Acids, Bases and Kw

Acid = Releases Protons 

Base = accepts Protons 

Bronsted Lowry Acids = Proton Donors - release hydrogen ions when mixed with water 

Bronsted Lowry Bases = Proton Acceptors - Grab hydrogen ions from water molecules when they are in solution 

Strong Acid = Dissociate almost completely in water - nearly all H+ ions will be released. E.g. Hydrochloric acid 

Strong Base = ionise almost completely in water 

Weak Acid = Only partially dissociate in water - only small numbers of H+ ions are formed. Equilibrium lies to the left 

Weak base = Only partially ionise in water. Equilibrium lies to the left

Protons can only be released if there is a base to accept them.

HA = Acid, B = Base 

HA (aq) + B (aq) ----> BH+ (aq) + A- (aq)

It is an equilibrium reaction. If more HA or B is added, equilibrium is shifted to the right

If more BH+ or A- is added, equilibrium is shifted to the left 

When an acid is added to water, water is the base and accepts the proton. 

Water dissociates in hydroxonium ions and hydroxide ions.

H2O ---> H+ + OH-

Water only dissociates a tiny amount - equilibrium lies left 

Equilibrium Constant - Kc = [H+] [OH-]

                                                   [H2O] 

Ionic Product of Water - Kw = [H+] [OH-]

pH = - log10 [H+]

 Example:

A solution of hydrochloric acid has a hydrogen ion concentration of 0.01 mol dm -3. What is the pH of this solution?

pH = - log10 [H+] = - log10 (0.01) = 2.0 

[H+] = 10-pH 

Example:

A solution of sulfuric acid has a pH of 1.5. What is the hydrogen ion concentration of this solution? 

[H+] = 10-pH = 10-1.5 = 0.03 mol dm-3 = 3 x 10-2 mol dm-3

Monoprotic acids sich as Hydrochloric acid and Nitric acid - each molecule of acid will release 1 proton when it dissociates. 

H+ concentration is the same as the acid concentration.

Example:

For 0.10 mol dm-3 HCl, [H+] is also 0.10 mol. So the pH = -log10 (0.10) = 1.00.

Or for 0.050 mol dm-3 HNO3, [H+] is also 0.050 mol dm-3, giving pH = -log10 (0.050) = 1.30

Strong diprotic acids such as sulfuric acid produce 2 moles of hydrogen ions for each mole of acid. 

Example:

For 0.10 mol dm-3 H2S04, [H+] is 0.20 mol dm-3. So the pH = -log10 [H+] = -log10 (0.20) = 0.70

Strong bases such as sodium hydroxide (NaOH) and potassium hydroxide (KOH) donate one mole of OH- ions per mole of base. 

The concentration of OH- ions is the same as the concentration of the base. 

To work out the pH you eed to know the [H+] - linked to the OH- through Kw.

Example: 

Find the pH of 0.10 mol dm-3 NaOH at 298K, given that Kw at 298K is 1.0 x 10-14 mol2 dm-6.

1. Put all the values you know into the expression for Kw -1.0 x 10-14 = [H+] x 0.10

2. Rearrange the expression to find [H+] - [H+] = 1.0 x 10-14 / 0.10 = 1.0 x 10-13 mol dm-3

3. Use the value for [H+] to find the pH of the solution - pH = -log10[H+] = -log10(1.0 x 10-13) = 13.00

Weak aqueous acid - HA(aq) ---> H+(aq) + A- (aq)

Weak acid = Ka = [H+]2 / [HA]

Example:

Calculate the hydrogen ion concentration and the pH of a 0.0200 mol dm-3 solution of proanoic acid at 298K. Ka for propanoic acid at 298K is 1.30 x 10-5 mol dm-3.

