Cambridge International
A2 Level
Chapter 20: Ionic Equilibria
CIE A2 Chemistry
, pH AND pOH
pH: the negative logarithm of [H+] (in
mol dm–3)
pH = –log [H+]
∴ [H+] = 10–pH
The negative log will give positive
number for pH
pOH: the negative logarithm of [OH–]
pOH = –log [OH–]
∴ [OH–] = 10–pOH
CIE A2 Chemistry
, pH
[H+][OH-] = Kw = 1×10-14 mol2 dm–6
∴ -log [H+] +(-log [OH-]) = -log (1×10-14M)
pH is also a way to express hydrogen
concentration
Acidic: [H+]>1×10-7 mol dm–3; pH < 7.00
Basic: [H+]<1×10-7 mol dm–3; pH > 7.00
Neutral: [H+]=1×10-7 mol dm–3; pH = 7.00
The relation can be described as
↓pH = -log [H+]↑
CIE A2 Chemistry
, pH AND pOH
In determining the pH of strong acid and
base, the [acid]initial and [base]initial is equal
to [dissociated ions]
Example: Determine the pH of 0.15moldm–3
Ba(OH)2 at 25oC.
Ba(OH)2 (aq) → Ba2+ (aq) + 2OH–
0.15 0.15 2(0.15)
Alternatively,
pOH = –log (0.3) ∴ pH = 14 – 0.52
= 0.52 = 13.48
CIE A2 Chemistry
A2 Level
Chapter 20: Ionic Equilibria
CIE A2 Chemistry
, pH AND pOH
pH: the negative logarithm of [H+] (in
mol dm–3)
pH = –log [H+]
∴ [H+] = 10–pH
The negative log will give positive
number for pH
pOH: the negative logarithm of [OH–]
pOH = –log [OH–]
∴ [OH–] = 10–pOH
CIE A2 Chemistry
, pH
[H+][OH-] = Kw = 1×10-14 mol2 dm–6
∴ -log [H+] +(-log [OH-]) = -log (1×10-14M)
pH is also a way to express hydrogen
concentration
Acidic: [H+]>1×10-7 mol dm–3; pH < 7.00
Basic: [H+]<1×10-7 mol dm–3; pH > 7.00
Neutral: [H+]=1×10-7 mol dm–3; pH = 7.00
The relation can be described as
↓pH = -log [H+]↑
CIE A2 Chemistry
, pH AND pOH
In determining the pH of strong acid and
base, the [acid]initial and [base]initial is equal
to [dissociated ions]
Example: Determine the pH of 0.15moldm–3
Ba(OH)2 at 25oC.
Ba(OH)2 (aq) → Ba2+ (aq) + 2OH–
0.15 0.15 2(0.15)
Alternatively,
pOH = –log (0.3) ∴ pH = 14 – 0.52
= 0.52 = 13.48
CIE A2 Chemistry
