Chapter 17 – Equlibria Involving Acids and Bases

Acidity of Solutions

• Bronsted-Lowry theory defines an acid as a proton donor and a a base as a proton acceptor.

• Acid-base reaction involves the transfer of one or more protons from an acid to a base.

• Some substances have the ability to both donate and accept protons…these are called amphiprotic.

• Water is an example of this.

Ionisation Constant of Water• Water will react with itself in a process called self ionisation:

H2O(l) + H2O(l) H3O+(aq) + OH-(aq) • At equilibrium, K = [H3O+][OH-] / [H2O]2

• In aqueous solutions, water is usually more abundant than any other substance present.

• The concentration of water in aqueous solutions is virtually constant at 56M.

• Therefore, the equilibrium law for the ionisation of water is:[H3O+][OH-] = K x [H2O]2 = a constant

• This can be written as:Kw = [H3O+][OH-]

• Kw is called the ionisation constant of water.• In pure water, at 25°C it has been found that the concentration of both

H3O+ ions and OH- ions is 10-7, therefore making Kw = 1.0 x 10-14M2

Acidic and Basic Solutions

• In solutions of acidic substances, H3O+ ions are formed bout from the reaction between the acid and the water and the self ionisation of water.

• Therefore the concentration of the H3O+ ions will be greater than 10-7M at 25°C.

• Since the product remains constant, the concentration of OH- ions in an acidic solution at 25°C must be less than 10-7M.

• It is the opposite in the case of basic solutions.

pH

• pH helps to measure the acidity.• It is defined as pH = -log10[H3O+]• Or it can be rearranged to give [H3O+] = 10-pH

• For pure water at 25°C, pH = 7• For acidic solutions, pH<7• For basic solutions, pH>7• Using the pH scale, more acidic solutions have pH

values slightly less than 0 and more basic solutions have values of about 14.

pH example40.0mL of 0.200M hydrochloric acid reacts with 20.0mL of 0.100M sodium hydroxide

solution. Calculate the pH of the resulting solution.HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Reactions between strong acids and strong bases can be regarded as virtually complete so:

n(HCl) = c x v = 0.200 x 0.0400 = 0.00800moln(NaOH) = c x v = 0.100 x 0.0200 = 0.00200mol

From the equation n(HCl) reacted with NaOH = n(NaOH) = 0.00200molSo: n(HCl) reacted = 0.00800-0.00200 = 0.00600mol

Since 1 mol of HCl gives 1 mole of H3O+, n(H3O+) = 0.00600mol. The total volume of the mixtue is 40.0+20.0 = 60.0mL

C(H3O+) = n/v = 0.006/0.0600 = 0.100MpH = -log100.100 = 1.00

Temperature and pH• The self ionisation of water is an endothermic reaction.• Equilibrium constants are temperature dependent.• If a reaction is endothermic, the equilibrium constant increases as

the temperature increases, and decreases as temp decreases.• As the temperature of the solution rises above 25°C, the position

of the equilibrium will favour the endothermic forward reaction.• This means that the concentration of the ions will increase,

causing Kw to rise.• As a consequence, the pH will decrease.• The opposite is true if the reaction occurs at a temperature below

25°C.• The pH of water only equals 7 at 25°C, however at other

temperatures the water is still described as neutral because the concentrations of the two ions are equal.

Acidity Constants

• Ka is known as the acidity constant.• The value of Ka for HCl is 107M at 25°C. This

means that in hydrochloric acid solutions, most of the HCl has been converted to ions.

• This is why HCl is considered as strong acid.• In contrast, ethanoic acid has an acidity

constant of 1.75 x 10-5M at 25°C. In a solution of ethanoic acid and water, the position of the equilibrium favours the reactants and there is a relatively small amount of products.

Acidity Constants

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

K = [H3O+][CH3COO-] / [CH3COOH][H2O]

Ka = [H3O+][CH3COO-] / [CH3COOH]

Ethanoic acid is classified as a weak acid because at equilibrium only a small proportion of the acid has been ionised.

The acidity constant can be used as a measure of an acid’s strength.

Calculate the pH and percentage hydrolysis of a 0.50M ethanoic acid solution, given that the K a for ethanoic acid is 1.75 x 10-5M.

The equation for the ionisation of ethanoic acid is:CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Ka = [H3O+][CH3COO-] / [CH3COOH] = 1.75x10-5M

From the equation for the reaction it can be seen that for every mole of ethanoic acid that ionises, one mole of H3O+ and one mole of CH 3COO- will be formed. So,

[H3O+] = [CH3COO-]

If we substitute this is the expression for Ka:[H3O+]2 / [CH3COOH] = 1.75x10-5M

We also know that [CH3COOH] + [CH3COO-] = 0.50M

However, ethanoic acid is a weak acid with a very small Ka, so we know it will only ionise to a very small extent. As the small value of Ka indicates that little CH3COO- is formed by ionisation, we can approximate the above expression to:

[CH3COOH] ≈ 0.50M

If we substitute this into the revised expression for Ka, we get:

[H3O+]2 / 0.5 = 1.75x10-5M

[H3O+]2 = 8.75x 10-6

[H3O+] = 2.96 x 10-3

pH = -log[2.96x10-3] = 2.5

We can measure the extent of the reaction by calculating the percentage hydrolysis, or percentage ionisation, the fraction of the acid that ionised. Percentage hydrolysis is given by the expression:

[CH3COO-] x 100 / [CH3COOH] = 2.96x10-3 x 100 / 0.5 = 0.59%

Acidity Constants cont…

• These ideas can be generalised to solutions of any weak acid represented by HA:

HA(aq) + H2O(l) A-(aq) + H3O+(aq)

• [H3O+] = [A-] and [HA] does not change during the ionisation (hydrolysis).

Buffers

• Buffer are solutions that can absorb the addition of acids or bases with little change of pH.

• They can be made by mixing a weak acid and salt of its conjugate base.

• Look at the example from a mixture of ethanoic acid and sodium ethanoate.

• The resulting solution will contain a mixture of CH3COOH, H3O+ and CH3COO-.

• The important feature of this solution is that it contains significant amounts of both weak acid and its conjugate base.

Buffers cont…• If a solution of strong acid such as HCl is added to the equilibrium

mixture, the pH will decrease, but much less than expected.• The addition of HCl solution disturbs the equlibrium.• Le Chatelier’s principle tells us that the system will respond to

oppose the change and restore equilbrium.• The addition of H3O+ causes a net back reaction creating more

CH3COOH.

• The addition of a strong bases consumes H3O+, causing a net forward reaction to produce more H3O+.

• In both cases the overall effect on [H3O+] is small and so the change

in pH is minimal.

pH in the Body

• A number of reactions within the body are acid-base reactions.• Without a means of controlling acidity, the pH could fluctuate

from highly basic to highly acidic.• Normal body functions could not be maintained with these

extreme changes, for example blood is maintained within a pH of 7.35-7.45.

• The presence of buffers maintains pH values within narrow limits in the body.

• Control of the pH of blood is achieved using different buffers.• One of the most important buffers is made up of carbonic acid

(H2CO3) and the hydrogen carbonate ion (HCO3-)

pH in the Body cont…

• The action of this buffer system become clear if we consider what happens when an acid or base is added to blood:– If H3O+ ions are added, a net back reaction occurs,

removing most of these ions.– If OH- ions are added, they react with H3O+ ions. A

net forward reaction results producing more H3O+ ions. The OH- ions have effectively reacted with the carbonic acid.