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 ]
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 ]
