Acids & Bases (Part III): pH, Indicators & Buffers

pH & pH Scale

The pH (hydrogen ions potential) is a measure of the acidity or alkalinity of a solution. The concept of pH was introduced by Sörensén in 1909 to bring about the convenience of working with very dilute solutions. To this effect, he developed a scale consisting of fifteen numbers (0 - 14), which is used in pH meter, for measuring the relative acidity or alkalinity in solutions. This scale is known as the pH scale. The numbers in the pH scale are the values of the negative logarithms of the hydrogen ions concentrations in such solutions.

From the above, we can define pH as the negative logarithm of the hydrogen ions concentration [H+] to base 10. Mathematically, this is given as:

pH = - log [H+] ..........................(i)

Alternatively, the above equation can be expressed as

pH = log 1/[H+] .........................(ii)

From equation (ii), we can also define pH as the logarithm of the reciprocal of the hydrogen ions concentration to base 10.

The pH scale is interpreted as follows:

    0 - 2 = strongly acidic
    3 - 4 = moderately acidic
    5 - 6 = weakly acidic
    7 - 7 = neutral
    8 - 9 = weakly alkaline
10 - 11 = moderately alkaline
12 - 14 = strongly alkaline

An analysis of the pH scale shows that acidity decreases with increase in pH value, and alkalinity increases as the pH value increases.

Examples of some substances with their pH values are

Milk of magnesia | 10.5
Pineapples | 3.6
Gastric juice | 1.0
Blood | 7.4
Urine | 4.7 - 8.1
Pure water | 7.0

pOH

Just like pH, pOH (hydroxide ions potential) is the negative logarithm of the hydroxide ions concentration [OH-] to base 10. It is mathematically expressed as:

pOH = - log [OH-] ........................(iii)

Alternatively, it can be defined as the logarithm of the reciprocal of the hydroxide ions concentration [OH-] to base 10, and expressed as:

pOH = log 1/[OH-] ........................(iv)

Calculations Involving pH & pOH

Dissociation of Water

The pH of pure or distilled water shows that it is neutral. This neutrality indicates that as water molecules dissociate into ions, they produce equal amounts of hydrogen and hydroxide ions, according to the equation:

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

Studies have shown that the dissociation constant of water, Kw  at 25°C equals 1.0 × 10^-14, and since water contains equal amounts of [H+] and [OH-], it implies that

[H+][OH-] = 1.0 × 10^-14 mol^2 dm^-6 .......................(v)

Hence,

[H+] = [OH-] = 1.0 × 10^-7 mol dm^-3

Taking the logarithm of both sides of equation (v) gives:

log [H+][OH-] = log 10^-14

Applying the laws of logarithm, we will obtain:

log [H+] + log [OH-] = - 14 log10    (but, log 10 = 1)

log [H+] + log [OH-] = - 14

Divide through by - 1

- log [H+] + (- log [OH-]) = 14 ................(vi)

Substituting equations (i) and (iii) into equation (vi) gives:

pH + pOH = 14 .............................(vii)

Using equations (i), (iii), (v) and (vii); we can calculate the pH, pOH, [H+] and [OH-] of any given solution.

Example 1

Calculate the pH of the following solutions:

(a) 0.005M H2SO4
(b) 0.015M aqueous NH3
(c) M/25 HCl

Answer

(a) Tetraoxosulphate (VI) acid ionizes according to the equation:

H2SO4 (aq) --------> 2H+(aq) + SO4––(aq)

The given concentration of the acid is the hydrogen ion concentration, i.e, [H+] = 0.005M or mol dm^-3

But, H2SO4 contains 2 moles of H+
Therefore, the total concentration of hydrogen ions in the acid, [H+] = 2 × 0.005 = 0.01M

Hence, applying equation (i) or (ii)

pH = - log [H+]
      = - log 0.01 = - log 10^-2
      = - (-2 log 10) = - (-2)
      = 2

(b) Aqueous ammonia dissociates weakly as shown in the equation:

NH4OH(aq) <-------> NH4+(aq) + OH-(aq)

Being a base, the 0.015M concentration in the question is that of the hydroxide ion, and not hydrogen ion; i.e, [OH-] = 0.015M or mol dm^-3.

Therefore, applying equation (iii) or (iv)

pOH = - log [OH-]
         = - log 0.015 = - log 15 × 10^-3
         = - (log 15 + log 10^-3)
         = - (1.1761 - 3) = - (-1.8239)
         = 1.8239

but

pH + pOH = 14
            pH = 14 - pOH
                  = 14 - 1.8239
                  = 12.1

(c) Hydrochloric acid ionizes as follows:

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

The concentration, M/25 = 1/25 M = 0.04M. Therefore, [H+] = 0.04M or mol dm^-3.

