Chemistry: Acid-Base Equilibria: Strong & Weak Acids

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The Bronsted-Lowry Theory

Bronsted-Lowry Acids and Bases

In 1923 physical chemists Johannes Nicolaus Bronsted in Denmark and Thomas Martin Lowry in England independently proposed the theory that carries their names. In the Bronsted-Lowry theory acids and bases are defined by the way they react with each other.

The defined an acid as a substance that can donate a proton, i.e. a proton donor (hydrogen ion, H+), and a base as a substance that can accept a proton, i.e. a proton acceptor.

We might expect that any substance containing hydrogen act as an acid. In practice, a substance behaves as an acid only if the hydrogen carries a slight positive charge. As an example, this is the case when it is bonded to a highly electronegative atom to the right of the Periodic Table, e.g. oxygen or halogen.

In order to accept a proton, a base has to contain a lone pair of electrons that it can use to form a dative covalent bond with the proton. So, a base must contain an atom to the right-hand side of the Periodic Table, and this is often oxygen.

Conjugate Acid-Base Pairs

When hydrogen chloride dissolves in water an equilibrium is established that can be represented by the following equation:

HCl(aq) + H2O(l) -><- H3O+(aq) + Cl-(aq)

In the forward reaction:

1. HCl is acting as an acid because it is donating a proton to H2O

2. H2O is behaving as a base as it is accepting a proton from HCl.

In the reverse reaction:

1. H3O+ is behaving as an acid because it is donating a proton by H2O

2. Cl- is behaving as a base because it is accepting a proton from H3O+

When the acid HCl loses a proton it forms a base, Cl-. These two species are called a conjugate acid-base pair.

When the acid H3O+ loses a proton it forms the base H2O. These two species also form a conjugate acid-base pair.

So the equilibrium mixture above contains two conjugate acid-base pairs.

Cl- is the conjugate base of HCl; H2O is the conjugate base of H3O+.

Hydrochloric acid can donate one proton - it is called a monoprotic or monobasic acid.

Sulphuric acid can donate two protons and is therefore called a diprotic or dibasic acid.

Step 1: H2SO4 -> H+ + (HSO4)-

Step 2: HSO4- -> H+ + (SO4)2-

Similarly, some bases such as the carbonate ion can accept more than one proton.

Step 1: (CO3)2- + H+ -> (HCO3)-

Step 2: (HCO3)- + H+ -> H2CO3

The carbonate ion is therefore described as a diprotic or diacidic base.

Another Example of a Conjugate Acid-Base Pair

When ammonia dissolves in water an equilibrium is established that can be represented by the following equation:

NH3(aq) + H2O(l) -><- NH4+(aq) + OH-(aq)

In the reverse equation:

1. NH3 is acting as a base because it is accepting a proton from N2O.

2. H2O is acting as an acid because it is donating a proton to NH3.

In the reverse reaction:

1. NH4+ is behaving as an acid because it is donating a proton to OH-.

2. OH- is behaving as a base because it is accepting a proton from NH4+.

Amphoteric Substances

In its reaction with HCl, H2O behaves as a base by accepting a proton. However, in its reaction with NH3, H2O behaves as an acid by donating a proton.

A substance that can act as either an acid or a base is described as being amphoteric.

You may also come across the term amphiprotic. An amphiprotic substance is one that can both donate and accept protons. So water is described as being amphiprotic. Other examples of amphiprotic substances are amino acids and the hydrogensulphate ion, (HSO4)-.

All amphiprotic substances are also amphoteric, but the reverse is not true. There are amphoteric substances, like aluminium oxide, that do not donate or accept protons when they act as acids and bases respectively. These substances fall into the category of acting as acids and bases according to the Lewis theory, which states that an acid is an electron pair accepter and a base is an electron pair donor.

The advantage of the Lewis theory is that it shows similarity between acid-base reactions that involve proton transfer and other acid-base reactions that do not. It is a useful way of explaining why some reactions in organic chemistry that are catalysed by acids can also be catalysed by some substances that can accept a pair of electrons. For example, the nitration of benzene is catalysed by sulphuric acid, but the reaction also takes place if nitric acid is used in conjunction with boron trifluoride, BH3.

Neither the term "amphiprotic" nor the Lewis theory of acids and bases is required learning for your A Level course.

