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Acids And Bases Flashcards

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1
Q

Define the term Brφnsted-Lowry acid.

A
  • A proton donor
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2
Q

What is meant by the term strong when describing an acid?

A
  • Fully dissociated
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3
Q

Give the value of the ionic product of water, Kw, measured at 298K, and state its units.

A
  • 1.0 x 10^-14
  • ( Kw = [ H^+ ] [ OH^- ] )
  • mol^2 dm^-6
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4
Q

At 298 K, 25.0 cm^3 of a solution of a strong acid contained 1.50 × 10^-3 mol of hydrogen ions.

Calculate the hydrogen ion concentration in this solution and hence its pH.

A
  • Conc = 1.50 x 10^-3 / 25 x 10^-3
  • = 0.06 mol dm^-3
  • pH = - log [ H^+ ]
  • pH = 1.2
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5
Q

Calculate the pH of the solution formed after the addition of 50.0 cm^3 of 0.150 M NaOH to the original 25.0 cm^3 of acid.

( At 298 K, 25.0 cm^3 of a solution of a strong acid contained 1.50 × 10^-3 mol of hydrogen ions. )

( M = concentration )

A
  • Moles OH^- = 0.15 x 50 x 10^-3
  • = 7.5 x 10^-3 ( in excess )
  • Moles H^+ = 1.5 x 10^-3
  • Conc OH^- = ( 7.5 x 10^-3 - 1.5 x 10^-3 ) / ( 25 + 50 ) x 10^-3
  • = 0.08
  • Kw = [ H^+ ] [ OH^- ]
  • 10^-14 = [ H^+ ] [ 0.08 ]
  • [ H^+ ] = 10^-14 / 0.08
  • [ H^+ ] = 1.25 x 10^-13
  • pH = - log [ H^+ ]
  • pH = 12.9
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6
Q

A solution of a strong acid was found to have a pH of 0.5

Calculate the hydrogen ion concentration in this solution.

A
  • pH = 0.5
  • [ H^+ ] = 10^-pH
  • [ H^+ ] = 10^-0.5
  • [ H^+ ] = 0.3 mol dm^-3
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7
Q

Calculate the volume of water which must be added to 25.0 cm^3 of this solution to increase its pH from 0.5 to 0.7.

A
  • pH = 0.5
  • [ H^+ ] = 10^-0.5
  • [ H^+ ] = 0.316227766
  • pH = 0.7
  • [ H^+ ] = 10^-0.7
  • [ H^+ ] = 0.1995262315
  • C1V1 = C2V2
  • 0.31… x 25 = 0.199…. x v
  • v = 39.62232982
  • Volume of water added = 39.622… - 25
  • = 14.62232981
  • = 14.6 cm^-3
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8
Q

The pH of a 0.15 M solution of a weak acid, HA, is 2.82 at 300 K.

Write an expression for the acid dissociation constant, Ka, of HA, and determine the value of Ka for this acid at 300 K, stating its units.

A
  • Ka = [ H^+ ] [ A^- ] / [ HA ]
  • Ka = [ H^+ ]^2 / [ HA ]
  • [ H^+ ] = 10^-2.82
  • [ H^+ ] =1.513561248 x 10^-3
  • Ka = [ 1.51… X 10^-13 ]^2 / [ 0.15 ]
  • = 1.527245102 x 10^-5
  • = 1.53 x 10^-5 mol dm^-3
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9
Q

The dissociation of HA into its ions in aqueous solution is an endothermic process.
How would its pH change if the temperature were increased? Explain your answer.

A
  • Decrease pH
  • Endothermic reaction is favoured, so more dissociation of HA, making pH smaller
  • ( More dissociation of HA makes more H^+ ions so a more acidic solution )
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10
Q

Solution A contains n moles of a different weak acid, HX.
The addition of some sodium hydroxide to A neutralises one third of the HX present to produce Solution B.

In terms of the amount, n, how many moles of HX are present in Solution B?

A
  • 2 / 3 n
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11
Q

Determine the ratio [ HX ] / [ X^- ] in Solution B.

( Solution A contains n moles of a different weak acid, HX.
The addition of some sodium hydroxide to A neutralises one third of the HX present to produce Solution B. )
( mole of HX = 2 / 3 n )

A
  • 2 / 3 n / 1 / 3 n = 2
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12
Q

Solution B has a hydrogen ion concentration of 4.2 × 10^-4 mol dm^-3.
Use this information and your answer to part ( b )( ii ) to determine the value of the acid dissociation constant of HX.

