Chemistry Ionization Constants of Weak Acids, Ionization of Weak Bases, Relation between K_a and K_b, Di- and Polybasic Acids and Di- and Polyacidic Bases, Factors Affecting Acid Strength.

### Topic to be covered

star Ionization Constants of Weak Acids.
star Ionization of Weak Bases.
star Relation between K_a and K_b.
star Di- and Polybasic Acids and Di- and Polyacidic Bases.
star Factors Affecting Acid Strength.

### Ionization Constants of Weak Acids

=> Consider a weak acid color{red}(HX) that is partially ionized in the aqueous solution. The equilibrium can be expressed by:

color{red}(HX (aq) + H_2O (l) ⇌ H_3 O^+ (aq) + X^- (aq))

color{green}("Initial concentration (M)")

color{red}(c \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ 0)

color{green}("Let α be the extent of ionization
Change (M)")

color{red}(-c alpha \ \ \ \ \ \ \ \ \ \ +c alpha \ \ \ \ \ \ \ \ \ \ +c alpha)

color{green}("Equilibrium concentration (M)")

color{red}(c- calpha \ \ \ \ \ \ \ \ \ \ c alpha \ \ \ \ \ \ \ \ \ \ c alpha)

Here, color{red}(c) = initial concentration of the undissociated acid, color{red}(HX) at time, color{red}(t = 0. α =) extent up to which color{red}(HX) is ionized into ions. Using these notations, we can derive the equilibrium constant for the above discussed acid dissociation equilibrium:

color{red}(K_a = c^2 alpha^2 // c (1-alpha) = c alpha^2 // 1- alpha)

color{red}(K_a) is called the dissociation or ionization constant of acid color{red}(HX). It can be represented alternatively in terms of molar concentration as follows,

color{red}(K_a = [H^+] [X^- ] // [HX]) ................(7.30)

=> At a given temperature color{red}(T, K_a) is a measure of the strength of the acid color{red}(HX) i.e., larger the value of color{red}(K_a), the stronger is the acid. color{red}(K_a) is a dimensionless quantity with the understanding that the standard state concentration of all species is 1M.

The values of the ionization constants of some selected weak acids are given in Table 7.6. The color{red}(pH) scale for the hydrogen ion concentration has been so useful that besides color{red}(pK_w), it has been extended to other species and quantities. Thus, we have:

color{red}(pK_a = -log(K_a)) .........(7.31)

Knowing the ionization constant, color{red}(K_a) of an acid and its initial concentration, color{red}(c), it is possible to calculate the equilibrium concentration of all species and also the degree of ionization of the acid and the color{red}(pH) of the solution.

=> A general step-wise approach can be adopted to evaluate the color{red}(pH) of the weak electrolyte as follows:

Step 1. The species present before dissociation are identified as Brönsted-Lowry acids / bases.

Step 2. Balanced equations for all possible reactions i.e., with a species acting both as acid as well as base are written.

Step 3. The reaction with the higher color{red}(K_a) is identified as the primary reaction whilst the other is a subsidiary reaction.

Step 4. Enlist in a tabular form the following values for each of the species in the primary reaction

color{red}(a) Initial concentration, color{red}(c).

color{red}(b) Change in concentration on proceeding to equilibrium in terms of color{red}(α), degree of ionization.

color{red}(c) Equilibrium concentration.

Step 5. Substitute equilibrium concentrations into equilibrium constant equation for principal reaction and solve for color{red}(α).

Step 6. Calculate the concentration of species in principal reaction.

Step 7. Calculate color{red}(pH = – log[H_3O^+])

Q 3180401317

The ionization constant of HF is 3.2 × 10^(–4). Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H_3O^+, F^– text(and) \ \ HF) in the solution and its pH.

Solution:

The following proton transfer reactions are possible :
1) HF + H_2O ⇌ H_3O^+ + F^–

K_a = 3.2 × 10^(–4)
2) H_2O + H_2O ⇌ H_3O^+ + OH^–

K_w = 1.0 × 10^(–14)
As K_a > > K_w, [1] is the principle reaction.

tt((HF , + , H_2O , ⇌ , H_3O^+ , +, F^-) , ("Initial concentration (M)" , , , , , , ) , ( 0.02 , , , , 0 , , 0) , ("Change (M)" , , , , , , ) , ( –0.02α , , , , +0.02α , , +0.02α) , (" Equilibrium concentration (M)" , , , , , , ) , ( 0.02 – 0.02 α , , , , 0.02 α , , 0.02α))
Substituting equilibrium concentrations in the equilibrium reaction for principal

reaction gives:
K_a = (0.02α)^2 // (0.02 – 0.02α)
= 0.02 α^2 // (1 –α) = 3.2 × 10^(–4)
We obtain the following quadratic equation:
α^2 + 1.6 × 10^(–2)α – 1.6 × 10^(–2) = 0
The quadratic equation in α can be solved and the two values of the roots are:
α = + 0.12 and – 0.12
The negative root is not acceptable and hence,
α = 0.12
This means that the degree of ionization, α = 0.12, then equilibrium concentrations of other species viz., HF, F^– and H_3O^+ are given by:

[H_3O^+] = [F^–] = cα = 0.02 × 0.12
= 2.4 × 10^(–3) M
[HF] = c(1 – α) = 0.02 (1 – 0.12)
= 17.6 × 10^(-3) M
pH = – log[H^+] = –log(2.4 × 10^(–3)) = 2.62
Q 3120601511

The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of species H^+ , A^– and HA at equilibrium. Also, determine the value of K_a and pK_a of the monobasic acid.

