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ENG CHEM > Chapter 5 Equilibra > Flashcards

Chapter 5 Equilibra Flashcards

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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
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
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
19
Q

A common buffer solution is an

A

aqueous mixture of ethanoic acid and sodium ethanoate

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

A
24
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)
25
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:
26
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
27
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

28
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)

29
Q

The Ksp is only useful for

A
  • sparingly soluble salts
  • The smaller the value of Ksp, the lower the solubility of the salt
30
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

31
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
32
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
33
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)
34
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
35
Q

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

A

intermolecular bonds between solute and solvent are

36
Q

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

A

polarity of the solute and solvent molecules

ENG CHEM (9 decks)

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