( Solution A contains n moles of a different weak acid, HX.
The addition of some sodium hydroxide to A neutralises one third of the HX present to produce Solution B. )
( Ratio of [ HX ] / [ X^- ] in solution B = 2 )

A
  • Ka = [ H^+ ] [ X^- ] / [ HX ]
  • Ka = [ H^+ ]^2 / [ HX ]
  • Ka = [ 4.2 x 10^-4 ] / [ 2 ]
  • Ka = 2.1 x 10^-4
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13
Q

Why is methyl orange not suitable as an indicator for the titration of HX with sodium hydroxide?

A
  • Weak acid / strong base
  • So pH is > 7
  • Methyl orange has a colour change when pH < 7
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14
Q

Solution B can act as a buffer.
Explain what this means and write an equation that shows how Solution B acts as a buffer if a little hydrochloric acid is added.

A

Meaning of buffer:

  • Buffer can resist change in pH on the addition of small amounts of H^+ ( or OH^- )

Equation of how solution B acts as a buffer:

  • H^+ ( aq ) + X^- ( aq ) < = > HX ( aq )
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15
Q

Explain the terms acid and conjugate base according to the Brønsted-Lowry theory.

A

Acid:

  • Proton donor

Conjugate base:

  • The substance formed when the acid has lost a proton
  • ( Substance that becomes an acid by gaining a proton )
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16
Q

For each of the following reactions, give the formula of the acid and of its conjugate base.

i ) NH3 + HBr → NH4^+ + Br^-

ii ) H2SO4 + HNO3 → HSO4^- + H2NO3^+

A

i )

Acid:

  • HBr

Conjugate base:

  • Br^-

ii )

Acid:

  • H2SO4

Conjugate base:

  • HSO4^-
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17
Q

Write an equation to represent the dissociation of water.

A
  • H2O < = > H^+ + OH^-
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18
Q

Give the expression for the equilibrium constant, Kc, for the reaction in ( c )( i ) and use this to derive the expression for the ionic product of water, Kw.

( H2O < = > H^+ + OH^- )

A
  • Kc = [ H^+ ] [ OH^- ] / [ H2O ]
  • [ H2O ] is effectively constant
  • Kc x [ H2O ] = [ H^+ ] [ OH^- ]
  • Kw = [ H^+ ] [ OH^- ]
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19
Q

The ionic product of water is 2.92 × 10^-14 mol^2 dm^-6 at 313K. Calculate the pH of water at this temperature.

A
  • Kw = [ H^+ ] [ OH^- ]
  • ( With water ), [ H^+ ] = [ OH^- ]
  • Kw = [ H^+ ]^2
  • 2.92 x 10^-14 = [ H^+ ]^2
  • [ H^+ ] = Root 2.92 x 10^-14
  • [ H^+ ] = 1.71 x 10^-7
  • pH = -log [ H^+ ]
  • pH = - log [ 1.71 x 10^-7 ]
  • pH = 6.77
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20
Q

Given that the pH of water is 7.00 at 298 K, state whether the dissociation of water is endothermic or exothermic.
Give a reason for your answer.

( H2O < = > H^+ + OH^- )

A
  • Endothermic

- There would be more dissociation of H^+ ions at higher temperatures

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21
Q

Give the Brønsted–Lowry definition of a base.
State the essential feature of an acid-base reaction in aqueous solution, writing an ionic equation to illustrate your answer.

A

Definition of a base:

  • Proton acceptor

Essential feature of an acid-base reaction:

  • Transfer of protons

Equation:

H^+ + OH^- < = > H2O

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22
Q

Explain what is meant by the term weak when applied to acids and bases.

A
  • Only partially dissociated in the solution
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23
Q

In aqueous solution, the weak acid propanoic acid, CH3CH2COOH( aq ), produces propanoate ions CH3CH2COO^- ( aq ).
Write an expression for the acid dissociation constant, Ka, of propanoic acid and state its units.

A

Expression for Ka:

  • Ka = [ H^+ ] [ CH3CH2COO^- ] / [ CH3CH2COOH ]

Units:

  • mol dm^-3
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24
Q

Explain what is meant by the term buffer solution.

