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Chemistry A Level CIE > Chapter 5 Equilibra > Flashcards

Chapter 5 Equilibra Flashcards

1
Q

A Brønsted acid

A
  • is a species that can donate a proton
    • For example, hydrogen chloride (HCl) is a Brønsted acid as it can lose a proton to form a hydrogen (H+) and chloride (Cl-) ion

HCl (aq) → H+ (aq) + Cl- (aq)

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2
Q

A Brønsted base

A
  • is a species that can accept a proton
    • For example, a hydroxide (OH-) ion is a Brønsted base as it can accept a proton to form water

OH- (aq) + H+ (aq) → H2O (l)

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3
Q

In an equilibrium reaction, the products are formed at the

A
  • same rate as the reactants are used
  • This means that at equilibrium, both reactants and products are present in the solution
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4
Q

A conjugate acid-base pair

A

is two species that are different from each other by an H+ ion

  • Conjugate here means related
  • In other words, the acid and base are related to each other by one proton difference
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5
Q

Conjugate acid-base pairs are a pair of reactants and products that are linked to each other by the

A

transfer of a proton

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6
Q

The pH indicates the

A
  • acidity or basicity of an acid or alkali
  • The pH scale goes from 0 to 14
    • Acids have pH between 0-7
    • Pure water is neutral and has a pH of 7
    • Bases and alkalis have pH between 7-14
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7
Q

calculation of pH

A
  • The pH can be calculated using: pH = -log10 [H+]

where [H+] = concentration of H+ ions (mol dm-3)

  • The pH can also be used to calculate the concentration of H+ ions in solution by rearranging the equation to:

[H+] = 10-pH

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8
Q

The Ka is the

A
  • acidic dissociation constant
    • It is the equilibrium constant for the dissociation of a weak acid at 298 K
  • For the partial ionisation of a weak acid HA the equilibrium expression to find Ka is as follows:

HA (aq) ⇌ H+ (aq) + A- (aq)

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9
Q

When writing the equilibrium expression for weak acids, the following assumptions are made:

A
  • The concentration of hydrogen ions due to the ionisation of water is negligible
  • The dissociation of the weak acid is so small that the concentration of HA is approximately the same as the concentration of A-
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10
Q

The value of Ka indicates the extent of

A

dissociation

  • A high value of Ka means that:
    • The equilibrium position lies to the right
    • The acid is almost completely ionised
    • The acid is strongly acidic
  • A low value of Ka means that:
    • The equilibrium position lies to the left
    • The acid is only slightly ionised (there are mainly HA and only a few H+ and A- ions)
    • The acid is weakly acidic
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11
Q

Since Ka values of many weak acids are high/low what values are used to compare

A
  • very low, pKa values are used instead to compare the strengths of weak acids with each other

pKa= -log10Ka

  • The less positive the pKa value the more acidic the acid is
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12
Q

The Kw is the

A
  • ionic product of water
    • It is the equilibrium constant for the dissociation of water at 298 K
    • Its value is 1.00 x 10-14 mol2 dm-6
  • For the ionisation of water the equilibrium expression to find Kw is as follows:

H2O (l) ⇌ H+ (aq) + OH- (aq)

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13
Q

Kw : As the extent of ionisation is

A
  • is very low, only small amounts of H+ and OH- ions are formed
  • The concentration of H2O can therefore be regarded as constant and removed from the Kw expression
  • The equilibrium expression therefore becomes:
  • Kw* = [H+] [OH-]
  • As the [H+] = [OH+] in pure water, the equilibrium expression can be further simplified to:
  • Kw* = [H+]2
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14
Q

Calculating [H+] & pH

A
  • If the concentration of H+ of an acid or alkali is known, the pH can be calculated using the equation:

pH = -log [H+]

  • Similarly, the concentration of H+ of a solution can be calculated if the pH is known by rearranging the above equation to:

[H+] = 10-pH

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15
Q

Strong acids are completely

A
  • ionised in solution

HA (aq) → H+ (aq) + A- (aq)