Solution:

pH = – log [H^+]

Therefore, [H^+] = 10^(-pH) = 10^(-4.50)

 = 3.16 xx 10^(-5)

[H^+] = [A^-] = 3.16xx10^(-5)

Thus, K_a = [H^+ ] [A^(-) ] //[HA]

[HA]_text(eqbm) = 0.1-(3.16 xx 10^(-5)) approx 0.1

K_a = (3.16xx10^(-5) )^2 // 0.1 = 1.0xx 10^(-8)

pK_a = - log (10^(-8)) = 8

Alternatively, “Percent dissociation” is another useful method for measure of strength of a weak acid and is given as:
Percent dissociation

 = [HA]_text(dissociated) // [HA]_text(initial) xx 100 %
Q 3170601516

Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization constant of the acid is 2.5 × 10^(–5) . Determine the percent dissociation of HOCl.

Solution:

tt((HOCl(aq) , +, H_2O (l), ⇌ , H_3O^+(aq), +, ClO^–(aq)) , ("Initial concentration (M)" , , , , , , ) , ( 0.08 , , , , 0 , , 0) , ("Change to reach equilibrium concentration (M)" , , , , , , ) , (-x , , , , +x , , +x) , ( "equilibrium concentartion (M)" , , , , , ,) , ( 0.08-x , , , , x , , x))

K_a = { [H_3O^+ ][ ClO^- ] // [HOCl ] }

As x < < 0.08, therefore 0.08 – x approx 0.08
x^2 / 0.08 = 2.5 × 10^(–5)

x^2 = 2.0 × 10^(–6), thus, x = 1.41 × 10^(–3)

[H^+] = 1.41 × 10^(–3) M.

Therefore, Percent dissociation

= { [HOCl]_text(dissociated) // [ HOCl]_text(initial) } xx 100

 = 1.41 xx 10^(-3) xx 10^2 //0.08 = 1.76 %

pH = -log ( 1.41xx10^(-3)) = 2.85

### Ionization of Weak Bases

=> The ionization of base color{red}(MOH) can be represented by equation:

color{red}(MOH (aq) ⇌ M^(+) (aq) + OH^(-) (aq))

In a weak base there is partial ionization of color{red}(MOH) into color{red}(M^+) and color{red}(OH^–), the case is similar to that of acid-dissociation equilibrium. The equilibrium constant for base ionization is called base ionization constant and is represented by color{red}(K_b). It can be expressed in terms of concentration in molarity of various species in equilibrium by the following equation:

color{red}(K_b = [M^+] [OH] // [MOH]) ...........(7.33)

Alternatively, if color{red}(c =) initial concentration of base and color{red}(α) = degree of ionization of base i.e. the extent to which the base ionizes. When equilibrium is reached, the equilibrium constant can be written as :

color{red}(K_c = ( c alpha)^2 //c (1-alpha) = c alpha^2//(1-alpha))

=> The values of the ionization constants of some selected weak bases, color{red}(K_b) are given in Table 7.7.

=> Many organic compounds like amines are weak bases. Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by another group. For example, methylamine, codeine, quinine and nicotine all behave as very weak bases due to their very small color{red}(K_b). Ammonia produces color{red}(OH^–) in aqueous solution :

color{red}(NH_3(aq) + H_2O(l) ⇌ NH_4^+ (aq) + OH^– (aq))

The color{red}(pH) scale for the hydrogen ion concentration has been extended to get:

color{red}(pK_b = - log (K_b)) ...............(7.34)

Q 3140701613

The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant K_b and pK_b.

Solution:

NH_2NH_2 + H_2O ⇌ NH_2NH_3^+ + OH^–
From the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:

[H^+] = antilog (–pH) = antilog (–9.7) = 1.67 ×10^(–10)

[OH^–] = K_w // [H^+] = 1 × 10^(–14) // 1.67 × 10^(–10)

= 5.98 × 10^(–5)

The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004M.

Thus,
K_b = [NH_2NH_3^+][OH^–] // [NH_2NH_2]
= (5.98 × 10^(–5))^2 / 0.004 = 8.96 × 10^(–7)

pK_b = –logK_b = –log(8.96 × 10^(–7)) = 6.04.
Q 3130801712

Calculate the pH of the solution in which 0.2M NH_4Cl and 0.1M NH_3 are present. The pK_b of ammonia solution is 4.75.

Solution:

NH_3 + H_2O ⇌ NH_4^+ + OH^–

The ionization constant of NH_3,K_b = antilog (–pK_b) i.e.