A
  • Resist change to pH

- On addition of small amounts of strong acid or base

25
Q

Identify two components that could be used to make a buffer solution.

A
  • HBr / Br^-
  • ( Any weak acid / co-base )
  • ( Or weak base / co-acid )
  • ( Co = Conjugative )
26
Q

Give an example of the use of a buffer solution.

A
  • To neutralise stomach acid?
27
Q

The graph below shows how the pH changes when 0.12 M NaOH is added to 25.0 cm3 of a
solution of a weak monoprotic acid, HA.

( X-axis is labelled “ pH “ )
( Y-axis is labelled “ Volume of 0.12 M NaOH / cm^3 “ )
( Line graph which starts at a “ pH “ of 1.6 )
( Both “ pH “ and “ Volume “ start from zero on the scale )
( Line graph shows a gradual increase in “ pH “ as “ Volume increase “ )
( Line graph spikes when “ Volume “ increases past 11 cm^3 )
( “ pH “ increases drastically when “ Volume “ remains at 11.8 cm^3 )
( Graph starts to plateau when “ pH “ reaches 10.8 )
( Graph reaches “ pH “ of 13.7 )

Use the graph to calculate the initial concentration of the acid HA.

A
  • 11.8
  • ( The volume at which the pH = 7 )
  • ( Point of neutralisation )
  • 0.12 x 11.8 = HA x 25
  • ( C1V1 = C2V2 )
  • HA = 0.12 x 11.8 / 25
  • HA = 177 / 3125 = 0.057
28
Q

Write an expression for the dissociation constant. Ka, of the weak acid HA.

A
  • Ka = [ H^+ ] [ A^- ] / [ HA ]
29
Q

Determine the volume of sodium hydroxide added when [ HA ] = [ A^- ] and use the graph to determine the pH at this point.

( Volume of NaOH at point of neutralisation = 11.8 )

A

Volume of NaOH added:

  • 11.8 / 2 = 5.9cm^3

pH:

  • 4.3 ( Taken from the graph )
30
Q

Use your answers to part ( a )( ii ) and part ( a )( iii ) to determine the value of Ka for the acid HA.

( Ka = [ H^+ ] [ A^- ] / [ HA ] )
( Volume of NaOH added = 5.9 cm^3 )
( pH = 4.3 )

A
  • ( As [ HA ] = [ A^- ] )
  • Ka = [ H^+ ]
  • pH = 4.3
  • Ka = 10^- 4.3
  • Ka = 5.0 x 10^- 5
31
Q

A buffer solution is formed, when approximately half of the original amount of the acid HA( aq ) has been neutralised by the base NaOH( aq ).
Explain how this buffer solution is able to resist change in pH when

i ) a small amount of NaOH( aq ) is added,

ii ) a small amount of HCl( aq ) is added.

A

i )

  • The added OH^- reacts with HA or H^+
  • The equilibrium HA < = > H^+ + A^-
  • Displaces to the right
  • ( We can say that OH^- reacts with HA, increasing the conc of HA, which shifts the equilibrium to the right )

ii )

  • The added H^+ reacts with A^-
  • The equilibrium HA < = > H^+ + A^-
  • Displaces to the left
32
Q

The graph below shows how the pH changes as 0.12 M HCl( aq ) is added to 25.0 cm^3 of a solution of sodium carbonate.
There are two end-points.
The second end-point is at 30.0 cm^3.

( “ pH “ on the y-axis )
( “ Volume of 0.12 M HCl / cm^3 “ on the x-axis )
( Line graph which goes has a weak negative correlation )
( Line graph continues downwards from around 10.2 in pH )
( End around at 0.19 in pH )
( Volume of HCl starts are 0 )
( Ends at 40 cm^3 )
( Graph has two dips representing the end-points of the reaction )
( The first being less than the second )
( Graph is labelled A, B, C, D, E and F )
( A and B is the first part of the graph, being a smooth decreasing straight line )
( B and C is a slight dip in the graph )
( C and D is another smooth decreasing straight line )
( D and E is a sharp dip in the graph, a straight line downwards )
( E and F is another smooth decreasing straight line but sharper than the other two )

Write equations for the reactions which occur in the solution between point A and point B on the graph and between point C and point D on the graph.