  • Therefore, the concentration of hydrogen ions ([H+]) is equal to the concentration of acid ([HA])
  • The number of hydrogen ions ([H+]) formed from the ionisation of water is very small relative to the [H+] due to ionisation of the strong acid and can therefore be neglected
  • The total [H+] is therefore the same as the [HA]
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16
Q

Strong alkalis are

A
  • completely ionised in solution

BOH (aq) → B+ (aq) + OH- (aq)

  • Therefore, the concentration of hydroxide ions ([OH-]) is equal to the concentration of base ([BOH])
    • Even strong alkalis have small amounts of H+ in solution which is due to the ionisation of water
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17
Q
  • The concentration of OH- in solution can be used to calculate the pH using the ionic product of water
A

Kw = [H+] [OH-]

  • Since Kw is 1.00 x 10-14 mol2 dm-6
  • Once the [H+] has been determined, the pH of the strong alkali can be founding using pH = -log[H+]
  • Similarly, the ionic product of water can be used to find the concentration of OH- ions in solution if [H+] is known
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18
Q

A buffer solution is a

A

solution in which the pH does not change a lot when small amounts of acids or alkalis are added

  • A buffer solution is used to keep the pH almost constant
  • A buffer can consists of weak acid - conjugate base or weak base - conjugate acid
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19
Q

A common buffer solution is an

A

aqueous mixture of ethanoic acid and sodium ethanoate

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20
Q

Ethanoic acid is a …acid

A
  • weak acid and partially ionises in solution to form a relatively low concentration of ethanoate ions
21
Q
  • When H+ ions are added:
A
  • The equilibrium position shifts to the left as H+ ions react with CH3COO- ions to form more CH3COOH until equilibrium is re-established
  • As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much as it reacts with the added H+ ions
  • As there is a large reserve supply of CH3COOH the concentration of CH3COOH in solution doesn’t change much as CH3COOH is formed from the reaction of CH3COO- with H+
  • As a result, the pH remains reasonable constant
22
Q
  • When OH- ions are added:
A
  • The OH- reacts with H+ to form water

OH- (aq) + H+ (aq) → H2O (l)

  • The H+ concentration decreases
  • The equilibrium position shifts to the right and more CH3COOH molecules ionise to form more H+ and CH3COO- until equilibrium is re-established

CH3COOH (aq) → H+ (aq) + CH3COO- (aq)

  • As there is a large reserve supply of CH3COOH the concentration of CH3COOH in solution doesn’t change much when CH3COOH dissociates to form more H+ ions
  • As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much
  • As a result, the pH remains reasonable constant
23
Q

When hydroxide ions are added to the solution, the hydrogen ions react with them to form water; The decrease in hydrogen ions would mean that the pH would increase however the equilibrium moves to the right to replace the removed hydrogen ions and keep the pH constant

24
Q

Uses of buffer solutions in controlling the pH of blood

A
  • In humans, HCO3- ions act as a buffer to keep the blood pH between 7.35 and 7.45
  • Body cells produce CO2 during aerobic respiration
  • This CO2 will combine with water in blood to form a solution containing H+ ions

CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

25
Q
  • This equilibrium between CO2 and HCO3- is extremely important
  • If the concentration of H+ ions is not regulated, the blood pH would
A

drop and cause ‘acidosis’

  • Acidosis refers to a condition in which there is too much acid in the body fluids such as blood
  • This could cause body malfunctioning and eventually lead to coma
26
Q
  • If there is an increase in H+ ions in the blood
A
  • The equilibrium position shifts to the left until equilibrium is restored

H+ (aq) + HCO3- (aq) → CO2 (g) + H2O (l)

  • This reduces the concentration of H+ and keeps the pH of the blood constant
27
Q

If there is a decrease in H+ ions in the blood

A
  • The equilibrium position shifts to the right until equilibrium is restored

CO2 (g) + H2O (l) → H+ (aq) + HCO3- (aq)

  • This increases the concentration of H+ and keeps the pH of the blood constant
28
Q

The pH of a buffer solution can be calculated using:

A
  • The Ka of the weak acid
  • The equilibrium concentration of the weak acid and its conjugate base (salt)
29
Q

To determine the pH, the concentration of hydrogen ions is needed which can be found using the equilibrium expression