Using equation (i) or (ii)

pH = - log [H+]
      = - log 0.04 = - log 4 × 10^-2
      = - (log 4 + log 10^-2)
      = - (0.6021 - 2) = - (-1.3979)
      = 1.4

Example 2

Calculate:

(a) the hydroxide ion concentration in 80.01M tetraoxosulphate (VI) acid.
(b) hydrogen ion concentration in 0.003M sodium hydroxide.

Answer

(a) We know that the ionization of tetraoxosulphate (VI) acid is given as:

H2SO4 (aq) --------> 2H+(aq) + SO4––(aq)

and the given concentration of the acid is the hydrogen ion concentration, i.e, [H+] = 80.01M or mol dm^-3;

but H2SO4 contains 2 moles of H+
Therefore, the total concentration of hydrogen ions in the acid, [H+] = 2 × 80.01 = 160.02M = 1.60 × 10^2 M

Applying equation (v) gives

[H+][OH-] = 1.0 × 10^-14 M^2
       [OH-] = 1.0 × 10^-14 M^2/[H+], but [H+] = 1.6 × 10^2 M
Therefore,
       [OH-] = (1.0 × 10^-14)/(1.6 × 10^2)
       [OH-] = 6.25 × 10^-17 M

(b) Sodium hydroxide ionizes as shown:

NaOH(aq) --------> Na+(aq) + OH-(aq)

The given concentration of 0.003 M is the hydroxide ion concentration. Therefore, [OH-] = 3.0 × 10^-3 M

Applying equation (v) gives

[H+][OH-] = 1.0 × 10^-14 M^2
        [H+] = 1.0 × 10^-14 M^2/[OH-], but [OH-] = 3.0 × 10^-3 M
Therefore,
        [H+] = (1.0 × 10^-14)/(3.0 × 10^-3)
        [H+] = 3.33 × 10^-12 M

The above solutions to Examples 2(a) & (b) show that irrespective of the concentration of a substance, it contains both hydrogen and hydroxide ions.

Example 3

If the dissociation constant, Kw, for water is 1.0 × 10^-14 at 25°C, calculate:

(a) the hydrogen ion concentration
(b) pH value of water

Answer

(a) Water dissociates according to the equation:

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

The dissociation constant, Kw, is given as:

[H+][OH-] = Kw; at 25°C, Kw = 1.0 × 10^-14 M^2 and [H+] = [OH-]
Therefore,
     [H+]^2 = 1.0 × 10^-14 M^2
         [H+] = √1.0 × 10^-14 M^2
                 = 1.0 × 10^-7 M

(b) pH = - log [H+]
            = - log 10^-7
            = - (-7 log 10)
            = - (-7) = 7

Therefore, the hydrogen ion concentration and pH value of water at 25°C are 1.0 × 10^-7 M and 7 respectively. We can infer from the above that if the hydrogen ion concentration of a substance, [H+] = 10^-x, then pH = x

Example 4

(a) The dissociation constant of ethanoic acid at 25°C is 1.8 × 10^-5. In a buffer solution, the concentration of CH3COO- and CH3COOH are 0.025 M and 0.05 M respectively. What is the pH of buffer solution?

Answer

The dissociation of ethanoic acid is as shown:

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

Its dissociation constant at 25°C, Ka, is given as:

Ka = [CH3COO-][H+]/[CH3COOH]
but Ka = 1.8 × 10^-5; [CH3COO-] = 0.025 M; [CH3COOH] = 0.05 M

1.8 × 10^-5 = 0.025 × [H+]/0.05
           [H+] = 0.05 × 1.8 × 10^-5/0.025
                   = (9.0 × 10^-7)/(2.5 × 10^-2)
                   = 3.6 × 10^-5 M

pH = - log [H+]
      = - log (3.6 × 10^-5)
      = - (log 3.6 + log 10^-5)
      = - (log 3.6 - 5 log 10)
      = - (0.5563 - 5)
      = - (-4.4437) = +4.4437
      = 4.4

(b) Calculate the hydrogen ion concentration and pH of a buffer solution, in which the concentration of Ca2+ is 0.02M and that of Ca(OH)2 in water is 0.04M, if the dissociation constant of calcium hydroxide, Kb = 5.5 × 10^-6 at 25°C.