A

Equation for reaction occuring between A and B:

  • Na2CO3 + HCl → NaHCO3 + NaCl

Equation for reaction occuring between C and D:

  • NaHCO3 + HCl → NaCl + CO2 + H2O
33
Q

Estimate the minimum volume of hydrochloric acid needed in this experiment for carbon dioxide to be produced from a well-stirred solution of sodium carbonate.

( “ pH “ on the y-axis )
( “ Volume of 0.12 M HCl / cm^3 “ on the x-axis )
( Line graph which goes has a weak negative correlation )
( Line graph continues downwards from around 10.2 in pH )
( End around at 0.19 in pH )
( Volume of HCl starts are 0 )
( Ends at 40 cm^3 )
( Graph has two dips representing the end-points of the reaction )
( The first being less than the second )
( Graph is labelled A, B, C, D, E and F )
( A and B is the first part of the graph, being a smooth decreasing straight line )
( B and C is a slight dip in the graph )
( C and D is another smooth decreasing straight line )
( D and E is a sharp dip in the graph, a straight line downwards )
( E and F is another smooth decreasing straight line but sharper than the other two )

A
  • 15cm^3

- ( Around the first end-point “ dip “ )

34
Q

Name an indicator which can be used to determine the end-point occurring between points D and E.
Explain why this indicator does not change colour between points B and C.

( “ pH “ on the y-axis )
( “ Volume of 0.12 M HCl / cm^3 “ on the x-axis )
( Line graph which goes has a weak negative correlation )
( Line graph continues downwards from around 10.2 in pH )
( End around at 0.19 in pH )
( Volume of HCl starts are 0 )
( Ends at 40 cm^3 )
( Graph has two dips representing the end-points of the reaction )
( The first being less than the second )
( Graph is labelled A, B, C, D, E and F )
( A and B is the first part of the graph, being a smooth decreasing straight line )
( B and C is a slight dip in the graph )
( C and D is another smooth decreasing straight line )
( D and E is a sharp dip in the graph, a straight line downwards )
( E and F is another smooth decreasing straight line but sharper than the other two )

A

Indicator:

  • Methyl orange

Explanation:

  • Methyl orange changes colour around a pH range of 3.2 - 4.4
35
Q

Use the end-point occurring between points D and E to calculate the concentration of sodium carbonate in the given solution.

( “ pH “ on the y-axis )
( “ Volume of 0.12 M HCl / cm^3 “ on the x-axis )
( Line graph which goes has a weak negative correlation )
( Line graph continues downwards from around 10.2 in pH )
( End around at 0.19 in pH )
( Volume of HCl starts are 0 )
( Ends at 40 cm^3 )
( Graph has two dips representing the end-points of the reaction )
( The first being less than the second )
( Graph is labelled A, B, C, D, E and F )
( A and B is the first part of the graph, being a smooth decreasing straight line )
( B and C is a slight dip in the graph )
( C and D is another smooth decreasing straight line )
( D and E is a sharp dip in the graph, a straight line downwards )
( E and F is another smooth decreasing straight line but sharper than the other two )

( Conc of HCl = 0.12 M )
( Volume of HCl at second end-point = 30 cm^3 )
( Volume of sodium carbonate = 25 cm^3 )

A
  • CO3^2 - + 2H^+ → H2O + CO2
  • ( Ionic equation of, NaHCO3 + HCl → NaCl + CO2 + H2O )
  • Moles H^+ = 30 x 0.12 x 10^-3 = 3.6 x 10^-3
  • Mole of CO3^2 - = 1.8 x 10^-3
  • Conc of CO3^2 - = 1.8 x10^-3 / 25 x 10^-3 = 0.072 M
36
Q

If the original solution had contained, in addition to sodium carbonate, an equal molar concentration of sodium hydrogen carbonate, at what volumes of hydrochloric acid would the two end-points have been detected?

( Conc NaOH = 0.12 M )
( Conc Na2CO3 = 0.12 M )
( Volume of HCl at first end-point = 15 cm^3 )
( Volume of HCl at second end-point = 30 cm^3 )

A

Volume of HCl( aq ) added for first end-point:

  • 15 cm^3

Volume of HCl( aq ) added for second end-point:

  • 45 cm^3
37
Q

Write an equation for the reaction between gaseous hydrogen chloride and water.
State the role of water in this reaction, using the Brønsted-Lowry definition.

