Ionic Equilibria (Acid/ Base)

 

 

There are two broad aspects of this chapter:

  • Acid/ Base Equilibria
  • Solubility Equilibria

Acid/ Base Equilibria

 

Important Relations

  • p = -log10

At 25°C:

  • pH + pOH = pKw = 14
  • pKa + pKb = 14
  • [H+][OH–] = 10-14
  • Ka x Kb = Kw = 10-14

Approach:

 

Determine what is present in solution: Acid/ Base/ Salt/ Buffer

Apply the corresponding equations:

 

Acid/ Base

 

 

Formula

Example

Strong Acid

[H+] = [HA]

pH = -log [HA]

pH of 0.100 moldm–3 HCl

 

pH = -log [0.100] = 1.0

Strong Base

[OH–] = [B]

pOH = -log [B]

pH of 0.100 moldm–3 NaOH

 

pOH = -log [0.100] = 1.0

pH = 14.0 - 1.0 = 13.0

Weak Acid

pH = -log [H+]

pH of 0.100 moldm–3 CH3COOH

(Ka = 1.8 x 10–5 moldm–3)

 

[H+] = [(1.8 x 10–5)(0.100)]1/2

        = 1.34 x 10–3 mol dm–3

pH = -log [1.34 x 10–3] = 2.9

Weak Base

pOH = -log [OH–]

pH of 0.100 moldm–3 NH3

(Kb = 1.8 x 10–5 moldm–3)

 

[OH–] = [(1.8 x 10–5)(0.100)]1/2

             = 1.34 x 10–3 mol dm–3

pOH = -log [1.34 x 10–3] = 2.9

pH = 14 - 2.9 = 11.1

 

Salt

 

Nature of Salt

Formed from

pH

Neutral

strong acid + strong base

= 7

Acidic

strong acid + weak base

Treat as weak acid;

use weak acid formula

Basic

weak acid + strong base

Treat as weak base;

use weak base formula

 

In calculating the pH of a salt solution, there are two common challenges:

 

1. Ka/ Kb of the salt (conjugate acid/ conjugate base) is not given.

Using Ka x Kb = Kw, find Ka/ Kb of the ion from Kb/ Ka of the parent base/ acid given.

 

2. finding concentration of salt.

[salt] = n (limiting reagent)/ Vtotal [Vtotal = Vacid + Vbase]

 

Buffer

 

Type

Formed from

pH

Acidic

weak acid + conjugate base

< 7

pH = pKa + log [salt]/[acid]

Basic

weak base + conjugate acid

> 7

pOH = pKb + log [salt]/[base]

 

How to calculate pH of a buffer on adding small amounts of H+ and OH–?

  1. Determine which species is reacted and which is formed.
  2. Calculate new amounts (in moles)
  3. Substitute into buffer equation

Note:

  • When using H-H equation, just calculate no. of moles of salt and acid/base since total volume is the same and cancels out.
  • Look out for [salt] = [acid] (max. buffering capacity) → pH = pKa (same for basic buffer)

Titration Curve

  • There are three important points on the titration curve:

Example: Titration of weak acid against strong base

Titration curve

What to observe

What can be found

1

Initial point

pH

[H+] due to dissociation of acid

2

Equivalence point

volume of base

pH

Given [acid], [base] can be found

Given [base], [acid] can be found

type of salt/ type of titration

3

Half equivalence point

pH

pKa of acid
  • Calculate pH at various points in titration curve

Strategy

  1. Determine what is in solution: Acid/ Base/ Salt/ Buffer
  2. Use the relevant equations.

Example: Titration of weak acid against strong base:

Titration curve

What is in solution

Initial point

Weak acid

Between initial point to equivalence point

Acidic buffer

Equivalence point

Basic Salt

Beyond equivalence point

Strong base