Nitric Acid as a Base

The most common method to nitrate benzene is to use the "nitrating mixture" of concentrated nitric acid and concentrated sulphuric acid. The mixture forms the equilibrium:

H2SO4 + HNO3 -><- (HSO4)- + H2NO3+

Acid 1 Base 2 Base 1 Acid 2

In this reaction:

1. H2SO4 is an acid; its conjugate base is (HSO4)-

2. HNO3 is a base; its conjugate acid is H2SO4+

It seems strange to refer to nitric acid as a base, but in this reaction that is exactly what it is behaving as.

Strong and Weak Acids

A strong acid is defined as one that is almost completely dissociated in aqueous solution.

In a dilute solution of hydrochloric acid, virtually all the hydrogen chloride molecules are dissociated. We usually represent this using a single direction arrow in the equation for the dissociation:

HCl(aq) -> H+(aq) + Cl-(aq)

By contrast, a weak acid is defined as one that is only partially dissociated (often less than 10%) in aqueous solution. Organic acids such as ethanoic acid are typically weak acids.

We represent partial dissociation by using the reversible arrow sign in the equation for the dissociation:

CH3COOH(aq) -><- CH3COO-(aq) + H+(aq)

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Hydrogen Ion Concentration and the pH Scale

Hydrogen Ion Concentration and pH

Strong Acids

As already mentioned, strong acids are assumed to be dissociated completely when they are dissolved in water. This means that the hydrogen ion concentration is related directly to the concentration of the acid.

For example, a solution of HCl of concentration 0.100moldm-3 will produce a hydrogen ion concentration 0.100moldm-3.

The pH of an aqueous solution is related to the hydrogen ion concentration by the following equation:

pH = -log([H+]) or pH = log(1/[H+])

The hydrogen ion concentration, [H+], is measured in moldm-3.

Which equation you decide to use is purely a matter of personal preference.

Calculating Hydrogen Ion Concentration from pH

It is simple matter to calculate the hydrogen ion concentration from a given pH value.

The following equation should be used:

[H+(aq)] = 10^-pH

Weak Acids

Determining the hydrogen ion concentration of an aqueous solution of a weak acid is more complicating because a significant amount of undissociated acid is present in solution. It is necessary to refer to the acid dissociation constant, Ka, for the acid.

Acid Dissociation Constant

If we use HA to represent a weak acid, then the equation for its dissociation in aqueous solution is:

HA(aq) -><- H+(aq) + A-(aq)

By applying the equilibrium law to this reaction we obtain:

a constant = ([H+(aq)][A-(aq)])/[HA(aq)]

This constant is called the acid dissociation constant and is given the symbol Ka.

We shall now calculate the hydrogen ion concentration of an aqueous solution of ethanoic acid of concentration 0.0500moldm-3. The value of Ka for ethanoic acid is 1.74x10^-5moldm-3 at 298K.

CH3COOH(aq) -><- CH3COO-(aq) + H+(aq)

Ka = ([CH3COO-(aq)][H+(aq)])/[CH3COOH(aq)]

Every time a molecule of CH3COOH dissociates, a CH3COO- ion and a H+ ion are formed.

This means that [CH3COO-(aq)] = [H+(aq)].

So, the expression for Ka can be simplified to:

Ka = ([H+(aq)])^2/[CH3COOH(aq)]

It is important to recognise at this stage that the concentrations in the expression of Ka are the equilibrium concentrations. However, if the value of Ka is very small (as in this case) then the concentration of the undissociated acid at equilibrium is very similar to the initial concentration of the acid. It is therefore reasonable to take the concentration at equilibrium as being the same as the initial concentration.

Therefore,

Ka = ([H+(aq)])^2/0.0500 = 1.74 x 10^-5 moldm-3.

[H+(aq)] = (o.0500 x 1.74 x 10^-5)^1/2 = 9.33 x 10^-4 moldm-3.

This gives a pH of 3.03.

If you are good as remembering equations, then the following equation can be used to calculate the hydrogen ion concentration of an aqueous solution of a weak monobasic acid.

[H+(aq)] = /(Ka x [acid])

where / = square root and Ka is the dissociation constant for the weak acid.

Ka and pKa Values

The larger the value of Ka the stronger the acid.

By contrast, the smaller the value of pKa the stronger the acid.

Calculating the pH of a Dibasic Acid

Sulphuric acid is the most common dibasic acid. It dissociates in two stages:

H2SO4(aq) -> H+(aq) + (HSO4)-(aq)

(HSO4)- -><- H+(aq) + (SO4)2-(aq)

H2SO4 is a strong acid and is therefore fully dissociated. (HSO4)- is a weak acid (Ka = 0.0100moldm-3).

