# 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)

## 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.