Acids And Bases Flashcards

1
Q

What is the Brønsted-Lowry definition of

a) an acid
b) a base

A

a) A proton donor

b) A proton acceptor

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2
Q

What does monoprotic mean

A

It donates 1 H+ per molecule

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3
Q

What does diprotic mean

A

It donates 2 H+ ions per molecule

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4
Q

What do strong acids and bases do

A

Completely dissociate

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5
Q

What equilibrium is set up when a weak acid dissolves in water

A

HA + H2O (equilibrium) H3O+ + A-

Or

HA (equilibrium) H+ + A-

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6
Q

Give the equation of NH3 (weak base) dissolving in H2O

A

NH3 + H2O (equilibrium) NH4+ + OH-

Or

NH3 + H+ (equilibrium) NH4

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7
Q

The weaker the base the ____ it dissociates and the more the equilibrium lies to the _____

A

Less

Left

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8
Q

Does water act as an acid or a base? What is this called

A

Both

Amphoteric (acts as both an acid and base)

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9
Q

Definition of pH

A

pH= -log[H+]

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10
Q

As the concentration of H+ increases, what happens to the pH

A

pH decreases

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11
Q

To what precision is pH given to

A

2 d.p.

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12
Q

Equation to convert pH to [H+]

A

[H+]=10^-pH

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13
Q

When converting from H2SO4 to H+ do you divide or multiply by 2

A

Multiply

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14
Q

What does Kw=

A

[H+][OH-]

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15
Q

What is Kw at 298K/ 25 degrees Celsius

A

1x10^-14 mol^2dm^-6

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16
Q

Why is [H2O] not included in the equation for Kw

A

Water is only weakly dissociated so equilibrium lies far over to the left and it’s concentration can be taken as constant

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17
Q

In pure water what must [H+]=

Therefore…?

A

[OH-]

Kw= [H+]^2

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18
Q

Is the dissociation of water exothermic?

A

No it is endothermic

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19
Q

Explain why pure water at 50 degrees celsius is not acidic despite being at a pH of 6.63

A

[H+]=[OH-]

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20
Q

Why does Kw increase with temperature

A

Equilibrium shifts to the right to decrease temperature
Degree of dissociation increases
Kw increases
pH increases

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21
Q

Define:

Monoprotic base

Diprotic base

A

Can accept 1 proton

Can accept 2 protons

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22
Q

Are hydroxides strong bases?

A

Yes

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23
Q

3 steps for calculating pH of diluted acid

A

1: calculate moles of H+
2: concert to concentration by dividing by new volume
3: pH=-log[H+]

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24
Q

Steps for working out pH of a strong base

A

1: calculate miles of OH-
2: convert to concentration by dividing by new volume
3: [H+]=Kw/[OH-]
4: pH=-log[H+]

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25
Q

How do you calculate pH of a mixture of acid and base

A
  • calculate moles of H+
  • calculate moles of OH-
  • work out what is in XS and convert to concentration by dividing by total volume

If H+ is in XS: pH=[H+]
If OH- is in XS: [H+]=Kw/[OH-]
pH=-log[H+]

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26
Q

Why must a pH meter be calibrated before use?

A

They do not give an accurate reading over time

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27
Q

How to calibrate a pH meter

A

Place meter in buffer solution of known pH and plot a calibration curve

Eg rinse probe with deionised water, dry with a paper towel and place in buffer
Wait for reading to steady then set meter to known pH of buffer

28
Q

What is a buffer solution

A

Maintains same pH despite the addition of small amounts of acid, base, or water

29
Q

Method for drawing a pH curve

A

Measure initial pH
Add solution from burette in small portions
Stir mixture
Record pH after each addition
Add solution dropwise as you near the end point
After end point add solution 1cm3 at a time until in XS

30
Q

What container do you use for a titration to find a pH curve

A

A beaker - NOT a conical flask

So both probe and burette can fit

31
Q

What is the equivalence point

A

When sufficient base has been added to neutralise the acid (or vice versa)

32
Q

Describe a pH curve of a:

Strong Base added to a strong acid

Strong base added to a weak acid

A

Very low to very high

Low to very high

33
Q

Describe the pH curve of a:

