Doc Brown's Chemistry Advanced A Level Notes - Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5

Part 5.5 Definition of a strong base and examples of pH and pK_w strong base calculations

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What is a strong base? How to calculate the pH of a strong base solution using Kw given the acid concentration.

Sub-index for Part 5

5.1 Equilibria:  Lewis and Bronsted-Lowry acid-base theories

5.2 Self-ionisation of water and pH scale

5.3 Strong acids - examples and pH calculations

5.4 Weak acids - examples & pH, K_a and pK_a calculations

5.5 Strong bases - examples and pH calculations (this page)

5.6 Weak bases - examples and pH, K_b and pK_b calculations

5.0 Basic notes and equations on acids, bases, salts, uses of acid-base titrations - upgrade from GCSE!

5.5 Definition, examples and pH and pK_w calculations of strong bases

• 5.5.1: Definition and examples of STRONG BASES

  • Strong bases are highly ionized in water.

  • 100% ionisation is assumed in the calculations and in reality strong bases have a very high K_b (compare K_b for weak bases).

  • Examples:

    • Gp 1 hydroxides e.g. sodium hydroxide: NaOH_(s) + aq ==> Na⁺_(aq) + OH^–_(aq)

    • Gp 2 hydroxides e.g. barium hydroxide: Ba(OH)_2(aq) + aq ==> Ba²⁺_(aq) + 2OH^–_(aq)

  • The high % of ionisation gives the maximum concentration of hydroxide ions and the minimum equilibrium concentration of hydrogen ions, therefore the most alkaline solutions of highest possible pH.

• 5.5.2: Strong base calculations – calculating the pH of a strong base

  • Calculation example 5.5.2a

    • Calculate the expected hydroxide and hydrogen ion concentrations and pH of a 0.50 mol dm^–3 sodium hydroxide solution.

    • [OH^–_(aq)] = [NaOH_(aq)] = 0.50 mol dm^–3

    • In base calculations you need to use the ionic product of water expression to calculate the H⁺ ion concentration.

    • K_w = [H⁺_(aq)] [OH^–_(aq)] = 1 x 10^–14 mol² dm^–6 and rearranging gives

    • [H⁺_(aq)] = Kw_/[OH^–_(aq)] = 1 x 10^–14_/0.50 = 2.0 x 10^–14 mol dm^–3

    • pH = –log(2.0 x 10^–14) = 13.7

    • Note: pOH = pK_w – pH = 14 – 13.7 = 0.3

  • Calculation example 5.5.2b

    • Calculate the expected hydroxide and hydrogen ion concentrations and pH of a 0.02 mol dm^–3 calcium hydroxide solution.

    • [OH^–_(aq)] = 2 x [Ca(OH)_2(aq)] = 0.04 mol dm^–3

    • K_w = [H⁺_(aq)] [OH^–_(aq)] = 1 x 10^–14 mol² dm^–6 and rearranging gives

    • [H⁺_(aq)] = Kw_/[OH^–_(aq)] = 1 x 10^–14_/0.04 = 2.50 x 10^–13 mol dm^–3

    • pH = –log(2.50 x 10^–13) = 12.6

WHAT NEXT?

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. K_c and K_p equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product K_sp, common ion effect, ion–exchange systems * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces watch out for sub-indexes to multiple sections or pages

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