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HSC Chemistry

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  1. Module 1: Properties and Structure of Matter
    1.1 Properties of Matter
  2. 1.2 Atomic Structure and Atomic Mass
  3. 1.3 Periodicity
  4. 1.4 Bonding
  5. Module 2: Introduction to Quantitative Chemistry
    2.1 Chemical Reactions and Stoichiometry
  6. 2.2 Mole Concept
  7. 2.3 Concentration and Molarity
  8. 2.4 Gas Laws
  9. Module 3: Reactive Chemistry
    3.1 Chemical Reactions
  10. 3.2 Predicting Reactions of Metals
  11. 3.3 Rates of Reactions
  12. Module 4: Drivers of Reactions
    4.1 Energy Changes in Chemical Reactions
  13. 4.2 Enthalpy and Hess's Law
  14. 4.3 Entropy and Gibbs Free Energy
  15. Module 5: Equilibrium and Acid Reactions
    5.1 Static and Dynamic Equilibrium
    5 Topics
  16. 5.2 Factors that Affect Equilibrium
    2 Topics
  17. 5.3 Calculating the Equilibrium Constant
    2 Topics
  18. 5.4 Solution Equilibria
  19. Module 6: Acid/Base Reactions
    6.1 Properties of Acids and Bases
    7 Topics
  20. 6.2 Using Brønsted–Lowry Theory
    2 Topics
  21. 6.3 Quantitative Analysis
    1 Topic
  22. Module 7: Organic Chemistry
    7.1 Nomenclature
    2 Topics
  23. 7.2 Hydrocarbons
    2 Topics
  24. 7.3 Products of Reactions Involving Hydrocarbons
  25. 7.4 Alcohols
    1 Topic
  26. 7.5 Reactions of Organic Acids and Bases
  27. 7.6 Polymers
    2 Topics
  28. Module 8: Applying Chemical Ideas
    8.1 Analysis of Inorganic Substances
    3 Topics
  29. 8.2 Analysis of Organic Substances
  30. 8.3 Chemical Synthesis and Design
  31. Working Scientifically
    Working Scientifically Overview
    1 Topic
Lesson 20, Topic 2
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Introduction to pOH

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The pOH of a solution is defined as:

\text{pOH} = -\text{log}_{10}[\text{OH}^-]

Acidic solutions have a pOH > 7. Alkaline solutions have a pOH < 7.

[H+], [OH] and Kw (water ionisation constant) are related:

K_{\text w} = 1.0 \times 10^{-14} = [\text H^+][\text{OH}^-]

pH and pOH are also related, as pH + pOH = 14.

Example

Q: Calculate the pH of a 0.00850 mol L-1 NaOH solution.


NaOH is a strong base and fully dissociates in water:

[\text{OH}^-]=[\text{NaOH}] = 0.00850 \space\text{mol} \space\text L^{-1}
\text{pOH} = -\text{log}_{10}[\text{OH}^-]=-\text{log}_{10}[0.00850]=2.07
\text{pH} = 14-\text{pOH} = 14-2.07=11.93