In a 0.0500moldm-3 aqueous solution of H2SO4 the contribution to the [H+(aq)] from the H2SO4 will be 0.0500moldm-3. If we assume that the contribution to the [H+(aq)] from the (HSO4)- ion is xmoldm-3, then we have the following relationship:

Ka((HSO4)-) = 0.0100 = (0.500-x)x/0.500

Solving this quadratic equation gives x=0.0098.

This gives a total [H+(aq)] of 0.05098moldm-3 with a subsequent pH of 0.293.

It is interesting to perform this calculation in order to recognise that the contribution to the [H+(aq)] from the (HSO4)- ions is negligible. This is because its dissociation is significantly reduced owing to the high [H+(aq)] from the full first ionisation of H2SO4.

A second interesting reason for performing this calculation is because many books state 0.5moldm-3 H2SO4(aq) can be used as the acid solution in a standard hydrogen electrode. As the standard hydrogen electrode requires a [H+(aq)] of 1.00moldm-3 (pH = 0.00), this is clearly incorrect.

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Ionic Product of Water, Kw

Dissociation of Water

Pure water has a slight electrical conductivity, so it must contain some ions. It self-ionises according to the following equation:

H2O(l) -- H+(aq) + OH-(aq)

If we apply the equilibrium law to this reaction we obtain:

[H+(aq)][OH-(aq)] / [H2O(l)] = a constant

As [H2O(l)] is constant at a given temperature, the expression may be simplified to:

[H+(aq)][OH-(aq)] = a constant

This constant is called the ionic product of water and is given the symbol Kw.

The value of Kw at 298K is 1.00 x 10^-14 mol2dm-6.

A neutral solution is defined as one in which the hydrogen ion concentration is equal to the hydroxide ion concentration. This is the case for pure water.

At 298K, [H+(aq)] = 1.00 x 10^-7 moldm-3.

So, the pH of pure water at 298K is 7.00 [-log(1.00 x 10^-7)]

Kw and pKw

The relationship between Kw and pKw is given by the following equation:

pKw = -logKw

As 298K, when Kw = 1.00 x 10^-14 mol2dm-6, pKw = 14.00

pH of Aqueous Solutions of Strong Bases

When an acid is dissolved in water, it produces so many hydrogen ions that the small contribution from the water is insignificant, unless the acid concentration is very small.

However, the fact that water ionises is the reason why even the most alkaline solutions contain some hydrogen ions.

Sodium hydroxide is a strong base, so in dilute aqueous solution we can consider its ions to be completely dissociated.

A sodium hydroxide solution of concentration 0.100moldm-3 therefore has a hydroxide ion concentration of 0.100moldm-3.

If [OH-(aq)] = 0.100moldm-3 and [H+(aq)][OH-(aq)] = 1.00 x 10^-14 mol2dm-6, then

[H+(aq)] = 1.00 x 10^-14 / 0.100 moldm-3.

= 1.00 x 10^-13 moldm-3.

The pH of this solution is therefore 13.0

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Analysing Data from pH Measurements

Comparing Solution Through pH Measurement

Strong and Weak Acids

The relative strengths of different acids can be determined by measuring the pH of equimolar aqueous solutions of the acids, at the same temperature.

The table shows the pH of 0.100moldm-3 aqueous solution of various acids at 298K.

Formula of acid HCl CHCl2COOH CH2ClCOOH HCOOH CH3COOH CH3CH2COOH

pH of 0.100moldm-3 1.00 1.14 1.93 2.38 2.87 2.93

aqueous solution

The higher the value of pH, the weaker the acid.

Strong and Weak Bases

The same method can be used to determine the relative strengths of bases.

The table below shows the pH of 0.100moldm-3 aqueous solution of various bases at 298K.

Formula of base NH3 CH3NH2 (CH3)2NH CH3CH2NH3 CH3CH2CH2NH2 NaOH

pH of 0.100moldm-3 11.13 11.82 11.83 11.84 11.86 13.00

aqueous solution

The higher the value of the pH, the stronger the base.

Salts

The table below shows the pH of aqueous solutions of various salts of concentration 0.100moldm-3 at 298K.

