ACIDS AND BASES

Question 1

Bronsted-Lowry defintion of an acid

Answer 1

Proton donator

Question 2

Bronsted-Lowry defintion of a base

Answer 2

Proton acceptor

Question 3

Acid–base equilibria (involving transfer of protons)

Answer 3

HCl (g) + H2O (l) –> H3O+
(aq) + Cl-
(aq)

Question 4

Calculating pH

Answer 4

pH = –log10[H+]

Question 5

Calculating Hydrogen Ion concentration [H+] from pH

Answer 5

[H+] = 10(to the power of) -pH

Question 6

pH value should always be given to what?

Answer 6

2 d.p

Question 7

what are monoprotic acids?

Answer 7

• each molecule of acid will release one proton when it dissociates, meaning 1 mole of acid produces 1 mole of hydrogen ions
  so H+ conc is same as acid conc
  e.g. HCl or HNO3

Question 8

what are diprotic acids?

Answer 8

each molecule of strong diprotic acid acid releases 2 protons when it dissociates

• diprotic acids produce 2 mol of Hydrogen ions for each mole of acid
  e. g. H2SO4

Question 9

what do strong and weak acids do in water?

Answer 9

• strong acids completely dissociate (or ionise) in water, nearly all H+ ions released
  HCl –> H+ + Cl-
  (reversible reaction but equilibrium lies very far to the right)
• weak acids dissociate partially in water, only small numbers of H+ formed

Question 10

what do strong and weak bases do in water?

Answer 10

• strong bases dissociate completely
  NaOH –> Na+ + OH-
• weak bases dissociate partially
  (just like weak acids, equilibrium lies well over to the left)

Question 11

water dissociation

Answer 11

H2O H+ + OH-
- water only disssociates partially so equilibrium lies well over to the left
-

Question 12

The ionic product of water constant

Answer 12

Kw= [H+][OH-]

Question 13

units of Kw (ionic product of water)

Answer 13

mol2 dm-6

Question 14

value of Kw changes as what else changes?

Answer 14

the temperature

Question 15

How is the ionic product of water constant changed when dealing with PURE WATER?

Answer 15

Kw= [H+]2 (sqaured)
b/c in pure water there’s always one H+ ion for each OH- ion
[OH-] = [H+]

Question 16

at 298K (25 C) what is the value of Kw for all aqueous solutions?

Answer 16

1x10-14 mol2dm-6

Question 17

where is Kw derived from?

Answer 17

Kc = [H+][OH-] / [H2O]

[H2O] is much bigger than conc of ions, so we assume its value is constant and make a new constant Kw

Question 18

Calculate the pH of water at 50ºC given that Kw =

5.476 x 10-14 mol2 dm-6 at 50ºC

Answer 18

SOLUTION:
[H+(aq) ] = √ Kw = √ 5.476 x 10-14 =  2.34 x 10-7 mol dm-3
pH = - log (2.34 x 10-7) = 6.6
It is still neutral though as 
[H+(aq) ] = [OH-(aq)]