This document discusses acids, bases, and pH. It defines acids as substances that produce hydrogen ions in water and bases as substances that produce hydroxide ions. pH is defined as the negative base-10 logarithm of the hydronium ion concentration and is a measure of acidity, with values below 7 indicating acidic solutions and above 7 indicating basic solutions. The autoionization of water is described by the ion product constant Kw, which relates the concentrations of hydronium and hydroxide ions in solution.

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