Formula of salt NaCl KNO3 CH3COONa NH4Cl CH3COONH4

pH of aqueous 7.00 7.00 8.88 5.13 7.00

solution

The pH of NaCl is 7.00 because the salt is made from a strong(HCl) and a strong a strong base (NaOH). The same is true for KNO3, which is a product of the strong HNO3, and the strong base KOH.

An aqueous solution of CH3COONa is alkaline because it is a product of a weak acid (CH3COOH) and a strong base (NaOH).

An aqueous solution of NH4Cl is acidic because it is a product of a strong acid (HCl) and a weak base (NH3).

An aqueous solution of CH3COONH4 is neutral (pH = 7.00) because it is a product of a weak acid (CH3COOH) and a weak base (NH3), and the relative strengths of the acid and base are the same.

This is shown by their dissociation constants.

CH3COOH: Ka = 1.74 x 10^-5 moldm-3.

NH3: Kb = 1.74 x 10^-5 moldm-3.

Summary Table for Aqueous Solution at 298K

salt of a strong acid and a strong base pH = 7 (solution is neutral)

salt of a weak acid and a strong base pH > 7 (solution is alkaline)

salt of a strong acid and a weak base pH < 7 (solution is acidic)

salt of a weak acid and a weak base pH depends on relative strength of acid and base;

if Ka = Kb, pH = 7

if Ka > Kb, pH < 7

if Ka < Kb, pH > 7

Effect of Dilution on the pH for Aqueous Solutions at 298K

Strong Acids

The table below shows the pH of five solutions of hydrochloric acid. In each case the acid has been diluted by a factor of ten from 1.00 (1.00 x 10^0) to 0.000100 (1.00 x 10^-4)moldm-3. All solutions have a temperature of 298K.

Concentration moldm-3 1.00 0.100 0.0100 0.00100 0.000100

pH 0.00 1.00 2.00 3.00 4.00

You will notice that the pH increases by a factor of one unit for each 10-fold decrease in concentration.

If we follow this to its logical conclusion, then an aqueous solution of hydrochloric acid of concentration 1.00 x 10^-8moldm-3 should have a pH of 8.00. This is clearly nonsense because this would mean that a solution of an acid was alkaline.

Earlier in this topic we mentioned the fact that the contribution to the hydrogen ion concentration from the dissociation of water can usually be ignored. When solutions are as dilute as 10^-8 moldm-3 this is no longer the case. The pH of 10^-8moldm-3 hydrochloric acid is very close to 7, as the contribution to the concentration of hydrogen ions from the water (10^-7moldm-3) is now greater than that of the acid.

Weak Acids

The table below shows the pH of five aqueous solutions of ethanoic acid (CH3COOH). In each case the acid has been diluted by a factor of ten from 1.00 to 0.000100 moldm-3. Once again, all solutions have a temperature of 298K.

Concentration moldm-3 1.00 0.100 0.0100 0.00100 0.000100

pH 2.38 2.88 3.38 3.88 4.38

With a weak acid the pH value increases by a factor of about 0.50 for each 10-fold decrease in concentration.

Determining Ka of a Weak Acid from Experimental Data

The following experiment can be performed to determine Ka of benzoic acid (C6H5COOH) - a weak monobasic acid.

1. Accurately weigh between 0.40 and 0.50g of benzoic acid and then dissolve water contained in a beaker. (Benzoic acid is not very soluble, so it may be necessary to warm the water to get it to dissolve. If so, allow the solution to cool before performing the next stage).

2. Transfer the solution to a 250cm3 volumetric flask. Add several washings from the beaker using deionised water, and then make up to the mark with deionised water.

3. Mix the solution by inverting the flask several times.

4. Withdraw a sample of the solution and place it in a small beaker.

5. Using a calibrated pH meter, measure the pH of the solution.

The result obtained is close to the accepted value, but that does not mean that we have performed an accurate experiment. There may have been a number of errors that have, by chance, cancelled out one another.

To start with, we have assumed that the [C6H5COOH(aq)] at equilibrium is identical to the original concentration of benzoic acid. Obviously this is not correct because it must have dissociated slightly to produce a solution of pH = 3.00.

The major uncertainty in this experiment is the measurement of the pH value. Even small error to a large discrepancy in the final answer. A pH of 3.10 would give a final answer of 3.92 x 10^-5, whereas a value of 2.90 would give 9.87 x 10^-5 for the value of Ka.

Transfer errors may also be significant when weighing out small amounts.