Chapter 15 Acids and Bases

1.  15.2 the Acids and Bases properties of water
 15.3 PH- a measure of acidity
Dr Laila Al-Harbi

2.  Acid: Substance that produces hydrogen ions in
water solution.
HCl (aq) → H+(aq) + Cl‐(aq)
 Base: Substance that produces hydroxide ions in
water solution.
NaOH (aq) → Na+(aq) + OH‐(aq)
 An acid neutralizes a base
H+(aq) + OH‐(aq) → H2O(ℓ)
Dr Laila Al-Harbi

3.  water is unique solvent , it can act as acid or base.
 In pure water, a few molecules act as bases and a few act as acids.
acid(1) + base (1) ⇄ acid(1) + base (1)
 This is referred to as autoionization of water
 The equilibrium expression for this process is
Kc = [H3O+] [OH−]
 This special equilibrium constant is referred to as the ion-product
constant for water, Kw.
At 25°C, Kw = 1.0  10−14

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)
Dr Laila Al-Harbi

4. In pure water,
Kw = [H3O+] [OH−] = 1.0  10−14
Because in pure water [H3O+] = [OH−],
[H3O+] = (1.0  10−14)1/2 = 1.0  10−7
In acidic solution
[H3O+] > [OH−]
In basic solution
[H3O+] < [OH−]
Dr Laila Al-Harbi

5.  Calculate the [H+] ions in
ammonia , [OH-] =0.0025 M
 Kw = [H3O+] [OH−]= 1.0  10−14
 [H3O+] = 1.0  10−14/ [OH−]
 [H3O+] = 1.0  10−14/ 0.0025
 [H3O+] = 4.0  10−12 M
 Calculate the [OH-] ions in a
1.3 M HCl.
 Kw = [H3O+][OH−]= 1.0  10−14
 [OH−] = 1.0  10−14/ [H3O+]
 [OH−] = 1.0  10−14/ 1.3
 [OH−] = 7.7  10−15 M
Dr Laila Al-Harbi

6.  pH is defined as the negative base-10 logarithm of the hydronium
ion concentration.
 In the same manner
 In the same manner
 In pure water,
pH + pOH = 14
pH = pOH = 7
pH = −log [H3O+] …. [H3O+] = 10-pH
pOH = −log [OH-] ….. [OH-] =10-pOH
pKw = −log [14×10-14 ] = 14
Dr Laila Al-Harbi

7. pH Range
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Neutral
[H+]>[OH-] [H+] = [OH-] [OH-]>[H+]
Acidic Basic
pH = pOH = 7 pH >7pH < 7
Increase the acidity Increase the basisty
Dr Laila Al-Harbi

8.  The [H+]=3.2 x 10–4 M.
 The [H+]=1.0 x 10–3 M.
What is the pH in the two
occasions.
 pH = −log [H3O+]
 pH = −log 3.2 x 10–4 = 3.49
 pH = −log 1.0 x 10–3 = 3.00
 [H3O+] increase ,pH
decrease >>> more acidic
 The [H+]=0.76 M, nitric
acid solution ,What is the
pH .
 pH = −log [H3O+]
 pH = −log 0.76 = 0.12
Dr Laila Al-Harbi

9.  The pH = 4.82 , What is the
[H+] of the rain water .
 [H3O+] = 10-pH
 [H3O+] = 10-4.82
 [H3O+] = 1.5 × 10–5M
 The pH = 3.33 , What is the
[H+] of orange juice
 [H3O+] = 10-pH
 [H3O+] = 10-3.33
 [H3O+] = 4.7 × 10–4M
Dr Laila Al-Harbi

10.  The [OH-]=2.9 x 10–4 M.
What is the pH of the NaOH
solution
 pOH = −log [OH-]
 pOH = −log 2.9 x 10–4 =
3.54
pH + pOH = 14
pH = 14 – pOH
= 14-3.54=10.46
 The [OH-]=2.5 x 10–7 M.
What is the pH of solution
the blood?
 pOH = −log [OH-]
 pOH = −log 2.5 x 10–7
=
pH + pOH = 14
pH = 14 – pOH
= 14-3.54=7.4
Dr Laila Al-Harbi

11. Acidic solution basic solution
[H3O+]
pH
[OH-]
pOH 14
1.0  10−14
Dr Laila Al-Harbi