Acids and Bases (1 of 2)

Part 2 of Acids and Bases is located here.

Definition of pH

Typically the hydrogen ion concentration of a solution is expressed in terms of pH.

pH is calculated as the negative log of a solutions hydrogen ion concentration

pH = -log10[H+]

If you plug the hydrogen ion concentration of water into this equation ([H⁺] of water = 1×10^-7 M) you get 7. This is known as neutral pH.

• High concentration of [H⁺] ions means a low pH (acidic)
• Low concentration of [H⁺] ions means a high pH (basic)

The pH Scale

Bronsted-Lowry

• Acid: Proton (H⁺) donor
• Base Proton (H⁺) acceptor

HCl + H2O → Cl- + H3O+

• In this reaction an extra electron from the H-Cl bond moves to the Cl, making it a Cl^–, and leaving the hydrogen as a H⁺
• The H₂O then acts as a Bronsted-Lowry base by accepting a proton (the H⁺)
• So in terms of Bronsted-Lowry, the acid is HCL, and the base is the Water
  • This means the conjugate base is the Cl^– and the conjugate acid is H₃O⁺

pKa and pKb Relationship (Weak Acids and Bases)

First, note that square brackets indicate the concentration of. For example: [H+] denotes the concentration of the hydrogen ion.

Now, in relation to weak acids and bases:

Ka = [products] / [reactants]

and:

Kb = [products] / [reactants]

Note also, this is only true for reactions in equilibrium. This means it can only be done for weak acids, or bases.

For example:

HA ^(aq) ↔ H^{+ (aq)} + A^{– (aq)}

∴ Ka = [H⁺] [A^–] / [HA]

And:

Kb = [HA] [OH^–] / [A^–]

-log Ka + -log Kb = 14

• -log Ka = pKa
• -log Kb = pKb

(aq) meaning it is an aqueous solution, which means dissociating in water

Weak Acid Equilibrium

A = Acid, for demonstration purposes

HA + H₂O ↔ H₃O⁺ + A^–

Ka = [H₃O⁺] [A^–] / [HA] [H₂O]

Example; with weak acids:

Note again that to calculate the pKa = -log Ka , and the lower the pKa value, the stronger the acid:

		Ka	pKa
HF	Hydrofluoric Acid	3.5 x10-4	3.46
CH3COOH	Acetic Acid	1.8 x10-5	4.74
CH3OH	Methanol	2.9 x10-16	15.54

Weak Base Equilibrium

B = Base for demonstration purposes

B + H₂O ↔ BH⁺ + OH^–

Kb = [BH⁺] [OH^–] / [B]

pKb = -log Kb

		Kb	pKb
NH3	Ammonia	1.8 x10-5	4.74
C6H5NH2	Aniline	4.3 x10-10	9.37

The lower the pKb value, the stronger the base.

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