Chapter 7.19 - Ionization Constants of Poly-Basic acids

In the previous section, we saw ionization constants of weak bases. We also saw relation between K_a and K_b. In this section, we will see Di- and Polybasic Acids and Di- and Polyacidic Bases. Later in this section, we will see factors affecting acid strength

• Our discussions so far, were related to acids like HCl, HNO3, CH₃COOH etc.,
    ♦ When these acids are added to water, each molecule can give only one H⁺ ion
        ✰ HCl ⇌ H⁺(aq) + Cl^-(aq)
        ✰ HNO₃ ⇌ H⁺(aq) + NO^3-(aq)
        ✰ CH₃COOH ⇌ H⁺(aq) + CH₃COO^-(aq)
    ♦ So they are called mono basic acids
• But some acids like H₂SO₄ (sulphuric acid), C₂H₂O₄ (oxalic acid) etc., can give more than one H⁺ ions
• We will now see some basic details about such acids. It can be written in 7 steps:
1. The dissociation of H₂SO₄ can be written as two stages:
Stage 1: H₂SO₄ dissociates to give one H⁺ and one HSO₄^-
H₂SO₄(aq) ⇌ H⁺(aq) + HSO₄^-(aq)
Stage 2: The HSO₄^- dissociates to give H⁺ and SO₄^2-
HSO₄^-(aq) ⇌ H⁺(aq) + SO₄^2-(aq)
2. If we add the two reactions, we will get the net reaction:
H₂SO₄(aq) + HSO₄^-(aq) ⇌ H⁺(aq) + HSO₄^-(aq) + H⁺(aq) + SO₄^2-(aq)
• HSO₄^-(aq) is common on both sides. So they will cancel out
• Thus the net reaction is:
H₂SO₄(aq) ⇌ 2H⁺(aq) + SO₄^2-(aq)
3. But for analyzing equilibrium, we must to consider the two stages individually
• The general form of the two stages are:
Stage 1: H₂X(aq) ⇌ H⁺(aq) + HX^-(aq)
Stage 2: HX^-(aq) ⇌ H⁺(aq) + X^2-(aq)
• Let us write the equilibrium constants:
    ♦ The equilibrium constant for stage 1 will be: $\mathbf\small{\rm{K_{a1}=\frac{[H^+][HX^-]}{[H_2X]}}}$
        ✰ K_a1 is called the first ionization constant of the acid H₂X
    ♦ The equilibrium constant for stage 2 will be: $\mathbf\small{\rm{K_{a2}=\frac{[H^+][X^{2-}]}{[HX^-]}}}$
        ✰ K_a2 is called the second ionization constant of the acid H₂X
◼ It is clear that, a tri basic acid like H₃PO₄ (phosphoric acid) will have a third ionization constant K_a3 also
4. Note the general forms:
    ♦ HX is the general form of mono basic acid
    ♦ H₂X is the general form of di basic acid
    ♦ H₃X is the general form of tri basic acid
    ♦ di and tri basic acids are called polybasic acids
         ✰ Poly basic acids are also called polyprotic acids
    ♦ K_a2 and K_a3 are called higher order ionization constants
5. For a polyprotic acid, K_a2 and K_a3 will be smaller than K_a2
◼ A smaller K_a2 indicates that: [H⁺] and [X^2-] will be smaller in stage 2. The reason can be explained as follows:
• The H⁺ and X^2- in the second stage are formed from HX^-
    ♦ It is easier to split H₂X into H⁺ and HX^-
        ✰ So in the first stage, [H⁺] and [HX^-] will be higher
    ♦ But it is difficult to split HX^- into H⁺ and X^2-
        ✰ This is due to the greater electrostatic attraction between H and X in the HX^-
        ✰ So in the second stage, [H⁺] and [X^2-] will be lower
An example:
    ♦ It is easier to separate H₂CO₃ into H⁺ and HCO₃^-
    ♦ But it is difficult to separate HCO₃^- into H⁺ and CO₃^2-
Another example:
    ♦ It is easier to separate H₂PO₄^- into H⁺ and HPO₄^-
    ♦ But it is difficult to separate HPO₄^- into H⁺ and PO₄^2-
◼ A smaller K_a3 indicates that: [H⁺] and [X^3-] will be smaller in stage 3. The reason can be explained as follows:
• The H⁺ and X^3- in the third stage are formed from HX^2-
    ♦ It is easier to split HX^- into H⁺ and X^2-
        ✰ So in the second stage, [H⁺] and [X^2-] will be higher
    ♦ But it is difficult to split HX^2- into H⁺ and X^3-
        ✰ This is due to the greater electrostatic attraction between H and X in the HX^2-
        ✰ So in the third stage, [H⁺] and [X^3-] will be lower
6. Thus we can write:
The H⁺ in the solution of a polyprotic acid mainly comes from stage 1
7. In a solution of a polybasic acid, all species that we write in the various stages will be present. This can be explained in 5 steps:
(i) Consider the dissociation of H₂X
• The two stages are:
Stage 1: H₂X(aq) ⇌ H⁺(aq) + HX^-(aq)
Stage 2: HX^-(aq) ⇌ H⁺(aq) + X^2-(aq)
(ii) The net reaction is:
H₂X(aq) ⇌ 2H⁺(aq) + X^2-(aq)
    ♦ When we write the net reaction, HX^- cancels out
    ♦ But in the solution, HX^- will be present at all times
(iii) This is because, in the forward direction, H₂X will not directly give 2H⁺ and one X^2-
    ♦ H₂X will first give one H⁺ and one HX^-
    ♦ The HX^- will dissociate to give one H⁺ and one X^2-
(iv) Similarly, in the backward direction, X^2- will not directly combine with two H⁺ to give H₂X
    ♦ X^2- will first combine with one H⁺ to give one HX^-
    ♦ This HX^- will combine with one H⁺ to give one H₂X
(v) Thus, HX^-, though absent in the net reaction equation, will be present in the solution at all times

