In the previous section, we saw Lewis acids and bases. In this section, we will see ionization constant of water and the pH scale
• We have seen that:
♦ In some situations, water act as acid
♦ In some other situations, water act as base
• We can now give a simple explanation for this dual behaviour. It can be written in 3 steps:
1. Donating or accepting proton
• Water can donate a proton
♦ When such a donation is made, it acts as an acid
• Water can accept a proton
♦ When such a acceptance is made, it acts as a base
2. When water is added to a strong acid:
• There will be a competition between two tendencies:
♦ Tendency of the acid to donate proton
♦ Tendency of water to donate proton
• Since the acid is strong, it will be the winner
♦ The water will be forced to accept the electron
♦ Thus water becomes the base
3. When water is added to a weak acid:
• There will be a competition between two tendencies:
♦ Tendency of the acid to donate proton
♦ Tendency of water to donate proton
• Since the acid is weak, it will be the loser
♦ The acid will be forced to accept the proton from the water
♦ Thus water becomes the acid
• When neither acid nor base is added to water, we can say that, water is in the pure form
• Pure water does not occur in nature. But for our present discussion, we can assume that, the sample of water is pure and so it contains only H2O molecules
• In such a pure sample, there will be both H3O+ ions and OH－ ions
• The equation can be written as:
H2O ⇌ H+ + OH－
• This is a reversible process
♦ The molecules will be continuously dissociating into ions
✰ This is the forward reaction
♦ The ions will be continuously combining to give back the molecules
✰ This is the backward reaction
♦ At equilibrium, rates of both forward and backward reactions will be the same
• The stoichiometric coefficients show that, equal quantities of H+ and OH－ will be present
♦ So a sample of pure water will be neither acidic nor basic
• We know that, every reversible reaction will have an equilibrium constant K
• Let us calculate the K for our present reaction which is: H2O ⇌ H+ + OH－
• It can be written in 5 steps:
1. We have: $\mathbf\small{\rm{K=\frac{[H^+][OH^-]}{[H_2 O]}}}$
2. The [H2O] will be a constant because, it is a pure liquid
• So the denominator is a constant
♦ That means, we have to multiply the constant K with another constant
♦ Multiplying a constant with another constant gives a third constant
• That means, K get modified
♦ We will denote the new K as Kw
♦ Kw is called the ionic product of water
• So we can write: Kw = [H+][OH－]
3. Experiments show that:
There will be 1.0 × 10-7 moles of H+ ions in one litre of pure water at 298 K
4. Since the stoichiometric coefficients are equal, the number of moles of OH－ ions will also be the same 1.0 × 10-7
• So we can write: [H+] = [OH－] = 1.0 × 10-7
5. Thus we get: Kw (at 298 K) = [H+][OH－] = (1.0 × 10-7)2 =1.0 × 10-14
• We just saw that, at equilibrium, there will be 10-7 moles of H+ ions in 1 L of water
• Let us calculate the number of moles of H2O in that same 1 L of water
• In this way, we can make a comparison between two items:
♦ Number of moles of H+ ions in 1 L water
♦ Number of moles of undissociated H2O molecules in the same 1 L water
• It can be written in 4 steps:
1. One mol water is 18 grams
2. 1 L of water is 1000 grams
• So number of moles of H2O in 1 L = 1000⁄18 = 55.55
3. If 10-7 mol H+ is present, that much mol H2O would have dissociated
• So the number of moles of H2O remaining = (55.55 - 10-7)
♦ 10-7 is very small when compared to 55.55
♦ So the subtraction will not make any difference
♦ We can write: (55.55 - 10-7) = 55.55
• That means, there are 55.55 moles of undissociated H2O molecules in 1 L water
4. Now we can make the comparison:
♦ Number of moles of H+ ions in 1 L water = 10-7
♦ Number of moles of undissociated H2O molecules in the same 1 L water = 55.55
• 10-7 is very small when compared to 10-7
• That means, in the reversible process H2O ⇌ H+ + OH－,
♦ Concentration on the left side
♦ is very large when compared to
♦ Concentrations on the right side
◼ That means, forward reaction is not favored
• On many occasions, we will want to know whether a given solution is acidic or basic in nature. The following 12 steps will help us in this regard:
1. A solution will be neutral if
♦ The number of moles of H+ in 1 L of the given solution
♦ is equal to
