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acid base Flashcards

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1
Q

Bronsted-Lowry acid

A

Substance that can donate a proton

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2
Q

A Bronsted-Lowry base

A

Substance that can accept a proton.

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3
Q

Conjugate acid is always

A

positive ion

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4
Q

Calculating pH

A

pH = - log [H+]

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5
Q

Calculating the pH of strong acids

A

Strong acids completely dissociate

The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid.

Always give pH values to 2d.p. in the exam

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6
Q

Finding [H+] from pH

A

10-pH

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7
Q

Ionic Product for water

A
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8
Q

Example 2 : Calculate the pH of water at 50ºC given that Kw = 5.476 x 10-14 mol2 dm-6 at 50ºC

A

[H+ (aq)] = √ Kw = √ 5.476 x 10-14 =2.34 x 10-7 mol dm-3

pH = - log 2.34 x 10-7 = 6.6

It is still neutral though as [H+ (aq)] = [OH-)

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9
Q

At different temperatures to 25oC the pH of pure water changes

A

Le Chatelier’s principle can predict the change. The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H+ ions and a lower pH.

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10
Q

Calculating the pH of a Strong Base

A

For bases, we are normally given the concentration of the hydroxide ion. To work out the pH we need to work out [H+ (aq)] using the Kw expression

Strong bases completely dissociate into their ions. NaOH —– Na+ + OH-

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11
Q

Example 3: Calculate the pH of the strong base 0.1 mol dm-3 NaOH Assume complete dissociation

A

Kw = [H+ (aq)][OH- (aq)] = 1x10-14 [H+ (aq)] = Kw/ [OH- (aq)] = 1x10-14 / 0.1 = 1x10-13 M

pH = - log[1x10-13] =13.00

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12
Q

Weak Acids

A

Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture

Ka=[H+ (aq)][A- (aq)] over [HA (aq)]

The larger ka the stronger the acid.

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13
Q

pKa

A

pKa = -log (Ka) so Ka = 10-pKa

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14
Q

Calculating pH of a Weak Acid

A
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15
Q

Example 5 What is the pH of a solution of 0.01 mol dm-3 ethanoic acid (ka is 1.7 x 10-5 mol dm-3 )?

A
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16
Q

Calculate the concentration of propanoic acid with a pH of 3.52 (ka is 1.35 x 10-5 mol dm-3 )

A
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17
Q

Strong Acid and Strong Base Neutralisations

A
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18
Q

Example 7 15cm3 of 0.5 mol dm-3 HCl is reacted with 35cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture

A
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19
Q

Example 8 45cm3 of 1 mol dm-3 HCl is reacted with 30cm3 of 0.65 mol dm-3 NaOH. Calculate the pH of the resulting mixture

A
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20
Q

Example 9 35cm3 of 0.5 mol dm-3 H2SO4 is reacted with 30cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture

A
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21
Q

Example 10 15cm3 of 0.5mol dm-3 HCl is reacted with 35cm3 of 0.45 mol dm-3 Ba(OH)2 . Calculate the pH of the resulting mixture

A
22
Q

Weak Acid and Strong Base Neutralisations

calculation pattern

A
23
Q

Example 11

55cm3 of 0.5 mol dm-3 CH3CO2H is reacted with 25cm3 of 0.35 mol dm-3 NaOH. Calculate the pH of the resulting mixture?

A
24
Q

Working out pH of a weak acid at half equivalence

assumptions

A
25
Q

Example 12

Calculate the pH of the resulting solution when 25cm3 of 0.1M NaOH is added to 50cm3 of 0.1M CH3COOH (ka 1.7 x 10-5 )

A
26
Q

Diluting an acid or alkali

A
27
Q

Example 13

Calculate the new pH when 50.0 cm3 of 0.150 mol dm-3 HCl is mixed with 500 cm3 of water.

A
28
Q

Buffer Solutions

A

A Buffer solution is one where the pH does not change significantly if small amounts of acid or alkali are added to it

29
Q

acidic buffer

A

An acidic buffer solution is made from a weak acid and a salt of that weak acid ( made from reacting to the weak acid with a strong base).

Example: ethanoic acid and sodium ethanoate CH3CO2H (aq) and CH3CO2 - Na+

30
Q

basic buffer

A

A basic buffer solution is made from a weak base and a salt of that weak base ( made from reacting to the weak base with a strong acid).