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Factors affecting acid strength

• We have seen that, some acids are strong while some others are weak
   ♦ We want to know what makes some acids stronger
• There are many complex factors which make some acids stronger. We will see those details in higher classes
• At present, we need to know some simple basics only. Those basics can be written in 8 steps
1. We know that, the general form of an acid is HX
• The X can be Cl, Br, S etc.,
2. There will be a bond between H and X
• If this bond is strong, it will be difficult to separate H and X
    ♦ Consequently, the number of H⁺ ions will be low
3. If the number of H⁺ ions is low, it will be a weak acid
◼  So we can write:
The H-X bond strength is an important factor which affects the strength of an acid
   ♦ If the bond strength is high, the acid will be weak
   ♦ If the bond strength is low, the acid will be strong
4. In addition to H-X bond strength, one more factor plays an important role. It is the ‘difference in electronegativity’. The following steps from (5) to (8)give an explanation:
5. From the data book, we have:
    ♦ Electronegativity (e_H) of H = 2.20
• Next we note down the electronegativity (e_X) of X
• Then we calculate the difference (e_X – e_H) = (e_X – 2.2)
6. If this difference is large, the shared pair of electrons will be shifted towards X
• Then there will be a significant ‘charge separation’ between H and X
7. If there is a significant 'charge separation', it will be easier to separate H and X
• So there will be a large number of H⁺ ions in the solution
8. If the number of H⁺ ions is large, it will be a strong acid
◼  So we can write:
The difference in electronegativity is an important factor which affects the strength of an acid
   ♦ If the difference is high, the acid will be strong
   ♦ If the difference is low, the acid will be weak

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• Based on the above discussion, we can now analyze the relation between the following two items:
    ♦ Strength of an acid HX
    ♦ Position of X in the periodic table
• The analysis can be written in 13 steps:
1. Consider two acids HX₁ and HX₂
• We want to know which one is stronger
2. Positions of X₁ and X₂ can be as follows:
    ♦ Let X₁ and X₂ belong to the same group of the periodic table
    ♦ Let X₁ be at a higher position than X₂
3. We know that, as we move down the group, the size of atom increases
• So we can write:
    ♦ In HX₁, the H atom will be nearer to the nucleus of X₁
        ✰ So HX₁ will have a greater bond strength
    ♦ In HX₂, the H atom will be further away from the nucleus of X₁
        ✰ So HX₂ will have a lesser bond strength
4. Thus dissociation of HX₂ is easier
• That means, HX₂ is stronger than HX₁
5. We can write:
As we move down a group in the periodic table, the strength of acid increases
• This is shown in fig.7.20(a) below:

Fig.7.20

6. Let us see an example. It can be written in 3 steps:
(i) F, Cl, Br and I belong to the same group of the periodic table
(ii) F is at the top. Below F come Cl, Br and I in order
(iii) So the order of acid strength will be: HF < HCl < HBr < HI
7. Another example can be written in 3 steps:
(i) O, and S belong to the same group of the periodic table
(ii) O is at the top. Below O comes S
(iii) So the order of acid strength will be: H₂O < H₂S
8. Next we consider position in periods
    ♦ Let Let X₁ and X₂ belong to a same row in the periodic table
    ♦ Let X₂ be on the right side of X₁
9. We know that, as we move from left to right along a period, electronegativity increases
• So we can write:
    ♦ The electronegativity e_X1 of X₁ will be lesser
    ♦ The electronegativity e_X2 of X₂ will be greater
10. Thus we get:
    ♦ The quantity (e_X1 - e_H) will be lesser
    ♦ The quantity (e_X2 - e_H) will be greater
11. So we can write:
    ♦ Charge separation will be lesser in HX₁
    ♦ Charge separation will be greater in HX₂
• Consequently,
    ♦ It is more easier to dissociate HX₂
• Thus we get: HX₂ is stronger than HX₁
12. So we can write:
As we move from left to right along a periodic table, the acid strength increases
• This is shown in fig. above
13. Let us see an example. It can be written in 3 steps:
(i) C, N, O and F belong to the same row of the periodic table
(ii) C is at the left. To the right come N, O and F in order
(iii) So the order of acid strength will be: CH₄ < NH₃ < H₂O < HF

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In the next section, we will see common ion effect


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