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A-LEVEL CHEMISTRY > Equilibria > Flashcards

Equilibria Flashcards

1
Q

A Brønsted acid can

A

donate a proton

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2
Q

A Brønsted base can

A

accept a proton

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3
Q

In an equilibrium reaction, the products are formed at the same rate as the

A

reactants are used

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4
Q

At equilibrium, both reactants and products are

A

present in the solution

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5
Q

Conjugate acid-base pairs are

A

a pair of reactants and products that are linked to each other by the transfer of a proton

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6
Q

The pH indicates

A

the acidity or basicity of an acid or alkali

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7
Q

pH =

A

-log10 [H+]

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8
Q

[H+] =

A

10^-pH

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9
Q

Ka is the

A

acidic dissociation constant
the equilibrium constant for the dissociation of a weak acid at 298 K

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10
Q

Ka=

A

[H+][A-]/[HA]
a more simplified version is:
[H+]^2/[HA]
because the ratio of H+ ions and A- ions are 1:1, therefore their concentration is the same

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11
Q

The value of Ka indicates

A

the extent of dissociation

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12
Q

The assumptions made when writing the equilibrium expression for weak acids

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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13
Q

A high value of Ka means that

A

-The equilibrium position lies to the right
-The acid is almost completely ionised
-The acid is strongly acidic

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14
Q

A low value of Ka means that

A

-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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15
Q

pKa values are used

A

to compare the strengths of weak acids with each other since Ka values of many weak acids are very low

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16
Q

pKa =

A

-log10 Ka

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17
Q

The less positive the pKa value

A

the more acidic the acid

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18
Q

Calculating the Ka & pKa of weak acids

A

Step 1: Write down the equation for the partial dissociation of weak acid
Step 2: Write down the equilibrium expression to find Ka
Step 3: Substitute the values into the expression to find Ka
Step 4: Determine the units of Ka
Step 6: Find pKa

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19
Q

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

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20
Q

Kw=

A

[H+][OH-]/[H2O]
[H+][OH-]
[H+]^2

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21
Q

Why can the concentration of H2O be regarded as a constant and be removed from the Kw expression

A

Because the extent of ionisation is very low, only small amounts of H+ and OH- ions are formed

22
Q

Kw/[OH-]=

23
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

24
Q

buffer solution can consist of

A

weak acid - conjugate base or
weak base - conjugate acid

25
Q

Name a common buffer solution

A

aqueous mixture of ethanoic acid and sodium ethanoate

26
Q

Ethanoic acid & sodium ethanoate as a buffer

A

-Ethanoic acid is a weak acid and partially ionises in solution to form a relatively low concentration of ethanoate ions
-Sodium ethanoate is a salt which fully ionises in solution
-The buffer solution contains relatively high concentrations of CH3COOH (due to ionisation of ethanoic acid) and CH3COO- (due to ionisation of sodium ethanoate)

27
Q

When H+ ions are added to sodium ethanoate buffer solution

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

28
Q

When OH- ions are added to sodium ethanoate buffer solution

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

29
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)
-This equilibrium between CO2 and HCO3- is extremely important
-If the concentration of H+ ions is not regulated, the blood pH would 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

-If there is an increase in H+ ions
-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

-If there is a decrease in H+ ions
-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

30
Q

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)

31
Q

[H+] in a buffer solution=

A

Ka x [acid]/[salt]

32
Q

pH in buffer solutions=

A

pKa x log10 [salt]/[acid]

33
Q

Solubility

A

the 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

34
Q

Ksp

A

-solubility product
-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

35
Q

C (s) ⇌ aA^x+ (aq) + bB^y- (aq)
Ksp =

A

[A^x+ (aq)]^a *[B^y- (aq)]^b

36
Q

When is equilibrium established for ions

A

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

37
Q

What happens when an equilibrium is established in ionic compounds

A

The ions move from the solid to the saturated solution at the same rate as they move from the solution to the solid

38
Q

Ksp is only useful for

A

sparingly soluble salts

39
Q

The smaller the value of Ksp

A

the lower the solubility of the salt

40
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

41
Q

Calculating the solubility product of a compound from its solubility

A

Step 1: Write down the equilibrium equation
Step 2: Write down the equilibrium expression
Step 3: Calculate the ion concentrations in the solution
Step 4: Substitute the values into the expression to find the solubility product
Step 5: Determine the correct units of Ksp

42
Q

Calculating the solubility of a compound from its solubility product

A

Step 1: Write down the equilibrium equation
Step 2: Write down the equilibrium expression
Step 3: Simplify the equilibrium expression (if possible)
Step 4: Substitute the value of Ksp into the expression to find the concentration

43
Q

solubility product can’t be used for

A

-Group 1 element salts
-All nitrates salts
-All ammonium salts
-Many sulfate salts
-Many halide salts (except for lead(II) halides and silver halides)

44
Q

saturated solution

A

a solution that contains the maximum amount of dissolved salt

45
Q

common ion effect

A

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

46
Q

How can solubility product be used to predict whether a precipitate will actually form or not

A

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

47
Q

partition coefficient (Kpc)

A

the 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)

48
Q

Kpc=

A

[concentration of solute in organic solvent]/[concentration of solute in aqueous solvent]

49
Q

Factors Affecting the Partition Coefficient

A

-The solubilities of the solute in the two solvents
-The degree of solubility of a solute is determined by how strong the intermolecular bonds between solute and solvent are
-The strength of these intermolecular bonds, in turn, depends on the polarity of the solute and solvent molecules

50
Q

When Kpc is < 1

A

the solute is more soluble in water than the organic solvent

51
Q

When Kpc is > 1

A

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

A-LEVEL CHEMISTRY (14 decks)

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