Answer

The dissociation of calcium hydroxide in water is as follows:

Ca(OH)2(aq) <-------> Ca2+(aq) + 2OH-(aq)

Its dissociation constant at 25°C, Kb, is given as:

Kb = [Ca2+][OH-]^2/[Ca(OH)2]
but Kb = 5.5 × 10^-6; [Ca2+] = 0.02 M; [Ca(OH)2] = 0.04 M

5.5 × 10^-6 = 0.02 × [OH-]^2/0.04
      [OH-]^2 = 0.04 × 5.5 × 10^-6/0.02
                   = (2.2 × 10^-7)/(2.0 × 10^-2)
                   = 1.1 × 10^-5 M^2
          [OH-] = √1.1 × 10^-5 M^2
                   = 3.3 × 10^-3 M

But

[H+][OH-] = 1.0 × 10^-14
         [H+] = 1.0 × 10^-14/[OH-]
                 = (1.0 × 10^-14)/(3.3 × 10^-3)
                 = 3.03 × 10^-12 M

pH = - log [H+]
      = - log (3.03 × 10^-12)
      = - (log 3.03 + log 10^-12)
      = - (log 3.03 - 12 log 10)
      = - (0.4814 - 12)
      = - (-11.5186) = +11.5186
      = 11.5

Therefore, the pH of the buffer solution is 11.5

Importance of pH

We cannot overstate the significance of pH in different areas of our lives such as agriculture, medicine, pharmacy, water purification and sewage treatment. In agriculture, a farmer must ensure the soil he intends to cultivate has the correct pH requirements for the crops he intends to plant on it. Studies have shown that soil pH values vary from 4 - 9, depending on the location, and most plants thrive in soils of pH values between 7 - 8. The pH of soils with excess acidity is usually corrected by using calcium hydroxide (slaked lime) to neutralize them.

In medicine and pharmacy, for our bodies to function properly, the body fluids must be maintained at the correct pH values, as enzymes, which aid in the metabolic processes in our bodies, are pH dependent. So, deviations from the normal pH values of our body systems are signs of ill health. When this occurs, the doctors prescribe drugs, which can be used to correct the anomaly. Meanwhile, when producing the drugs, the pharmacists ensure that they are produced under the right pH conditions, without which their efficacy of the active ingredients cannot be guaranteed.

In water purification, it is important to ensure that the water distributed for town supply has a neutral pH to prevent the growth of microbes within a very short period. Similarly, in sewage treatment, the waste products must be made harmless and of neutral pH before they are finally released into the environment, especially, water bodies. If this is not done, it will foster the growth of bacteria and other micro-organisms, which compete for oxygen with the fishes and other aquatic animals in the water bodies. This leads to low oxygen level in the water bodies and eventual death of the water organisms.

Indicators

Indicators are weak organic acids and bases that ionize slightly in solution to produce different colours at different pH of the solution.

During ionization of indicators, there is always an equilibrium mixture of undissociated molecules and dissociated ions, as represented in the equation:

HIn(aq) <------------------> H+(aq) + In-(aq)

The undissociated molecules, HIn, always produce different colours from the dissociated ions. Therefore, the colour of the indicator in a given solution will depend on the ratio of the undissociated molecules and dissociated ions present. Let's take a look at how methyl orange and phenolphthalein behave in acidic and alkaline solutions.

Methyl orange: Let us represent the unionized molecules of methyl orange as HMe, and the dissociated ions as Me–. In solution, HMe is red, while Me– is yellow, and dissociates as follows:

HMe(aq) <------------------> H+(aq) + Me–(aq)

When an acid is added to a solution of methyl orange, the red colour becomes prominent because of the introduction of more hydrogen ions from the acid. This destabilizes the equilibrium of the system, which is forced to adjust according to Le Chatelier's Principle. The excess hydrogen ions combine with the dissociated ions to form the unionized molecules, and this causes the equilibrium position of the reaction to shift to the left. This shift is indicated by the increased intensity of the red colour of the solution.

However, when an alkaline is added to the methyl orange solution, the yellow colour becomes dominant. This is because the hydroxide ions from the alkaline combine with the hydrogen ions from the indicator to form water, thereby, removing them from the equilibrium mixture. This causes the equilibrium position to shift to the right, by forcing the dissociation of more undissociated molecules to replace the lost hydrogen ions. This also increases the concentration of dissociated ions in the solution, which is marked by the yellow colour.

