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A-LEVEL CHEMISTRY > Equilibria > Flashcards

Equilibria Flashcards

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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

Study These Flashcards
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

Study These Flashcards
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

Study These Flashcards
A

weak acid - conjugate base or
weak base - conjugate acid

25
Name a common buffer solution
aqueous mixture of ethanoic acid and sodium ethanoate
26
Ethanoic acid & sodium ethanoate as a buffer
-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
When H+ ions are added to sodium ethanoate buffer solution
-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
When OH- ions are added to sodium ethanoate buffer solution
-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
Uses of buffer solutions in controlling the pH of blood
-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
pH of a buffer solution can be calculated using
-The Ka of the weak acid -The equilibrium concentration of the weak acid and its conjugate base (salt)
31
[H+] in a buffer solution=
Ka x [acid]/[salt]
32
pH in buffer solutions=
pKa x log10 [salt]/[acid]
33
Solubility
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
Ksp
-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
C (s) ⇌ aA^x+ (aq) + bB^y- (aq) Ksp =
[A^x+ (aq)]^a *[B^y- (aq)]^b
36
When is equilibrium established for ions
When an undissolved ionic compound is in contact with a saturated solution of its ions,
37
What happens when an equilibrium is established in ionic compounds
The ions move from the solid to the saturated solution at the same rate as they move from the solution to the solid
38
Ksp is only useful for
sparingly soluble salts
39
The smaller the value of Ksp
the lower the solubility of the salt
40
Calculations involving the solubility product (Ksp) may include
-Calculating the solubility product of a compound from its solubility -Calculating the solubility of a compound from the solubility product
41
Calculating the solubility product of a compound from its solubility
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
Calculating the solubility of a compound from its solubility product
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
solubility product can't be used for
-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
saturated solution
a solution that contains the maximum amount of dissolved salt
45
common ion effect
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
How can solubility product be used to predict whether a precipitate will actually form or not
A precipitate will form if the product of the ion concentrations is greater than the solubility product (Ksp)
47
partition coefficient (Kpc)
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
Kpc=
[concentration of solute in organic solvent]/[concentration of solute in aqueous solvent]
49
Factors Affecting the Partition Coefficient
-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
When Kpc is < 1
the solute is more soluble in water than the organic solvent
51
When Kpc is > 1
the solute is more soluble in the organic solvent than the water

A-LEVEL CHEMISTRY (14 decks)

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