Weak base added to a strong acid

A weak base added to a weak acid

A

Very low to quite high

Quite low to quite high

34
Q

How to use pH curve to find concentration of acid or base

A

Use volume at end point (vertical line) to find concentration of substance added

35
Q

What does pKa equal

36
Q

What is the half neutralisation point

A

Half way between the zero and equivalence point

37
Q

What does Ka equal

A

[H+][A-]
————-
[HA]

38
Q

Why can Ka be simplified to Ka=[H+] at half neutralisation point

A

Half the acid has been neutralised so [HA]=[A-]

39
Q

How to tell if it’s a half neutralisation point calculation

A

If one reactant has half the volume of the other but equal concentration

Ie if one has half the moles of the other

40
Q

What is an acid-base indicator

A

A water soluble, weak organic acid, whose acid form (HA) and base form (A-) have different colours

41
Q

Why can only 2-3 drops of an acid-base indicator be added to a solution

A

It is acidic so it will affect pH otherwise

42
Q

What happened when the indicator is placed in acidic solution

A

The extra H+ ions will push the equilibrium to the LHS so the colour of HA will be seen

43
Q

What happened when the indicator is placed in alkaline solution

A

The extra OH- ions will react with the H+ ions, pushing the equilibrium to the RHS and so the colour of A- will be seen

44
Q

What properties must a suitable indicator have

A

Colour change must be sharp rather than gradual at end point

End point of the titration given by the indicator must be the same as the equivalence point

Indicator should give a distinct colour change

45
Q

When will change point for an indicator be

A

When [HA]=[A-] and neither colour dominates

At this point [H+]=Ka

46
Q

pH at which the colour of the indicator changes =

A

pKa of the indicator

47
Q

How do you calculate Ka from a titration pH curve?

A

From the curve, read off the volume at equivalence
Divide this volume by 2 to find volume at half equivalence
Read pH at the half equivalence volume - this will be your pKa value
Calculate the 10^(-pKa) to find Ka

48
Q

What is the molar ratio of a carbonate to a monoprotic acid?

49
Q

Define buffer solution

A

A solution that resists change in pH when small amounts of acid, base or water are added

50
Q

What can Ka be useful for?

A

To confirm an acid’s identity

51
Q

The weaker the acid, the _____ it dissociates and the more the equilibrium lies to the ____

How about bases?

A

Less

LHS

Same for bases

52
Q

Does water strongly dissociate?

What does this mean for the equilibrium?

A

No- weakly dissociated

Lies far to the left and conc of water is taken as constant

53
Q

What is the ionic product of water

A

Kw

Kc when water is taken as constant - [H+][OH-]

54
Q

How does temperature affect Kw?

Why?

A

Kw increases with temperature

It involves breaking covalent bonds (inputting energy) and is an endothermic process

55
Q

What are the two types of acidic buffer?

A

Weak acid and one of its salts

Weak acid and strong acid (salt is made as a product of the reaction)

56
Q

What is sodium ethanoate used for in an acidic buffer?

A

Acts as a source of conjugate base, CH3COO-

57
Q

What is added to an alkaline buffer containing NH3

58
Q

What happens in the equilibrium of blood when
An Acid

A Base

Is added

A

a) extra H+ shift equilibrium to left to remove H+

b) extra OH- remove H+ by reacting to form water. Equilibrium shifts left to replace H+

59
Q

Give example of buffers in everyday life

Why are buffers needed in them

A

Detergent

Shampoo

Wrong pH could damage fabric or skin and hair

60
Q

For acidic what must you say after explains buffer action?

A

The ratio of
[HA]
————
[A-]

Remains constant

61
Q

How many moles of Fe2+ react with one chromate ion?

62
Q

Is there any point rearranging equations before putting in numbers?

A

Yes - you get a mark for it

63
Q

Why is the gradient of a PH curve initially near 0?

A

Buffer formed (excess weak acid and strong base)

64
Q

Describe pH curve of strong base to weak acid

A

Initial ‘ramp up’

Near horizontal until endpoint where curve is vertical and then levels out

65
Q

What must you dip the probe in to calibrate it?

A

Buffer solutions (pH 4, pH 7 and pH10)

66
Q

Can you swirl when doing investigate pH

A

No - mixture in a beaker so must stir

67
Q

How to work out original [HX] from incomplete pH curve

A

At equivalence point
n OH- = n HX

So divide volume added at equivalence point by original volume of HX