♦ The number of moles of OH－ in that 1 L
2. A solution will be acidic if
♦ The number of moles of H+ in 1 L of the given solution
♦ is greater than
♦ The number of moles of OH－ in that 1 L
3. A solution will be basic if
♦ The number of moles of H+ in 1 L of the given solution
♦ is less than
♦ The number of moles of OH－ in that 1 L
4. We know that, the number of moles in 1 L is represented using square brackets. It is called concentration
• So we can write:
♦ A solution will be neutral if [H+] = [OH－]
♦ A solution will be acidic if [H+] > [OH－]
♦ A solution will be basic if [H+] < [OH－]
5. It is clear that, we will want to report the concentrations [H+] and [OH－] on many occasions. We can decide upon the acidic nature or basic nature only after studying those reported values
6. It is not convenient to use statements like these:
♦ Concentration [H+] = 10-7 mol L-1
♦ Concentration [H+] = 10-5 mol L-1
♦ Concentration [OH－] = 10-9 mol L-1
• So the Danish scientist Sorensen introduced the pH scale
• The pH scale is used to express the concentration of H+ ions in a convenient way
• The following steps from (7) to (10) will explain the pH scale
7. For finding the pH, the first step is to write the [H+]
• Let the [H+] of the given solution be 10-5
♦ That is., 10-5 moles of H+ are present in 1 L of the given solution
8. The second step is to write the logarithm (to base 10) of that [H+]
◼ Logarithm can be explained in 2 steps:
(i) First let us see some examples from math classes:
• Since 8 = 23, log2(8) = 3
♦ That is., logarithm of '8' to the 'base 2' is 3
♦ This is same as: log2(23) = 3
• Since 1024 = 45, log4(1024) = 5
♦ That is., logarithm of '1024' to the 'base 4' is 5
♦ This is same as: log4(45) = 5
• Since 1000 = 103, log10(1000) = 3
♦ That is., logarithm of '1000' to the 'base 10' is 3
♦ This is same as: log10(103) = 3
• Since 1⁄10000000 = 10-7, log10(1⁄10000000) = -7
♦ That is., logarithm of '1⁄10000000' to the 'base 10' is -7
♦ This is same as: log10(10-7) = -7
(ii) In step (1), we wrote that, [H+] = 10-5
• So logarithm (to the base 10) of this concentration = log10([H+]) = log10(10-5) = -5
9. The next step is to take the negative of this logarithm
• The logarithm obtained is -5. So the negative of the logarithm is 5
10. This negative of the logarithm is the pH of the solution
• That is., the pH of the given solution is 5
10. So now we can define the pH:
◼ pH of a solution is:
♦ the negative of the logarithm (to base 10)
♦ of the [H+] of that solution
11. Let us see an example:
• Suppose that some HCl molecules are dissolved in 1 L of water
• Let us assume that, in the resulting solution, there are 0.004 moles of H+ ions
• Since the volume of water is 1 L, we can write: [H+] = 0.004 M
• So pH of the solution = -log10([H+]) = -log10(0.004) = 2.4
12. Another example:
• Let the [H+] of a solution be 10-4 M
• So pH of the solution = -log10([H+]) = -log10(10-4) = 4
• We have seen the basics of pH scale
• Next we will see some important features of the scale. It can be written in steps:
1. Consider a sample of pure water
• We have seen that, 1 L of that sample will contain 10-7 moles of H+ ions
♦ That is., [H+] of pure water is 10-7
• So pH of pure water = -log10([H+]) = -log10(10-7) = 7
2. In pure water, there will be equal number of H+ and OH－ ions
• So pure water will be neutral
• Thus we can write: pH of a neutral solution will be 7
3. Let us see the situations in which the pH takes values other than 7
It can be explained in steps:
(i) Consider the reversible reaction: H2O ⇌ H+ + OH－,
♦ We have seen that, Kw of this reaction (at 298 K) is 10-14
♦ This Kw is a constant. It will not change unless we change the temperature
(ii) At 298 K, let us add some H+ ions
• We have seen that Kw =10-14 = [H+][OH－]
• In order to keep Kw constant, some of the newly added H+ ions will combine with OH－ ions and become H2O molecules
• In effect,
♦ There will be an increase in the number of H+ ions
♦ There will be a decrease in the number of OH－ ions
♦ But Kw will remain constant
(iii) But the increase in number of H+ ions will increase the [H+]
So the pH will change
(iv) Suppose that, the number of moles of H+ increased from 10-7 to 10-5
(Remember that, 10-5 is greater than 10-7)
• So the new [H+] = 10-5
♦ So the new pH = -log10([H+]) = -log10(10-5) = 5
(v) We see that: when the number of H+ increases, the pH decreases from 7
