little change for several additions.
  6. Draw a graph of pH against total volume of base added
57
Q

Describe a pH titration curve

A
  1. Steep stair shaped- vertical section in the middle
  2. When the base is first added the acid is in great excess and the pH increases very slowly, as the vertical section is approached the pH starts to increase more quickly as the acid is used up more quickly.
  3. Eventually the pH increases rapidly during the addition of a very small volume of base, producing the vertical section.
  4. After the vertical section, the pH will rise very slightly as the base is now in great excess.
  5. Can be plotted with an acid adding to a base and is same shape just opposite way round - Starts high
58
Q

Describe what the equivalence point is

A
  1. The point in a titration at which the volume of one solution has reacted exactly with the volume of the second solution.
  2. The equivalence point is the centre of the vertical section of the pH titration curve.
59
Q

Describe what the end point is

A
  1. The point in a titration where the indicator changes colour- Shows equal concentration of the weak acid and conjugate base.
60
Q

Describe why an indicator changes colour

A
  1. An indicator is a weak acid , the equilibrium is shifted towards the weak acid in acidic conditions or towards the conjugate base in basic conditions, changing the colour as it does.
  2. e.g Methyl orange
    a) If a strong base is added to a strong acid, methyl orange is initially red as the presence of H+ ions forces the equilibrium position to the left.
    b) as OH- ions are added the OH- react with the H+ in the indicator
    c) The weak acid dissociates shifting the equilibrium position to the right.
    d) The colour changes first to orange at the end point and finally to yellow as the equilibrium position is shifted to the right.
61
Q

What happens at the end point

A
  1. At the end point pKa value of HA= pH
62
Q

Draw the curves for strong acid- base, weak acid-base, weak acid-strong base and strong acid-weak base

A
  1. Strong acid and base-curve starts at around 1 and ends around 12. Equivalence point is around 7,
  2. Weak acid and weak base- starts at around 3 and ends at around 9, very small vertical section
  3. Weak acid and strong base- starts at around 3 and ends at around 13, equivalence point is around 9
  4. Strong acid and weak base- starts at around 1 ends at around 9, equivalence point at around 5
63
Q

What makes an indicator suitable for a titration

A
  1. Must have indicator that has a colour change that coincides with the vertical section of the pH titration curve
  2. Ideally the endpoint and equivalence point would coincide- but not always possible and nay difference in volume would be very small as vertical section is so narrow.

A level- Chemistry (31 decks)

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