Oceans - Acid/Base Equilibria

Question 1

what is an acid?

Answer 1

a substance that donates a hydrogen ion in solution

Question 2

what is a base?

Answer 2

a substance that can accept a H+ in solution

Question 3

what is a conjugate base?

Answer 3

an acid that has donated a proton but then can re-accept a proton back

Question 4

what is a conjugate acid?

Answer 4

a base that has accepted a proton but could donate it back into solution

Question 5

how can water act as an acid?

Answer 5

H₂O -> OH^- + H⁺

Question 6

how can water act as a base?

Answer 6

H₂O + H⁺ -> H₃O⁺

Question 7

what is an alkali?

Answer 7

a base dissolved in water

Question 8

what is an oxonium ion?

Answer 8

H₃O⁺

Question 9

what are examples of acids that don’t dissolve in water?

Answer 9

benzoic acid

carboxylic acids

Question 10

what is the equation for the dissociation of a strong acid?

Answer 10

HA -> H⁺ + A^-

Question 11

what is pH

Answer 11

the representation of protons in solution

Question 12

what is the equation for calculating pH?

Answer 12

pH = -log₁₀[H⁺]

Question 13

what is a strong acid?

Answer 13

an acid that dissociates fully in solution

Question 14

what are examples of strong acids?

Answer 14

hydrochloric acid

sulfuric acid

Question 15

what is the relationship between concentration of the acid and pH in strong acids?

Answer 15

1 x 10^-x molar solution = pH x

eg

0.001 molar solution = pH 3

because -log₁₀[0.001] = 3

Question 16

what is the equation for calculating the concentration of hydrogen ions from pH?

Answer 16

[H⁺] = 10^-pH

Question 17

what is a weak acid?

Answer 17

an acid that doesn’t dissociate fully in solution

Question 18

what is the equation for the dissociation of a weak acid?

Answer 18

HA ⇔ H⁺ + A^-

Question 19

what is an example of a weak acid?

Answer 19

carboxylic acids

Question 20

what is Ka?

Answer 20

the acid dissociation constant

similar to Kc

Question 21

what assumptions can be made when calculating the Ka for weak acids?

Answer 21

1. conc of conjugate base formed = conc of hydrogen ions present
2. initial conc of acid = conc of acid at equilibria

Question 22

when is the assumption that the conc of the conjugate base is equal to the conc of the hydrogen ions in weak acids less valid?

Answer 22

in dilute acids - more dissociation of water into hydroxide + hydrogen

Question 23

when is the assumption that the initial conc of a weak acid is equal to the conc of the acid at equilibrium less valid?

Answer 23

when the acid is stronger / when Ka is bigger

Question 24

how can Ka be calculated using the example of the dissociation of ethanoic acid?

Answer 24

[CH₃COO^-] [H⁺]

[CH₃COOH]

Question 25

what is the dissociation of a strong base?

Answer 25

dissociates completely

Question 26

what is Kw?

Answer 26

ionisation product of water

Question 27

what is the ionisation product of water?

Answer 27

the product of the conc of H+ ions and the conc of hydroxide ions from the dissociation of water

Question 28

how can Kw be calculated?

Answer 28

Kw = [H⁺] [OH^-] also given on datasheet

Question 29

what is the expression for the Ka of water?

Answer 29

Question 31

what is a buffer solution?

Answer 31

a solution that resists change in pH for **small** additions of acid or alkali

Question 32

how are simple buffer solutions made?

Answer 32

weak acid mixed with salt of a strong base and weak acid

Question 33

if ethanoic acid is used in a buffer, how will it resist hydroxide ions being added to the solution?

Answer 33

ethanoic acid deprotonated by reacting with hydroxide ions to form ethanoate ions and water

Question 34

if ethanoic acid is used in a buffer solution, how will it resist the change in pH from hydrogen ions being added?

Answer 34

ethanoate ions from the salt will react with the hydrogen ions, producing more ethanoic acid

Question 35

what assumptions can be made about simple buffers, using example HA ⇔ H⁺ + A^-

Answer 35

all A^- ions come from the salt because the weak acid has low levels of dissociation almost all HA molecules put into the buffer remain unchanged

Question 36

what is the equation for calculating the Ka of a buffer solution?

Answer 36

Question 38

what is the solubility product / Ksp?

Answer 38

represents conditions for equilibrium between sparingly soluble solid and its saturated solution

Question 39

using example: CaCO_3(s) ⇔ Ca²⁺_(aq) + CO₃^2-_(aq) give the expression for the Ksp of calcium carbonate

Answer 39

Ksp (CaCO_3(s)) = [Ca²⁺_(aq)] [CO₃^2-_(aq)]

Question 40

what will happen when solids are added to equilibria involving solubility products?

Answer 40

the equilibrium will **not** shift because the solution is saturated - causing the equilibrium to be unaffected by any solids present

Question 41

what can Ksp be used to predict?

Answer 41

whether a precipitate will be formed from a solution

Question 42

using example: CaCO_3(s) ⇔ Ca²⁺_(aq) + CO₃^2-_(aq) explain when Ksp can be used to predict a formation of a precipitate

Answer 42

if [Ca²⁺_(aq)] x [CO₃^2-_(aq)] gives a value in excess of Ksp, then CaCO₃ will precipitate out of the solution

Question 43

using example: CaCO_3(s) ⇔ Ca²⁺_(aq) + CO₃^2-_(aq) explain how Ksp can be used to predict that a precipitate will not be formed

Answer 43

if [Ca²⁺_(aq)] x [CO₃^2-_(aq)] is \< or = Ksp, a calcium carbonate precipitate will not be formed - ions will stay in solution