topic 12 - acid base equilibria

Question 1

acid

Answer 1

proton donor
positive charge/becomes more negative
higher pH weaker the acid
carboxylic - weak
others strong

Question 2

base

Answer 2

proton acceptor
lone pair/becomes more positive
higher pH stronger base
oh - strong base
NH3 - weak base

Question 3

conjugate base pair

Answer 3

strong reactactant forms weak opposite product

Question 4

monoprotic/monobasic

Answer 4

donates/accepts 1 proton

Question 5

diprotic/dibasic

Answer 5

donates/accepts 2 protons
e.g sulfuric acid

Question 6

amphoteric

Answer 6

acts as either a base or an acid

Question 7

strong acid

Answer 7

completely dissociates in an aqueous solution
conc = [H+]
pH = -lg[H+]
[H+] = 10^-pH
pH increases by a factor of one for each 10x decrease of conc

Question 8

weak acid

Answer 8

partly dissociated <10%
HA ⇌ H+ + A-
[H+] = square root(Ka x [acid])
pH=-lg[H+]
pH increases 0.5 for each 10x decrease in conc

Question 9

dissociation of water

Answer 9

H2O ⇌ H+ + OH-
Kw = [H+][OH-]
pKw = -lgKw

Question 10

equivalence point

Answer 10

when an acid and base have reacted together in the exact proportions as dictatef by the stoichiometric equation

Question 11

strong acid strong base

Answer 11

equivalence 7

Question 12

weak acid strong base

Answer 12

equivalence 8-9
conjugate base makes it alkaline

Question 13

strong acid weak base

Answer 13

equivalence <7
- conjugate acid makes it acidic

Question 14

weak acid weak base

Answer 14

point of inflexion

Question 15

indicators

Answer 15

either weak acids or weak bases
The best indicator for a particular titration is the one with pKIn close to pH at equivalence

Question 16

buffer solution

Answer 16

A buffer solution’s pH is almost unchanged on the addition of small amounts of acid or base
Buffer solutions can be made by mixing:
A buffer solution can be made from a weak acid and a salt of the acid.
The acid is partially dissociated and the salt fully dissociated
The resulting solution has a high concentration of undissociated acid molecules, CH3COOH(aq), and a high concentration of the conjugate base, CH3COO-(aq)

Question 17

calculating the ph of buffer sol

Answer 17

use Ka and conc of acid and salt
Assume the dissociation of the acid is negligible
When mixed the concentrations of each solution are halved as the volume doubles.
pH = pKa (-lgKa) +log [salt]/ [acid]
[H+]=Ka x [acid]/[salt]
pH=-lg[H+]

Question 18

making buffer

Answer 18

To make a buffer with a pH < 7 - use a mixture of a weak acid and its conjugate base
e.g - sodium hydroxide and ethanoic acid
To make a buffer with a pH > 7 - use a mixture of a weak base and its conjugate acid

Question 19

eg strong acids

Answer 19

HCl

Question 20

strong base

Answer 20

NaOH

Question 21

eg weak acids

Answer 21

CH3COOH
-carboxylic

Question 22

eg weak bases

Answer 22

NH3

Question 23

phenolphthalein indicator

Answer 23

acid-colourless
neutral - pale pink
base-pink

Question 24

methyl orange

Answer 24

acid - red
neutral - orange
base - yellow
only with strong acids

Question 24

when can you calculatte Kc using equilibrium amounts

Answer 24

when moles are equal on both sides as volume cancels

Question 25

enthlapy change of reaction of diff conc of acid

Answer 25

strong acids have high values
dibasic acids have 2x enthlapy change for 1 mol as 2 mols are formed
weak acids are lower as some energy is required to break o-h bonds to realse h+ ions as includes dissociation

Question 26

good indicator

Answer 26

ph at equivilence point is very close to pka of indicator
ph region of indication lies completely in vertical region on curve

Question 27

buffer action

Answer 27

carbonate ions react with h+ ions
hydrogencarbonate ions react with oh- ions

Question 27

strong base

Answer 27

Kw = [H+][OH-]=1x10^-14
[H+