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Revision notes on chemical equilibrium - Explaining what a strong base is and pH calculations

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 pKw 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

5.5 Definition, examples and pH and pKw 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 Kb (compare Kb 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.

• Kw = [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 = pKw – 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

• Kw = [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. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product Ksp, 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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