1. Write down the expression and rearrange to find [H+]. 

Ka = [H+]2 / [CH3CH2COOH] = 1.30 x 10-5 x 0.0200 = 2.60 x 10-7

        [H+] = Square root (2.60 x 10-7) = 5.10 x 10-4 mol dm-3

Now use the value for [H+] to find pH:

pH = -log10 (5.10 x 10-4) = 3.292

Example: 

The pH of an ethanoic acid solution was 3.02 at 298K. Calulate the molar concentration of this solution. Ka of ethanoic acid is 1.75 x 10-5 mol dm-3 at 298K.

Use pH to find [H+] = 10-pH = 10 -3.02 = 9.55 x 10-4 mol dm-3

Rearrange the expression for Ka and plug in values to find [CH3COOH]: 

Ka = [H+]2 / [CH3COOH] ---> [CH3COOH] = [H+]2 / Ka = (9.55 x 10-4)2 / 1.75 x 10-5 - 0.0521 mol dm-3 

PKa = -log10 Ka and Ka = 10-pKa

Example:

If an acid has a Ka value of 1.50 x 10-7 mol dm-3, what is the PKa?

PKa = -log1- (1.50 x 10-7) = 6.824

What is the Ka value of an acid if its PKa is 4.32?

Ka = 10-4.32 = 4.8 x 10-5 mol dm-3

To work out concentrations or pH - convert it to Ka and use the Ka expression.

Example:

Calculate the pH of 0.0500 mol dm-3 methanoic acid. Methanoic acid has a pKa of 3.75 at this temperature. 

Ka = 10-pKa = 10-3.75 = 1.8 x 10-4 mol dm-3

Ka = [H+]2 / [HCOOH] --> [H+]2 = Ka x [HCOOH] = 1.78 x 10-4 x 0.0500 = 8.90 x 10-6

[H+] = SQUARE ROOT(8.90 x 10-6) = 2.98 x 10-3 mol m-3

pH = -log10 (2.98 x 10-3) = 2.526

Strong acid/strong base - pH starts around 1, excess of strong acid - finishes around pH 13, excess of strong base.

Strong acid/weak base - pH starts around 1, excess of strong acid - finishes around pH 9, excess of weak base 

Weak acid/strong base - pH starts around 5, excess of weak acid - finishes around pH 13, excess of strong base

Weak acid/weak base - pH starts around 5, excess of weak acid - finshes around pH 9, excess of weak base 

pH curves can help you decide which indicator to use 

Methyl Orange - Low pH = red - Approx pH of colour change = 3.1 - 4.4 - Colour at high pH = Yellow

Phenolphthalein - Low pH = Colourless - Approx pH of colour change = 8.3 - 10 - Colour at high pH = Pink 

For weak acid/weak base - there are no sharp pH changes so neither indicators will work - no indicators work, have to use pH meter. 

Buffers resist changes in pH when small amounts of acid or base are added or when it is diluted

Acidic buffers =  pH of less than 7 - made by mixing a weak acid with one of its salts e.g Ethanoic acid and Sodium ethanoate

Basic buffers =  pH of greater than 7 - made by mixing a weak base with one ofits salts e.g. a solutionof ammonia and ammonium chloride

Need to know the Ka of the weak acid and the concentrations of the weak acid and its salts 

Example:

A buffer solution contains 0.40 mol dm-3 methanoic acid HCOOH, and 0.60 mol dm-3 sodium methanoate,HCCO-Na+. For methanoic acid, Ka = 1.6 x 10-4 mol dm-3. What is the pH of this buffer?

1. Write the expression for Ka of the weak acid. 

Ka = [H+(aq)]x[HCOO-(aq)] / [HCOOH(aq)

2. Then rearrange the expression and stick the data in to calculate [H+(aq)]

[H+(aq)] = Ka x [HCOOH(aq)]/[HCOO-(aq)] 

So [H+(aq)] = 1.6 x 10-4 x (0.40 / 0.60) = 1.07 and 10-14 mol dm-3

3. Convert [H+(aq)] to pH

pH = -log10 [H+(aq)] =-log10 (1.07 x 10-4) = 3.971