O - THE OCEANS

Acids, Equilibria and Buffer Solutions

Acid-Base Equilibria

The definition of an acid and a base are inseparable. An acid is a hydrogen ion, H⁺, donor. A base is a hydrogen ion acceptor. Hydrochloric acid ionises in water to produce oxonium ions, H₃O⁺.

HCl(aq) + H₂O(l)

H₃O⁺(aq) + Cl^-(aq)

The presence of oxonium ions gives rise to the familiar acidic properties of all aqueous solutions of acids.

Hydrochloric acid is an acid because it loses an H⁺ ion to form a Cl^- ion. Water is a base because it accepts H⁺ ions to form H₃O⁺ ions.

The H₃O⁺ ion can itself act as an acid; it can donate H⁺ and turn into a water molecule. This is shown in the reverse of the equation above.

Strong and Weak Acids

Acids vary in strength. Different acids donate H⁺ ions to differing extents. Strong acids have a strong tendency to donate H⁺ ions. The donation of H⁺ ions is essentially complete. We say these acids are 100% ionised. Examples of strong acids are: hydrochloric acid, HCl; sulphuric acid, H₂SO₄; nitric acid, HNO₃ and phosphoric acid, H₃PO₄. The reaction with water can be regarded as going to completion. No unreacted HA remains in solution (where A represents the other species involved e.g. Cl in the case of HCl).

HA(aq) + H₂O(l)

H₃O⁺(aq) + A^-(aq)

This can be simplified to:

HA(aq)

H⁺(aq) + A^-(aq)

If a substance is a weak acid, its tendency to donate H⁺ ions is weaker and the reaction with water is incomplete; an equilibrium is set up. Some H⁺ ions are formed but there is still some unreacted acid in solution.

HA(aq)

H⁺(aq) + A^-(aq)

Examples of weak acids are organic acids, ethanoic acid, citric acid, oxalic acid and benzoic acid.

The pH Scale

The strength of an acid can be determined by measuring its pH. The pH is a measure of the concentration of H⁺ ions in solution.

pH is defined as: pH = -log₁₀ [H⁺(aq)]

pH runs in the opposite direction from concentration of H⁺ ions. A low pH corresponds to a high concentration of H⁺ ions.

pH can be measured by using universal indicator paper or more accurately by using a pH meter, which is a special glass electrode connected to a voltmeter.

Calculating pH Values

Strong acids

As strong acids are 100% ionised, the concentration of H⁺ ions is effectively equal to the concentration of acid (HA) put into solution.

Therefore for a 0.001 mol dm^-3 solution, [H⁺(aq)] = 0.001 mol dm^-3

pH	= -log₁₀ (0.001)
	= 3

For 0.02 mol dm^-3 H₂SO₄, [H⁺(aq)]	= 2 x 0.02
	= 0.04 mol dm^-3

pH	= -log₁₀ (0.04)
	= 1.4

Weak acids

For weak acids the concentration of H⁺ ions is not equal to the concentration of acid (HA) put in.

HA(aq)

H⁺(aq) + A^-(aq)

An equilibrium is set up, and an equilibrium constant can be written:

K_a is called the acidity constant or ionisation constant. The value of K_a gives an indication of the strength of an acid. The smaller the value of K_a, the weaker the acid. Often, pK_a is used instead of K_a for ease. The larger the value of pK_a, the weaker the acid.

pK_a = -log₁₀ K_a

Typical values of K_a and pK_a are shown in this table:

Acid	Formula	K_a	pK_a
methanoic acid	HCO₂H	1.6 x 10^-4	3.8
ethanoic acid	CH₃CO₂H	1.7 x 10^-5	4.8
benzoic acid	C₆H₅CO₂H	6.3 x 10^-5	4.2

Calculating pH

The pH of a weak acid can be calculated if the concentration and K_a is known for the acid. To determine pH we need to know [H⁺(aq)].

Consider the general reaction:

HA(aq)

H⁺(aq) + A^-(aq)

At equilibrium:

[H⁺(aq)] = [A^-(aq)]

Also the concentration of HA at equilibrium is approximatly equal to the concentration of HA put into the solution. We can ignore the fraction of HA molecules which have lost H⁺, because we are dealing with a weak acid and this fraction is very small.

Hence for a concentration of 0.1 mol dm^-3, [HA(aq)] = 0.1 mol dm^-3

Example 1. Calculate the pH of a 0.001M solution of glycine, where K_a = 1.7 x 10^-10 mol dm^-3.

K_a	=
[H⁺(aq)]²	= K_a x molarity
[H⁺(aq)]²	= 1.7 x 10^-10 x 0.001
	= 1.7 x 10^-13
[H⁺(aq)]	= 4.12 x 10^-7
pH	= -log₁₀ (4.12 x 10^-7)
	= 6.38

Example 2. The same expression can be used to calculate K_a for an acid knowing the concentration and the pH of the acid.

Calculate K_a for 0.01M ethanoic acid whose pH is 3.4.

pH	= 3.4 = -log₁₀ [H⁺(aq)]
[H⁺(aq)]	= 10^(-3.4)
	= 3.98 x 10^-4
K_a	=
	=
	= 1.58 x 10^-5 mol dm^-3

Ionisation of Water

Water is ionised to a very small extent:

H₂O(l)

H⁺(aq) + OH^-(aq)

An equilibrium constant can be written:

Since the degree of ionisation is very small, the concentration of unionised water is constant and the equation can be re-written:

K_w = [H⁺(aq)] [OH-(aq)]

K_w is known as the ionic product for water.

