Summary Acids and Bases Notes - AQA Chemistry A Level

Acids and Bases
 Bronsted-Lowry acid – proton donor
 Bronsted-Lowry base – proton acceptor
 Bronsted-Lowry acid-base reaction – reaction involving the transfer of a proton
 HCl + NaOH -> NaCl + H2O
o HCl donates H+
o NaOH accepts H+

pH of Strong Acids
 Monoprotic acid – acid that releases one hydrogen ion per molecule
o HCl, HNO3, CH3COOH
 Diprotic acid – acid that releases two hydrogen ions per molecule
o H2SO4, H2C2O4
 pH = -log[H+]
 [H+] = 10–pH
 Always give pH to two decimal places
 For diprotic acids, multiply the concentration of acid by 2 to find the concentration of H +

Dilution of a Strong Acid
𝑜𝑙𝑑 𝑣𝑜𝑙𝑢𝑚𝑒
 [H+] in diluted solution = [H+] in original solution x
𝑛𝑒𝑤𝑣𝑜𝑙𝑢𝑚𝑒
The Ionic Product of Water (Kw)
 H2O ⇌ H+ + OH–

 Kc = ¿ ¿
+ –
o Kc[H2O] = [H ][OH ]

o Kw = [H+][OH–]
 As [H2O] is very much greater than [H+] and [OH–], then [H2O] is effectively a constant number
 At 298K, Kw is equal to 1.0x10-14 mol2 dm–6

The Effect of Temperature on the pH of Water and Neutrality of Water
 If temperature is increased, the equilibrium shifts to the right (endothermic direction) to oppose
the increase in temperature
 [H+] and [OH–] increase, and therefore Kw increases
 As [H+] increases, pH decreases
 The water is still neutral as [H+] = [OH–]

Calculating the pH of Water
 In pure water [H+] = [OH–]
 Kw = [H+]2
 [H+] = √Kw

, pH of Strong Bases
 Calculate [OH–]
 Use Kw to find [H+]
𝐾𝑤
o [H+] =
[ 𝑂 𝐻− ]
 Calculate pH

Dilution of a Strong Base

𝑜𝑙𝑑 𝑣𝑜𝑙𝑢𝑚𝑒
 [OH–] in diluted solution = [OH–] in original solution x
𝑛𝑒𝑤 𝑣𝑜𝑙𝑢𝑚𝑒
 Use Kw to find [H+]
 Calculate pH

pH of Mixtures of Strong Acids and Strong Bases
 Calculate moles of H+
 Calculate moles of OH–
 Calculate the moles of excess H+ or OH–
 Calculate excess [H+] or [OH–]
 If [OH–] is in excess, use Kw to find [H+]
 Calculate the pH

Weak Acids
 Strong Acid – Fully dissociates in solution
o HA -> H+ + A–
 Weak acid – Partially dissociates in solution
o HA ⇌ H+ + A–
o Carboxylic Acids

Acid Dissosiation Constant (Ka)

 Ka = ¿ ¿
 pKa = -log(Ka)

 Ka = 10–pKa
 Ka has the units mol dm–3
 The larger the value of Ka, the stronger the acid, more dissociation
 The smaller the value of pKa, the stronger the acid, more dissociation
 In a solution of a weak acid in water with nothing else added:
o [H+] = [A–]
o [HA] at equilibrium is virtually the same as at the start as very little had dissociated
o Ka = ¿ ¿ ¿
o [H+] = √ 𝐾𝑎 × [ 𝐻𝐴 ]