A
  • To simplify the calculations, logarithms are used such that the expression becomes:
  • Since -log10 [H+] = pH, the expression can also be rewritten as:
30
Q

Solubility is defined as the

A

number of grams or moles of compound needed to saturate 100 g of water, or it can also be defined in terms of 1 kg of water, at a given temperature

  • For example, sodium chloride (NaCl) is considered to be a soluble salt as a saturated solution contains 36 g of NaCl per 100 g of water
31
Q

The solubility product (Ksp) is

A
  • The product of the concentrations of each ion in a saturated solution of a relatively soluble salt
  • At 298 K
  • Raised to the power of their relative concentrations

C (s) ⇌ aAx+ (aq) + bBy- (aq)

Ksp = [Ax+ (aq)]a [By- (aq)]b

32
Q

When an undissolved ionic compound is in contact with a saturated solution of its ions, an (what?)

A
  • equilibrium is established
  • The ions move from the solid to the saturated solution at the same rate as they move from the solution to the solid
    • For example, the undissolved magnesium chloride (MgCl2) is in equilibrium with a saturated solution of its ions

MgCl2 (s) ⇌ Mg2+ (aq) + 2Cl- (aq)

33
Q

The Ksp is only useful for

A
  • sparingly soluble salts
  • The smaller the value of Ksp, the lower the solubility of the salt
34
Q

The general equilibrium expression for the solubility product (Ksp) is:

A

C (s) ⇌ aAx+ (aq) + bBy- (aq)

Ksp = [Ax+ (aq)]a [By- (aq)]b

35
Q

Calculations involving the solubility product (Ksp) may include:

A
  • Calculating the solubility product of a compound from its solubility
  • Calculating the solubility of a compound from the solubility product
36
Q

A saturated solution is a solution that contains the

A
  • maximum amount of dissolved salt
  • If a second compound, which has an ion in common with the dissolved salt, is added to the saturated solution, the solubility of the salt reduces and a solid precipitate will be formed
  • This is also known as the common ion effect
37
Q

The solubility product can be used to predict whether a precipitate will

A

l actually form or not

  • A precipitate will form if the product of the ion concentrations is greater than the solubility product (Ksp)
38
Q

Common ion effect in silver chloride

A
  • When a KCl solution is added to a saturated solution of AgCl, an AgCl precipitate forms
  • In a saturated AgCl solution, the silver chloride is in equilibrium with its ions

AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)

  • When a solution of potassium chloride is added:
    • Both KCl and AgCl have the common Cl- ion
    • There is an increased Cl- concentration so the equilibrium position shifts to the left
    • The increase in Cl- concentration also means that [Ag+ (aq)] [Cl-(aq)] is greater than the Ksp for AgCl
    • As a result, the AgCl is precipitated
39
Q

The partition coefficient (Kpc) is the

A

ratio of the concentrations of a solute in two different immiscible solvents in contact with each other when equilibrium has been established (at a particular temperature)

40
Q

A separating funnel

A
  • is shaken with the organic solvent and aqueous methylamine
  • The methylamine is soluble in both solvents, so when the mixture is left to settle an equilibrium is established
    • The rate of methylamine molecules moving from the organic layer into the aqueous layer is equal to the rate of molecules moving from the aqueous layer to the organic layer

CH3NH2(aq) ⇌ CH3NH2(organic solvent)

41
Q

The value of its equilibrium constant is also called the

A

partition coefficient

42
Q

The partition coefficient is the ratio of methylamine molecules in the organic and aqueous layer once equilibrium has been established

43
Q

The partition coefficient (Kpc) for a system

A

in which the solute is in the same physical state in the two solvents can be calculated using the equilibrium expression

44
Q

The partition coefficient (Kpc) depends on the

A

solubilities of the solute in the two solvents

45
Q

The degree of solubility of a solute is determined by how strong the

A

intermolecular bonds between solute and solvent are

46
Q

The strength of these intermolecular bonds, in turn, depends on the

A

polarity of the solute and solvent molecules

47
Q

When Kpc is < 1

A
  • the solute is more soluble in water than the organic solvent
48
Q

When Kpc is > 1

A

the solute is more soluble in the organic solvent than the water

Chemistry A Level CIE (18 decks)

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