( Brønsted-Lowry definition = Proton donor )

A

Equation:

  • HCl( g ) + H2O( l ) < = > H3O^+( aq ) + Cl^-( aq )

Role of water:

  • Base
38
Q

Write an equation for the reaction between gaseous ammonia and water.
State the role of water in this reaction, using the Brønsted-Lowry definition.

( Brønsted-Lowry definition = Proton donor )

A

Equation:

  • NH3( g ) + H2O( l ) < = > NH4^+( aq ) + OH^-( aq )

Role of water:

  • Proton donor ( acid )
39
Q

The ion H2NO3^+ is formed in the first stage of a reaction between concentrated nitric acid and an excess of concentrated sulphuric acid.
In this first stage the two acids react in a 1 : 1 molar ratio.
In the second stage, the H2NO3^+ ion decomposes to form the nitronium ion, NO2^+.
Write equations for these two reactions and state the role of nitric acid in the first reaction.

A

Equation for the formation of H2NO3^+:

  • H2SO4 + HNO3 < = > HSO4^- + H2NO3^+

Role of nitric acid:

  • Base ( proton acceptor )

Equation for the formation of NO2^+:

  • H2NO3^+ - > NO2^+ + H2O
40
Q

Explain the term weak acid.

A
  • Acid is only partially dissociated in aqueous solution
41
Q

Write an expression for the acid dissociation constant, Ka, of HA, a weak monoprotic acid.

A
  • Ka = [ H^+ ] [ A^- ] / [ HA ]
42
Q

The value of the acid dissociation constant for the monoprotic acid HX is 144 mol dm^-3.
What does this suggest about the concentration of undissociated HX in dilute aqueous solution?

A
  • Not very big

- ( Acid has been dissociated by quite a big amount )

43
Q

State whether HX should be classified as a strong acid or a weak acid.
Justify your answer.

( HX has an acid dissociation of 144 mol dm^-3 )
( Concentration of undissociated HX is small )

A

Nature of HX:

  • Strong acid

Justification:

  • Although HX isn’t fully dissociated, the relative concentration of undissociated acid HX is very small
44
Q

For acid HA, Ka = 2.00 × 10–4 mol dm^-3.

Write an equation for the reaction of HA with NaOH.

A
  • NaOH + HA - > NaA + H2O
45
Q

A solution was formed by adding 15 cm^3 of 0.34 M NaOH to 25 cm^3 of 0.45 M HA.

Calculate the number of moles of A^–( aq ) and HA( aq ) in this solution.
( You should neglect the small number of moles of A^–
( aq ) formed by ionisation of the remaining HA( aq ).)

A

Number of moles of A^-( aq ):

  • Moles of A^- = Moles of NaOH
  • Moles of NaOH = 0.34 x 15 x 10^-3
  • = 5.1 x 10^-3

Number of moles of HA( aq ):

  • Inital moles of HA = 0.45 x 25 x 10^-3
  • = 0.01125
  • Moles of remaining HA = 0.01125 - 5.1 x 10^-3
  • ( NaOH was added and we are working out the moles of HA in the final solution )
  • = 6.15 x 10^-3
46
Q

Calculate the concentration of A^-( aq ) and HA( aq ) in this solution.

( Moles A^- = 5.1 x 10^-3 )
( Moles HA = 6.15 x 10^-3 )
( Volume A^- = 15 cm^-3 )
( Volume HA = 25 cm^-3 )

A

Cocentration A^-:

  • 5.1 x 10^-3 / ( 15 + 25 ) x 10^-3
  • = 5.1 x 10^-3 / 40 x 10^-3
  • = 0.1275

Concentration HA:

  • 6.15 x 10^-3 / 40 x 10^-3
  • = 0.15375
  • = 0.1538
47
Q

Using an expression for Ka, calculate the pH of this solution.

( Concentration A^- = 0.1275 )
( Concentration HA = 0.1538 )
( Ka = 2.00 × 10^-4 mol dm^-3 )

A
  • Ka = [ H^+ ] [ A^- ] / [ HA ]
  • [ H^+ ] = [ 2.00 x 10^-4 ] [ 0.1538 ] / [ 0.1275 ]
  • [ H^+ ] = 2.41254902 x 10^-4
  • pH = - log [ H^+ ]
  • pH = - log [ 2.41254902 x 10^-4 ]
  • pH = 3.62
48
Q

Write an expression for the acid dissociation constant, Ka, for phenol.