Example :ammonia and ammonium chloride NH3 and NH4+Cl

31
Q

If small amounts of acid is added to the buffer:

A

Equilibrium will shift to the left removing nearly all the H+ ions added

CH3CO2 - (aq) + H+ (aq) —- CH3CO2H (aq)

As there is a large concentration of the salt ion in the buffer the ratio [CH3CO2H]/ [CH3CO2 -] stays almost constant, so the pH stays fairly constant

32
Q

If small amounts of alkali is added to the buffer

A

The OH - will react with H+ ions to form water

H+ + OH - —-H2O

The Equilibrium will then shift to the right to produce more H+ ions

CH3CO2H (aq)—- CH3CO2 - (aq) + H+ (aq)

Some ethanoic acid molecules are changed to ethanoate ions but as there is a large concentration of the salt ion in the buffer the ratio [CH3CO2H]/ [CH3CO2 -] stays almost constant, so the pH stays fairly constant

33
Q

Calculating the pH of Buffer Solutions

A
34
Q

Example 14:

making a buffer by adding a salt solution

Calculate the pH of a buffer made from 45cm3 of 0.1 mol dm-3 ethanoic acid and 50cm3 of 0.15 mol dm-3 sodium ethanoate (Ka = 1.7 x 10-5 )

A
35
Q

Example 15 :

making a buffer by adding a solid salt

A buffer solution is made by adding 1.1g of sodium ethanoate into 100 cm3 of 0.4 mol dm-3 ethanoic acid. Calculate its pH. (Ka =1.7 x10-5 )

A
36
Q

Example 16

55cm3 of 0.5 mol dm-3 CH3CO2H is reacted with 25cm3 of 0.35 mol dm-3 NaOH. Calculate the pH of the resulting buffer solution

A
37
Q

Calculating the change in pH of buffer on the addition of a small amount of acid or alkali

A

If a small amount of alkali is added to a buffer then the moles of the buffer acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of pH can be done with the new values. CH3CO2H (aq) +OH-  CH3CO2 - (aq) + H2O (l)

If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount so a new calculation of pH can be done with the new values. CH3CO2 - (aq) + H +  CH3CO2H (aq)

38
Q

Example 17:

0.005 mol of NaOH is added to 500cm3 of a buffer where the concentration of ethanoic acid is 0.200 mol dm-3 and the concentration of sodium ethanoate is 0.250 mol dm-3. (Ka = 1.7 x 10-5 )

Calculate the pH of the buffer solution after the NaOH has been added.

A
39
Q

Diluting a buffer solution

A

Diluting a buffer solution with water will not change its pH

This is because in buffer equation below the ratio of [HA]/[A-] will stay constant as both concentrations of salt and acid would be diluted by the same proportion

H+=Ka times HA over A-

40
Q

Constructing a pH curve

A
  1. Transfer 25cm3 of acid to a conical flask with a volumetric pipette
  2. Measure the initial pH of the acid with a pH meter
  3. Add alkali in small amounts (2cm3 ) noting the volume added
  4. Stir the mixture to equalize the pH
  5. Measure and record the pH to 1 d.p.
  6. Repeat steps 3-5 but when approaching endpoint to add in smaller volumes of alkali
  7. Add until alkali in excess
41
Q

How to calibrate a meter

A

Calibrate meter first by measuring the known pH of a buffer solution. This is necessary because pH meters can lose accuracy in storage. Most pH probes are calibrated by putting the probe in a set buffer (often pH 4) and pressing a calibration button/setting for that pH. Sometimes this is repeated with a second buffer at a different pH

Can also improve accuracy by maintaining a constant temperature

42
Q

Strong acid – Strong base

A
43
Q

Weak acid – Strong base

A
44
Q

Strong acid – Weak base

A
45
Q

Weak acid weak base

A
46
Q

How indicators work

A
47
Q

Choosing an Indicator

A

Indicators can be considered as weak acids. The acid must have a different color to its conjugate base

An indicator changes color from HIn to In- over a narrow range.

Different indicators change colors over different ranges

The end-point of a titration is defined as the point when the color of the indicator changes color The end-point of a titration is reached when [HIn] = [In-]. To choose a correct indicator for a titration one should pick an indicator whose end-point coincides with the equivalence point for the titration.

An indicator will work if the pH range of the indicator lies on the steep part of the titration curve. In this case the indicator will change colour rapidly and the colour change will correspond to the neutralisation point.

48
Q

phenolphthalein

A

Only use phenolphthalein in titrations with strong bases but not weak bases

Colour change: colourless acid —- pink alkali

49
Q

methyl orange

A

Use methyl orange with titrations with strong acids but not weak acids

Colour change: red acid—– yellow alkali (orange end point)

50
Q
A

Chemistry aqa (21 decks)

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