module 5 - acids, bases and buffers

Question 1

bronsted-lowry acid

Answer 1

a proton donor

Question 2

bronsted-lowry base

Answer 2

a proton acceptor

Question 3

strong and weak acids

Answer 3

fully v partially dissociate

-strong acids exist manly as ions and weak acids mainly as molecules

Question 4

conjugate-acid base pairs

Answer 4

• contains two species that can be interconverted by transfer of a proton
• they differ by H+

Question 5

hydronium ion

Answer 5

• dissociation doesn’t take place unless water is present
• HCl + H2O –> H3O+ + Cl-
• the hydronium ion is a conjugate acid

Question 6

monobasic, dibasic…

Answer 6

refers to the total number of hydrogen ions that can be replaced per molecule in an acid-base reaction

Question 7

redox equation for acid and a metal

Answer 7

2H+ + Zn –> Zn2+ + H2

Question 8

ionic equation for acids and carbonates

Answer 8

2H+ + CO3-2 –> H2O + CO2

Question 9

ionic equation for acid and metal oxides

Answer 9

2H+ + MgO –> Mg2+ + H2O

Question 10

PH of strong acids

Answer 10

[H+] = [HA]

-therefore can be calculated directly

Question 11

Ph of diluted strong acids

Answer 11

• find the PH before and after dilution

- look at change in PH

Question 12

the acid dissociation constant (Ka)

Answer 12

Ka = [H+] [A-] / [HA]

• units: moldm-3
• changes with temperature
• the larger the Ka value, the further the equilibrium is to the right, the stronger the acid

Question 13

relationship between pKa and Ka

Answer 13

pKa = -log(Ka)
Ka = 10 -pKa
-pKa values are more manageable than Ka values, making it easier to compare acid strengths

Question 14

rules about Ka and pKa

Answer 14

• the stronger the acid, the larger the Ka value and the smaller the pKa value
• the weaker the acid, the smaller the Ka value and the larger the pKa value

Question 15

approximations for PH of weak acids

Answer 15

1. HA dissociates to produce equal equilibrium concentrations of H+ and A-. The dissociation of water is negligible
2. [HA eqm] = [HA start] as the dissociation of weak acids is small

Question 16

when do these approximations break down?

Answer 16

• if there is a very dilute solution (PH > 6)
• if you have a stronger weak acid as H+is too big to be considered negligible
• therefore it is true for weak acids with small Ka values

Question 17

Kw

Answer 17

the ionic product of water (the ions in water multiplied together)

• it varies with temperature
• it has a value of 1x10-14 at 25 degrees

Question 18

prove water has a PH of 7

Answer 18

• On dissociation, water is neutral, it produces the same number of OH- and H+ ions
• Kw = [H+][OH-]
• [H+] = square root of 1x10-14 = 1x10-7
• PH = -log[H+] = 7

Question 19

strong bases

Answer 19

• they fully dissociate so the [OH-] = [NaOH]

- rearrange Kw = [H+][OH-]

Question 20

example of weak base

Answer 20

• a weak base is also an alkali
• ammonia gas is an example of a weak base
• an equilibrium is set up positioned well the left.

Question 21

buffer solution

Answer 21

a system that minimises PH changes when small amounts of an acid or base are added
-they contain a weak acid to remove added alkali and a conjugate base to remove added acids

Question 22

preparing a buffer from a weak acid and its salt

Answer 22

• mix a weak acid with a solution of one of its alts
• when added to water, ethanoic acid partially dissociates, producing a small amount of ethanoate ions. This is the buffers solution source of weak acid.
• salts are a source of conjugate base. when added to water, it completely dissociates
• we need to add the salt as the weak acid only partially dissociates so there isn’t much conjugate base
• this creates large reservoirs of weak acid and conjugate base

Question 23

preparation by partial neutralisation

Answer 23

-the weak acid is partially neutralised by the alkali forming the conjugate base
some of the weak acid is left over which remain unreacted

Question 24

how does a buffer neutralise PH change when acid is added?

Answer 24

• on addition of an acid [H+] increases
• H+ ion react with the conjugate base A-
• the equilibrium position shifts to the left, removing most of the H+ ions and resulting in minimal PH change

Question 25

how does a buffer neutralise the PH change when alkali is added?

Answer 25

- on addition of alkali, [OH-] increases - the small concentration of H+ ions react with the OH- ions to form water - HA dissociates, shifting the equilibrium position to the right to restore most of the H+ ion, minimising change in PH.

Question 26

when are buffers most effective?

Answer 26

when there are equal concentrations of the weak acid and conjugate base.

Question 27

what holds true when [HA] = [A-] in a buffer?

Answer 27

- the PH of the buffer solute is the same as the pKA value of HA - the operating PH is typically over about two PH units, centred at the PH of the pKa value

Question 28

assumptions about calculating PH from buffers

Answer 28

- CH3COOH has not fully dissociated [HA]start = [HA]equilibrium - CH3COOH has fully dissociated [CH3COONa] = [CH3COO-]

Question 29

how do you calculate PH of a buffer solution?

Answer 29

[H+] = Ka x [HA]/[A-]

Question 30

why does blood act as a buffer solution?

Answer 30

- blood PH needs to be maintained at a PH between 7.35 and 7.45 - the carbonic acid-hydrogen carbonate (H2CO3/ HCO3-) is the most important - if the PH slips out of this range, people can develop acidosis which can be fatal. If it goes above this PH range, people can develop alkalosis - the b body produces more acidic material than alkaline which the conjugate base converts to H2CO3. - The body prevents it from building up by converting H2CO3 in to carbon dioxide gas which is exhaled by the lungs.

Question 31

equations for blood buffer

Answer 31

On addition of an acid H2CO3 H+ + HCO3- On addition of an alkali H2CO3 H+ and HCO3-