( C6H5OH( aq ) + H2O( l ) < = > H3O^+( aq ) + C6H5O^-( aq ) )

A
  • Ka = [ H3O^+ ] [ C6H5O^- ] / [ C6H5OH ]

- ( [ H^+ ] is accepted )

49
Q

Write an expression linking Ka with pKa.

Ka = [ H3O^+ ] [ C6H5O^- ] / [ C6H5OH ]

A
  • pKa = - log ( Ka )

- ( Not square brackets )

50
Q

The value of the acid dissociation constant, Ka, for phenol is 1 × 10^-10 mol dm^-3.
Calculate the pKa value of phenol.

( pKa = - log( Ka ) )

A
  • pKa = - log( 1 x 10^-10 )

- pKa = 10

51
Q

Ethanoic acid is a stronger acid than phenol.
State whether the pKa value for ethanoic acid will be greater or smaller than that of phenol

( pKa = - log( Ka ) )
( Phenol has a Ka value of 1 x 10^-10 )
( Phenol has a pKa value of 10 )

A
  • pKa value will be lower for ethanoic acid
52
Q

The indicator phenolphthalein is a weak acid which can be represented by the formula HIn.
It dissociates in solution and has a pKa value of 9.3.

HIn( aq ) < - > H^+( aq ) + In^-( aq )
colourless red

Suggest and explain, with reference to the pKa value, the pH range of phenolphthalein.

A
  • At the end point, pH = pKa = 9.3
  • A colour change is detectable over a range of 2 pH units
  • So range = 8.3 - > 10.3
53
Q

State the colour change that would be observed at the end point in an acid-base titration using phenolphthalein if sodium hydroxide solution were being added from the burette.
Explain, in terms of the species present, why this colour is formed.

( The indicator phenolphthalein is a weak acid which can be represented by the formula HIn.
It dissociates in solution and has a pKa value of 9.3.

HIn( aq ) < - > H^+( aq ) + In^-( aq )
colourless red )

A

Colour change observed at the end point due to phenolphthalein:

  • Colourless to red

Reason:

  • [ In^- ] = > [ HIn ]
54
Q

State why phenolphthalein is unsuitable for a titration between a strong acid and a weak base.

A
  • The end point of the titration will be below a pH of 7

- Which is lower than the range of the colour change of phenolphthalein ( 8.3 - 10.3, from previous quesiton )

55
Q

Define the term pH

A
  • pH = - log [ H^+ ]
56
Q

For HX, Ka = 4.25 × 10^-5 mol dm^-3.

Calculate the pH of a 0.45 M solution of this acid.

A
  • Ka = [ H^+ ] [ X^- ] / [ HX ]
  • Ka = [ H^+ ]^2 / [ HX ]
  • 4.25 x 10^-5 = [ H^+ ]^2 / [ 0.45 ]
  • [ H^+ ] = Root( 4.25 x 10^-5 [ 0.45 ] )
  • [ H^+ ] = 4.373213921 x 10^-3
  • pH = - log( 4.373213921 x 10^-3 )
  • pH = 2.36
57
Q

In a 0.25 M solution, a different acid HY is 95% dissociated.
Calculate the pH of this solution.

A
  • [ H^+ ] = 0.25 x 95 x 10^-2
  • [ H^+ ] = 19 / 80
  • = 0.24
  • pH = - log( 19 / 80 )
  • pH = 0.62
58
Q

Calculate the value of Ka for the acid HY.

( In a 0.25 M solution, a different acid HY is 95% dissociated. )

( [ H^+ ] = 19 / 80 )

A
  • [ H^+ ] = [ Y^- ]
  • Ka = [ H^+ ]^2 / [ HY ]
  • [ HY ] = 0.05 x 0.25
  • ( 100 - 95 = 5% )
  • [ HY ] = 1 / 80 = 0.0125
  • Ka = [ 19 / 80 ]^2 / [ 1 / 80 ]
  • Ka = 361 / 80 = 4.5
59
Q

11a

A

Chemistry (13 decks)

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