Back to Course
HSC Chemistry
0% Complete
0/47 Steps
-
Module 5: Equilibrium and Acid Reactions5.1 Static and Dynamic Equilibrium5 Topics
-
5.2 Factors that Affect Equilibrium2 Topics
-
5.3 Calculating the Equilibrium Constant2 Topics
-
5.4 Solution Equilibria
-
Module 6: Acid/Base Reactions6.1 Properties of Acids and Bases7 Topics
-
6.2 Using Brønsted–Lowry Theory2 Topics
-
6.3 Quantitative Analysis1 Topic
-
Module 7: Organic Chemistry7.1 Nomenclature2 Topics
-
7.2 Hydrocarbons2 Topics
-
7.3 Products of Reactions Involving Hydrocarbons
-
7.4 Alcohols1 Topic
-
7.5 Reactions of Organic Acids and Bases
-
7.6 Polymers2 Topics
-
Module 8: Applying Chemical Ideas8.1 Analysis of Inorganic Substances3 Topics
-
8.2 Analysis of Organic Substances
-
8.3 Chemical Synthesis and Design
-
Working ScientificallyWorking Scientifically Overview1 Topic
Participants2
-
EduKits Education
-
Michael
Lesson 6, Topic 1
In Progress
Introduction to pH
Lesson Progress
0% Complete
The pH of a solution is defined as:
\text{pH} = -\text{log}_{10}[\text H^+] = \text{log}_{10}[\text H_3 \text O^+]
Acidic solutions have a pH < 7. Alkaline solutions have a pH > 7.
Example
Q: Calculate the pH of a 0.00250 mol L-1 nitric acid solution.
Nitric acid is a strong monoprotic acid and completely dissociates in water:
\text{HNO}_3 (aq) + \text H_2 \text O (l) \rightarrow \text H_3\text O^+ (aq) + \text{NO}_3^- (aq)
[\text H_3 \text O^+] = [\text {HNO}_3] = 0.00250 \space \text{mol} \space\text L^{-1}
\text{pH} = \text{log}_{10}[0.00250]=2.60