STPM Chemistry Form 6 Notes – Ionic Equilibrium (Part 2)

by BerryBerryTeacher

in Berry Reference (Notes)

Now that you have been introduced to Ionic Equilibrium from the previous part of Berry Berry Easy’s STPM Form 6 (Tingkatan 6) concise study notes on Ionic Equilibrium, it is time to move on to the next part, Part 2. In Part 1, you learned all about the Brønsted-Lowry Acid-Base Theory, strong and weak acids and bases (limited to common strong acids and common strong bases). This is a direct continuation and will focus on common weak acids, common weak bases and Ostwald Dilution Law.

None of them are particularly hard but you would need to be able to distinguish (or memorise) which of the acids and bases are strong/weak before you can even try to understand the concepts from Part 1 and 2.

(Tips: Read back Part 1 and identify the strength of the acids and bases. Also recall the definition of ionic equilibrium and use keywords like proton donor and proton acceptor, or similar phrases. Don’t be put off by the ‘difficult’ and long chemical formulas.)

STPM Chemistry Form 6 Notes – Ionic Equilibrium (Part 2)

 

Natural Universal Indicator_Purple cabbage

Natural Universal Indicator_Purple cabbage

III) Common weak acids

  • ethanoic acid / acetic acid (CH3COOH)
  • fluoroacetic acid (C2H3FO2)
  • formic acid (HCHO2)
  • phosphori(V) acid (H3PO4)
  • boric acid (H3BO3)

Example: Ethanoic acid partially ionises in water.

CH3COOH(aq) + H2O(l) <—-> CH3COO- + H3O+

  • About 5% of the ethanoic acid ionises and 95% remains in the molecular form
  • Ethanoic acid is a weak acid
  • Water acts as a base by accepting the proton from the ethanoic acid

Calculating ion concentration in solution

Example: Ka expression for the weak acid is

Ka = ([H3O+] [CH3COO-]) / [CH3COOH] = 1.8 x 10-5

[CH3COOH] = molar concentration of CH3COOH at equilibrium not initially.
[H2O] does not appear in Ka expression because it is a constant.
[H3O+] = [CH3COO-] = α

* CH3COOH ionises partially to form hydronium ion and acetate ion, so the amount of ethanoic acid remaining at equilibrium as the initial amount, 2.0 M, minus the amount that ionises, α.

* α is very small, so that 2.0 – α is approximately equal to 2.0 = equilibrium concentration of the weak acid with its initial concentration.

  [CH3COOH] [CH3COO-] [H3O+]
Initial (M) 2.0 0 0
Equilibrium (M) 2.0 – α 2.0 (α) 2.0 (α)

Ka = 1.8 x 10-5 = ([α] [α]) / [2.0] = [α]2 / [2.0]

[α]2 = (1.8 x 10-5) [2.0]
[H3O+] = [CH3COO-] = α = 6.0 x 10-3

 

 

IV) Common weak bases

  • ethylamine (CH3CH2NH2)
  • Ammonium hydroxide / ammonia water (NH4OH)
  • Ammonia (NH3)

Example: Ammonia partially ionises in water.

NH3(g) + H2O(l) <—-> NH4+ + OH-

  • About 5% of the ammonia ionises and 95% remains in the molecular form
  • Ammonia is a weak base
  • Water acts as a acid by donating the proton from the ammonia

Calculating ion concentration in solution

Example: Kb expression for the weak base is

Kb = ([OH-] [NH4+]) / [NH3] = 1.8 x 10-5

[NH3] = molar concentration of NH3 at equilibrium not initially.
[H2O] does not appear in Kb expression because it is a constant.
[NH4+] = [OH-] = β

* NH3 ionises partially to form hydroxide ion and ammonium ion, so the amount of ammonia remaining at equilibrium as the initial amount, 2.0 M, minus the amount that ionises, β.

* β is very small, so that 2.0 – β is approximately equal to 2.0 = equilibrium concentration of the weak base with its initial concentration.

  [NH3] [NH4+] [OH-]
Initial (M) 2.0 0 0
Equilibrium (M) 2.0 – β 2.0 (β) 2.0 (β)

Kb = 1.8 x 10-5 = ([β] [β]) / [2.0] = [β]2 / [2.0]

[β]2 = (1.8 x 10-5) [2.0]
[OH-] = [NH4+] = β = 6.0 x 10-3

 

 

Ostwald Dilution Law

Ka = (cα)(cα) / c(1-α) = α2c / (1-α)

[H3O+] = √(Ka x c)

Kb = (cβ)(cβ) / c(1-β) = β2c / (1-β)

[OH-] = √(Kb x c)

This is the end of Part 2. Be prepared to face the onslaught of the three theories: Arrhenius Theory, Brønsted-Lowry Acid-Base Theory, Lewis Theory in the next part, Part 3 of Berry Berry Easy STPM Form 6 notes on Ionic Equilibrium.

{ 2 comments… read them below or add one }

July 9, 2011
July 9, 2011

Previous post:

Next post: