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Revision notes on chemical equilibrium - Explaining what a strong base is and pH calculations - for Advanced Level Theoretical-Physical Chemistry

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Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 5.5

5.5 Definition of a strong base and examples of pH & pKw strong base calculations

 

What is a strong base?

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

Chemical Equilibrium Notes Parts 5 & 6 Index



 

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

  • –


Chemical Equilibrium Notes Parts 5 & 6 Index


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