At 298K, K_w = 10^-14 mol² dm^-6. For pure water [H⁺(aq)] = [OH^-(aq)]

K_w	= 10^-14 = [H⁺(aq)]²
Therefore [H⁺(aq)]	= 10^-7 mol dm^-3
pH	= 7

K_w allows us to calculate the pH of a solution of a strong base. This is one in which the production of hydroxide ions is complete. An example is sodium hydroxide which is completely ionised in solution.

What is the pH of 0.001 M sodium hydroxide solution?

[OH^-(aq)]	= 0.001 mol dm^-3
K_w	= 10^-14 = [H⁺(aq)] x 0.001
[H⁺(aq)]	=
	= 10^-11
pH	= 11

An alternative way of doing this calculation is to use the expression:

pOH	= -log₁₀ [OH-(aq)]
	= -log₁₀ (0.001)
	= 3
pH + pOH	= 14
Hence pH	= 14 - 3
	= 11

Buffer Solutions

The progress of many chemical reactions is affected critically by the pH of the solution in which they take place, and it is often necessary to control the pH in a specific manner. The pH of an aqueous solution is usually sensitive to addition of small amounts of acids and alkalis. A buffer solution is able to counteract the effect of relatively small quantities of acid or alkali on the pH of the solution. Such solutions have important applications in biochemistry and chemistry, e.g. the human blood is a natural buffer system which maintains a pH of 7.4.

An acidic buffer solution (pH < 7) is prepared by mixing together definite amounts of a weak acid and the sodium or potassium salt of the same acid, e.g. ethanoic acid and sodium ethanoate.

An alkaline buffer solution (pH > 7) is prepared by mixing a weak base and a soluble salt of the base, e.g. ammonia solution and ammonium chloride.

The mixture of ethanoic acid and sodium ethanoate acts as a buffer as follows:

Ethanoic acid takes part in its usual equilibrium in solution.
CH₃CO₂H(aq)

CH₃CO₂-(aq) + H⁺(aq)

We assume that approximately all the ethanoic acid molecules put into the buffer remain unchanged, and that the sodium ethanoate, being an ionic salt, is 100% ionised in solution.

CH₃COONa(s) + aq

CH₃CO₂-(aq) + Na+(aq)

We assume that all the ethanoate ions come from the salt. The weak acid will supply very few CH₃CO₂- ions when compared with the salt.

The resulting mixture contains a large concentration of ethanoate ions from the salt and a large concentration of ethanoic acid.

If (acidic) H⁺ ions are added to the buffer, the equilibrium is disturbed and moves to the left. Some ethanoate ions from the salt react with the extra H⁺ ions to form ethanoic acid. This means that overall the pH hardly changes.

If (alkaline) OH^- ions are added to the buffer, they react with the H⁺ ions in the equilibrium to form water.

H⁺(aq) + OH^-(aq)

H₂O(l)

The equilibrium is disturbed and moves to the right to restore the H⁺ ions so pH hardly changes.

The presence of both a weak acid and its salt are necessary for a buffer to work. A buffer solution will contain a large concentration of HA from the acid and a large concentration of A^- from the salt.

An alkaline buffer acts in a similar way. If ammonia is mixed with ammonium chloride the following equilibrium is established:

NH₄⁺(aq)

NH₃(aq) + H⁺(aq)

There is a high concentration of NH₄(aq) from the ammonium chloride and a high concentration of NH₃(aq) from the ammonia.

If acid is added to the buffer, the equilibrium shifts to the left and the pH hardly changes. If alkali is added to the buffer, it reacts with H⁺ ions to produce water. The equilibrium shifts to the right to restore the concentration of H⁺ ions. The pH hardly changes.

Calculations with Buffer Solutions

The equation for a weak acid is as follows:

HA(aq)

H⁺(aq) + A^-(aq)

But [HA(aq)] = [acid] and [A^-(aq)] = [salt]

Hence:

The value of pH depends on two factors:

K_a. This provides a 'coarse tuning' of the buffer's pH. K_a values normally lie in the range 10^-4 mol dm^-3 to 10^-10 mol dm^-3, giving pH in region 4 to 10.
Ratio of [salt] : [acid]. This provides a 'fine tuning' of the buffer's pH. Changing the ratio from about 3:1 to about 1:3 changes [H⁺(aq)] by a factor of 9 hence pH is altered by about 1 unit. The pH of a buffer solution is not affected by dilution since addition of water will reduce the concentrations of acid and salt equally.

Example. Calculate the pH of a solution which is 0.5 mol dm^-3 with respect to ethanoic acid and 0.2 mol dm^-3 with respect to sodium ethanoate. (K_a for ethanoic acid is 1.7 x 10-5 mol dm^-3).

K_a	= [H⁺(aq)] x
1.7 x 10^-5	= [H⁺(aq)] x
[H⁺(aq)]	= 1.7 x 10^-5 x
	= 4.25 x 10^-5
pH	= 4.37

The oceans are effective at controlling the pH as they act as a giant buffer solution. In simple terms we can regard dissolved carbon dioxide as equivalent to an acid, carbonic acid.

CO₂(aq) + H₂O(l)

H₂CO₃(aq)

This ionises in solution:

H₂CO₃(aq)

H⁺(aq) + HCO₃^-(aq)

H₂CO₃ is a weak acid hence there is a high concentration available. HCO₃^- are available in almost limitless supply. Therefore this reaction acts as a buffer.