(vi) Also, increase in number of H+ makes the solution acidic
• So we can write: If a solution is acidic, it's pH will be less than 7
(vii) Suppose that, the [H+] is very high, say 10-3
• Then the pH will be 3
• Also, the solution will be more acidic than when the pH is 5
♦ This is because:
✰ When pH is 5, there are 10-5 moles of H+
✰ When pH is 3, there are 10-3 moles of H+
✰ 10-3 is greater than 10-5
(viii) So we can write:
♦ Greater the acidity of the solution
♦ greater will be the deviation
♦ of the pH value from 7
• As the acidity increase, the pH will move in the direction:
7 → 6 → 5 → 4 → 3 → 2 → 1
4. In the above step (3), we saw the situations where pH becomes lesser than 7
• Let us now see the situations where the pH becomes greater than 7
• It can be explained in steps:
(i) Consider the reversible reaction: H2O ⇌ H+ + OH－,
♦ We have seen that, Kw of this reaction (at 298 K) is 10-14
♦ This Kw is a constant. It will not change unless we change the temperature
(ii) At 298 K, let us add some OH－ ions
• We have seen that Kw =10-14 = [H+][OH－]
• In order to keep Kw constant, some of the newly added OH－ ions will combine with H+ ions and become H2O molecules
(ii) In effect,
♦ There will be an increase in the number of OH－ ions
♦ There will be a decrease in the number of H+ ions
♦ But Kw will remain constant
(iii) But the decrease in number of H+ ions will decrease the [H+]
• So the pH will change
(iv) Suppose that, the number of moles of OH－ increased from 10-7 to 10-6
(Remember that, 10-6 is greater than 10-7)
• We can calculate the new [H+] using the relation: Kw = 10-14 = [H+][OH－]
♦ In our present case, we get: 10-14 = [H+] × 10-6
♦ So the new [H+] = 10-8
♦ (This 10-8 is lesser than the original 10-7)
• So the new pH = -log10([H+]) = -log10(10-8) = 8
(v) We see that: when the number of H+ decreases (same as increase in number of OH－), the pH increases from 7
(vi) But decrease in number of H+ (same as increase in number of OH－) makes the solution basic
• So we can write: If a solution is basic, it's pH will be greater than 7
(vii) Suppose that, the [OH－] is very high, say 10-2
• Then we get: [H+] = $\mathbf\small{\rm{\frac{10^{-14}}{10^{-2}}}}$ = 10-12
♦ Then the pH will be 12
• Also, the solution will be more basic than when the pH is 8
♦ This is because:
✰ When pH is 8, there are 10-6 moles of OH－
✰ When pH is 12, there are 10-2 moles of OH－
✰ 10-2 is greater than 10-6
(viii) So we can write:
♦ Greater the basicity of the solution
♦ greater will be the deviation
♦ of the pH value from 7
• As the basicity increase, the pH will move in the direction:
7 → 8 → 9 → 10 → 11 → 12 → 13 → 14
An important note can be written in 3 steps:
1. To calculate pH, we use the [H+]. We do not use [OH－]
♦ This is because, pH = -log10([H+])
2. If we know the [H+], we can directly calculate pH
3. If it is [OH－] that we know, we can calculate [H+] using the relation:
[H+][OH－] = 10-14
• Now we will see how pH is measured for practical applications
• To find the pH of a given solution, we use the pH indicator strips
♦ They are strips of paper, which are coated with a suitable chemical substance
♦ The following 4 steps will help us to understand the procedure
1. Fill half of a test tube with the given solution
2. Take a strip and dip it in the solution
• The solution will react with the substance coated on the paper
• As a result, the color of the paper will change
3. Compare the new color of the paper with the colors in the standard chart
• Find the matching color on the chart
(Some images of the color chart and strips can be seen here)
4. The number on the chart corresponding to the matching color is the pH
• Another method is to use pH meters
• They work by measuring the electrical potential between two electrodes
• Accurate values of pH can be obtained just by dipping the electrodes in the solution
An interesting point can be written in 3 steps:
1. From the above discussion, it is clear that, pH of pure water is 7
• But it is important to remember that, this pH of 7 is applicable only when the water is at 298 K
2. If the water is at 310 K, the ionic product [H+][OH-] will be 2.7 × 10-14 instead of 10-14
• So we get: [H+] = [OH-] = √(2.7 × 10-14) = 1.643 × 10-7
3. Thus pH of water at 310 K = log10([H+]) = log10(1.643 × 10-7) = 6.78
In the next section, we will see some solved examples related to pH
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment