Topic 13: Acids, Bases, Buffers and Solubility 1

Question 1

Bronsted-Lowry acid and base

Answer 1

Acid- proton donor
Base- proton acceptor

Question 2

HA (aq) + H2O (l) <=>

Answer 2

H3O^+ (aq) + A^- (aq)

Question 3

Ka =
If acid concentration is not too low, assume

Answer 3

[H3O^+][A^-]/[HA]
[B][H3O^+]/[BHA^+]

[H3O^+]^2/[HA]

Question 4

B (aq) + H2O (l)

Answer 4

BH^+ (aq) + OH^- (aq)

Question 5

Kb =

Answer 5

[BH^+][OH^-]/[B]
[AH][OH^-]/[B]

Question 6

Conjugate base of an acid

Answer 6

The species formed when an acid donates a proton

Question 7

Conjugate acid of a base

Answer 7

The species formed when a base gains a proton

Question 8

A^- (aq) + H2O (l)

Answer 8

AH (aq) + OH^- (aq)

Question 9

BH^+ (aq) + H2O (l) <=>

Answer 9

B (aq) + H3O^+ (aq)

Question 10

Kw is the … constant of water
The equation is =

Answer 10

autoprotolysis
[H3O^+][OH^-] = 1.00 x 10^-14 at room temp (equilibrium on left)

Question 11

pKa =

Answer 11

-log10Ka

Question 12

Ka (from pKa) =

Answer 12

10^-pKa

Question 13

pKa of super acids - strong acids

Answer 13

-10 - -5

Question 14

pKa of weak acid/weak bases

Answer 14

3-6

Question 15

pKa of weak acid/medium base

Answer 15

8-10

Question 16

pKa of strong base

Answer 16

15-20

Question 17

Strong acid
pKa =
Ka =

Answer 17

Almost completely dissociates into ions in water
<0
»1

Question 18

Weak acid
pKa =
Ka =

Answer 18

Partially dissociates into water
>0
«1

Question 19

Concentration

Answer 19

number of atoms or molecules in solution

Question 20

Strength of acid

Answer 20

Degree of dissociation of ions as the acid dissolves in water which is related to the proton affinity of the conjugate base

Question 21

[A^-] = [H3O^+] when

Answer 21

No buffer is present

Question 22

[H3O^+] =

Answer 22

Root (Ka[HA])

Question 23

3 types of Bronsted acids

Answer 23

aqua acids
hydroxoacids
oxoacids

Question 24

All Bronsted acids have a … group attached to the … .The … on this group is an …

Answer 24

OH x
central atom
proton
acidic proton

Question 25

Many metal salts create ... solutions when they are dissolved in water. The ions in water become ...

Answer 25

acidic hydrated

Question 26

The acidic proton of an aqua acid is on the ... molecule coordinated to the ...

Answer 26

water central metal ion

Question 27

[M(H2O)n]^n+ <=> The water is behaving as a ... due to the ... of the ... bonds. The H2O bonds to the ... via the ... .Charge is drawn towards the ... which ... This is a type of ...

Answer 27

[M(H2O)n-1(OH)]^(n-1)+ + H^+ Bronsted acid polarisation O-H M^n+ oxygen metal centre weakens the O-H bond hydrolysis

Question 28

Higher charge density (higher charge/smaller ion) = ... hydrolysis and ... acidic

Answer 28

more more

Question 29

In hydroxoacids and oxoacids, the ... proton is on a ... group. In a hydroxoacid, this is ... a neighbouring ... group. In an oxoacid, this is ... a neighbouring ... group. Neighbouring groups attached to the same atom

Answer 29

acidic hydroxyl without =O with =O

Question 30

Name 4 examples of mononuclear oxoacids

Answer 30

carbonic acid nitric acid phosphoric acid sulfuric acid

Question 31

E(O)x(OH)y (aq) + H2O (l) <=>

Answer 31

E(O)x(OH)y-1(O^-) (aq) + H3O^+ (aq)

Question 32

Can substitute oxyacids by replacing the ... with another atom. This can effect the acid's ...

Answer 32

OH group strength

Question 33

You can increase the acids strength by ...

Answer 33

adding an electron withdrawing group such as F/CF3 which withdraws electron density from the central atom making it more electropositive

Question 34

You can decrease the acids strength by ...

Answer 34

adding an electron donating group such as NH2 which donates electron density to the central atom making it more electronegative

Question 35

pKa of the first deprotonation of E(O)x(OH)y pKa of the second + subsequent deprotonations of E(O)x(OH)y

Answer 35

pKa1 = 8 - 5x pKa2 = pKa1 + 5

Question 36

Higher oxidation state means ... acid

Answer 36

stronger

Question 37

Buffers control the ...

Answer 37

pH

Question 38

Henderson-Hasselbalch equation

Answer 38

pH = pKa + log10 [A^-]/[HA]

Question 39

Derive Henderson-Hasselbalch from Ka

Answer 39

Ka = [H3O^+][A^-]/[HA] log10Ka = log10 ([H3O^+][A^-]/[HA]) = log10[H3O^+] + log10[A^-]/[HA] -log10Ka = -log10[H3O^+] - log10[A^ ]/[HA] pKa = pH - log10[A^ ]/[HA] pH = pKa + log10[A^ ]/[HA]

Question 40

pH =

Answer 40

-log10[H^+]

Question 41

pKa + pKb =

Answer 41

14

Question 42

Ka x Kb =

Answer 42

Kw

Question 43

Equivalence point

Answer 43

quantity of titrant is precisely correct for a given stoichiometric reaction with analyte

Question 44

Estimating pH of a weak acid [𝐻3𝑂+] =

Answer 44

root (𝐾𝑎[𝐻𝐴]𝑎𝑞,𝑖𝑛𝑖𝑡𝑖𝑎𝑙)

Question 45

If acid conc not too low, assume

Answer 45

[HA] = [HA] aq, initial