Starter question.

Consider propanoic acid (CH3CH2COOH) reacting with water…..

1. Write a symbol equation for the equilibrium that is established.

2. Identify the conjugate base and acid on the right hand side of the equation.

3. Write a Kc expression for the equilibrium.

Starter answers

CH3CH2COOH (aq) + H20(l) ↔ H3O+(aq) + CH3CH2COO-(aq)

Conjugate base….CH3CH2COO-

Conjugate acid H3O+

Ka = [H+(aq)][CH3CH2COO-(aq)]

[CH3CH2COOH (aq)]

The pH scale

The pH scale

Devised by a Danish chemist called Soren Sorensen.

It indicates how much acid or alkali is present in a solution.

The ‘p’ in pH stands for ‘potens’, which is latin for power.

pH = -log[H+ (aq)]

pHScale

Calculating the pH of a strong acid.

Calculating the pH of a strong acid

The pH of strong acids are relatively simple to calculate.

If we assume that the acid molecules fully dissociate then [H+] = [HA]

Therefore pH = -log [HA]

E.g. 1 mol HCl, pH = –log[1] = 0

E.g. 0.1 mol HCl, pH = -log[0.1] = 1

Calculating the pH of a weak acid.

Calculating the pH of a weak acid. An example.

What is the pH of 0.1moldm-3 ethanoic acid, which has a Ka of….1.74X10-

5moldm-3?)

You need to… 1. Write out the equilibrium expression

(Ka) 2. convert hydrogen concentration [H+]

to pH

Step 1 the Ka expression.

CH3COOH (aq) ↔ H+ (aq) + CH3COO-(aq)

Equilibrium equation.

Ka = [CH3COO-(aq)][H+(aq)]

[CH3COOH(aq)]

So the general Ka expression will be……

But we need to consider when full equilibrium has been reached and take account of our initial concentration of acid.

Step 2 the equilibrium position.

CH3COOH H+ CH3COO-

Start Conc. 0.100 0 0

At equilibrium. 0.100- [CH3COO-] [H+] [CH3COO-]

CH3COOH ↔ H+ + CH3COO-

CH3COOH

H+ CH3COO-

For every ethanoic acid molecule that dissociates an ethanoate ion

and a H+ also forms.

Writing the new Ka expression at equilibrium.

Ka = [CH3COO-(aq)]eq[H+ (aq)]eq

0.100 – [CH3COO-(aq)]eq

As [CH3COO-] = [H+] at equilibrium we can write…

Ka = [H+]2

0.100 – [H+]

Initial concentration minus the dissociation concentration of the

ions.

Because the disassociation of the weak acid is so small we assume 0.100 – [H+] = 0.100

Plugging the numbers in..

Substituting in the values gives….

1.74 X 10-5 = [H+]2

0.100 [H+]2= 0.100 X 1.74 X 10-5

[H+] = √(1.74 X 10-6) [H+]= 1.319 X 10-3moldm-3

= pH = 2.88 (using –log)

Comparing strong and weak acids

Factors Strong Acid Weak Acid

Acid diluted by a factor of 100

[H+ (aq)] becomes 100 x smaller

[H+ (aq)] becomes 10 x smaller

pH increases by 2 pH increases by 1

Conductivity Higher as more free ions Lower as less free ions.

Reaction Rate Reacts far quicker due to greater dissociation

Reacts slower as relies on H+ being removed from equilibrium to completely dissociate acid molecules.

Strength vs. Concentration

Concentration is a measure of the amount of substance in a given volume of solution.

Strength is a measure of the extent to which an acid can donate H+. Measured as pKa values:

pKa = -log Ka

Calculating the pH of a strong base.

Calculating pH of a strong base.

To begin with we need to understand another term, Kw, this is the ionic product of water.

For water, Ka = [H+(aq)] [OH-(aq)] / [H2O(l)] As water is always present in excess we ignore

the term, [H2O(l)] to give:

Kw = [H+(aq)] [OH-(aq)]

At 298K Kw = 1x10-14 mol2 dm-6

Calculating pH of a strong base.

Kw = [H+(aq)] [OH-(aq)]

At 298K Kw = 1x10-14 mol2 dm-6

Kw = 1x10-14 = [H+]2

Therefore [H+] = 1x10-7

=pH7 at 298K, pH will fall as temperature increases.

Calculating pH of a strong base.

What is the pH of a 0.1 mol solution of sodium hydroxide?

Kw = 1 x 10-14 = [H+] x [OH-]

Kw = 1 x 10-14 = [H+] x [0.1] [H+] = 1 x 10-14 / 0.1 [H+] = 1 x 10-13

pH = -log [1 x 10-13] = 13