Phenolphthalein: Let the unionized molecules of phenolphthalein be HPh, and the dissociated ions be Ph–. In solution, HPh is colourless, while Ph– is pink, and dissociates as follows:

HPh(aq) <------------------> H+(aq) + Ph–(aq)

In an acidic solution, phenolphthalein is colourless because of the introduction of more hydrogen ions from the acid. The excess hydrogen ions combine with the dissociated ions to form the undissociated molecules, and this causes the equilibrium position to shift to the left. This change is accompanied by the colourless nature of the solution. However, in an alkaline solution, the pink colour is predominant, and the reason is the same as that for methyl orange above.

Indicators & Acid-Base Titration

Indicators are very useful in acid-base titrations to determine the endpoints of the reactions. An endpoint is a stage in titration in which a change in colour of the indicator shows that no more titrant should be added. This is because of the colourless nature of the acid and base solutions. Different indicators give different colours at endpoints depending on the pH of the mixture, which is somewhat dependent on  the types of acids and bases used. The list below shows the colour changes of some common indicators in acidic, basic and neutral solutions and their suitability in a given acid-base titration.

Indicator: Methyl orange
Colour in Solutions: Acid - red (pH: 1.0-3.2); Neutral - orange (pH: 3.2-6.1); Basic - yellow (pH: 6.1-14.0)
Titration Suitability: Strong Acid + Weak Base; Strong Acid + Strong Base

Indicator: Litmus
Colour in Solutions: Acid - red (pH: 1.0-6.0); Neutral - purple (pH: 6.0-8.5); Basic - blue (pH: 8.5-14.0)

Indicator: Phenolphthalein
Colour in Solutions: Acid - colourless (pH: 1.0-8.9); Neutral - pale pink (pH: 8.9-11.0); Basic - pink (pH: 11.0-14.0)
Titration Suitability: Weak Acid + Strong Base; Strong Acid + Strong Base

The pH of a solution can also be measured by using a universal indicator. Though not accurate, it can be used to detect values within the range of 2 - 12, through a series of colour changes. It is obtained by mixing different indicators together. These indicators work at different pH ranges. The pH values obtained are easily determined by comparing the colours obtained with the given standards.

Buffers

A buffer is a solution that shows resistance to changes in pH on dilution or when a little quantity (drops) of acid or base is added to it. It is made up of a mixture of a weak acid and its salt or a weak base and its salt. Examples of buffers are mixtures of:

a) ethanoic acid and potassium ethanoate (CH3COOH/CH3COOK)

b) aqueous ammonia and ammonium chloride (NH4OH/NH4Cl)

c) trioxocarbonate (IV) acid and sodium hydrogen trioxocarbonate (IV) (H2CO3/NaHCO3)

d) calcium hydroxide and calcium chloride (Ca(OH)2/CaCl2)

The concept of buffers and pH is very important in biochemical activities, because most metabolic processes in the human body are pH specific. This implies that a change in the pH of the system by about +/-0.5 may lead to a fatal outcome, if not carefully handled.

For example, the pH of the blood is about 7.4, and if an injection is to be administered to a patient intravenously (through the veins), such injection must be buffered to prevent it from altering the pH of the blood, since it is going directly into the bloodstream. Otherwise, the effect can be fatal.

Similarly, most of the processed foods and drinks we take are buffered. Different parts of the human digestive system have different pH. The mouth, where the digestion of food starts, is alkaline, the stomach is acidic and the small intestine, where the digestion ends, is also alkaline. Therefore, it is necessary that the foods and drinks we take are buffered to prevent unnecessary changes in their pH, as they pass through our alimentary canal. This also prevents complications that may arise due to indigestion.

Do These

Question 1
(a) Define pH in terms of hydrogen ions concentration and give two uses of pH.
(b) What is a buffer solution? Give two examples.

Question 2
The concentration of hydrogen ions in three solutions are:
(a) 2 × 10^-3 mol dm^-3
(b) 7 × 10^-9 mol dm^-3
(c) 5 X 10^-8 mol dm^-3
What is the pH of each solution?

Question 3
(a) A solution has a pH of 4.5. What is its hydrogen ions concentration?
(b) The pH values of two solutions are 5.0 and 10.5. What will be the colours of methyl orange, litmus and phenolphthalein indicators in these solutions?

Question 4
Which of the following pairs is/are buffer solution(s)?
(a) Hydrochloric acid and sodium chloride
(b) Ethanoic acid and sodium ethanoate
(c) Calcium chloride and potassium trioxocarbonate (IV)
(d) Potassium hydroxide and potassium chloride
(e) Aqueous ammonia and ammonium tetraoxosulphate (VI)

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