K_b = 10^(–4.75) = 1.77 × 10^(–5) M

 tt((NH_3 , +, H_2O , ⇌ , NH_4^(+) , + , OH^(–)) , ("Initial concentration (M)" , , , , , , ) , (0.1,
, , , 0.20 , , 0) ,( "Change to reach equilibrium (M)" , , , , , , ) , (–x , , , , +x , , +x) ,( "At equilibrium (M)" , , , , , ,) , (0.10 – x , , , , 0.20 + x , , x))
K_b = [NH_4^+][OH^–] // [NH_3]

= (0.20 + x)(x) /((0.1 – x)) = 1.77 × 10^(–5)

As K_b is small, we can neglect x in comparison to 0.1M and 0.2M. Thus,

[OH^–] = x = 0.88 × 10^(–5)

Therefore, [H^+] = 1.12 × 10^(–9)

pH = – log[H^+] = 8.95.

### Relation between K_a and K_b

=> As seen, color{red}(K_a) and color{red}(K_b) represent the strength of an acid and a base, respectively. In case of a conjugate acid-base pair, they are related in a simple manner so that if one is known, the other can be deduced. Considering the example of color{red}(NH_4^+) and color{red}(NH_3)

we see,

color{red}(NH_4^+ (aq) +H_2O (l) ⇌ H_3O^+ (aq) + NH_3(aq))

color{red}(K_a = [H_3O^+][ NH_3] // [NH_4^+] = 5.6 × 10^(–10))

color{red}(NH_3(aq) + H_2O(l) ⇌ NH_4^+(aq) + OH^(–)(aq))

color{red}(K_b =[ NH_4+][ OH^–] // NH_3 = 1.8 × 10^(–5))

Net: color{red}(2 H_2O(l) ⇌ H_3O^+(aq) + OH^–(aq))

color{red}(K_w = [H_3O^+][ OH^– ] = 1.0 × 10^(–14) M)

Where, color{red}(K_a) represents the strength of color{red}(NH_4^+) as an acid and color{red}(K_b) represents the strength of color{red}(NH_3) as a base.

It can be seen from the net reaction that the equilibrium constant is equal to the product of equilibrium constants color{red}(K_a) and color{red}(K_b) for the reactions added. Thus,

color{red}(K_a × K_b = {[H_3O^+][ NH_3] // [NH_4^+ ]} × {[NH_4^+ ][ OH^–] // [NH_3]})

color{red}(= [ H_3 O^+ ] // [OH] = K_w)

color{red}(= (5.6x10^(–10)) × (1.8 × 10^(–5)) = 1.0 × 10^(–14) M)

This can be extended to make a generalisation. The equilibrium constant for a net reaction obtained after adding two (or more) reactions equals the product of the equilibrium constants for individual reactions:

color{red}(K_(NET) = K_1 xx K_2 xx ...............) (3.35)

Similarly, in case of a conjugate acid-base pair

color{red}(K_a xx K_b = K_w) ..................(7.36)

Knowing one, the other can be obtained. It should be noted that a strong acid will have a weak conjugate base and vice-versa.

Alternatively, the above expression color{red}(K_w = K_a × K_b), can also be obtained by considering the base-dissociation equilibrium
reaction:

color{red}(B (aq) + H_2O (l) ⇌ BH^+(aq) + OH^–(aq))

color{red}(K_b = [BH^+] [OH^-] // [B])

As the concentration of water remains constant it has been omitted from the denominator and incorporated within the
dissociation constant. Then multiplying and dividing the above expression by color{red}([H^+]), we get:

color{red}(K_b = [ BH^+] [OH^-] [H^+] // [B] [H^+])

color{red}( = { [OH^-][H^+]}{[BH^+]//[B][H^+]})

 color{red}(= K_w / K_a)

or color{red}(K_a xx K_b = K_w)

It may be noted that if we take negative logarithm of both sides of the equation, then color{red}(pK) values of the conjugate acid and base are related to each other by the equation:

color{red}(pK_a + pK_b = pK_w = 14) (at 298K)

Q 3110001819

Determine the degree of ionization and pH of a 0.05M of ammonia solution. The ionization constant of ammonia can be taken from Table Also, calculate the ionization constant of the conjugate acid of ammonia.

Solution:

The ionization of NH_3 in water is represented by equation:

NH_3 + H_2O ⇌ NH_4^+ + OH^–

We use equation (7.33) to calculate hydroxyl ion concentration,

[OH^–] = c α = 0.05 α
K_b = (0.05 α^2) / (1 – α)
The value of α is small, therefore the quadratic equation can be simplified by neglecting α in comparison to 1 in the denominator on right hand side of the equation,

Thus,
K_b = c α^2 or alpha = √ (1.77 × 10–5 / 0.05)
= 0.018.
[OH^–] = c α = 0.05 × 0.018 = 9.4 × 10^(–4)M.
[H^+] = K_w / [OH^–] = 10^(–14) // (9.4 × 10^(–4))
= 1.06 × 10^(–11)
pH = –log(1.06 × 10^(–11)) = 10.97.
Now, using the relation for conjugate acid-base pair,
K_a × K_b = K_w
using the value of K_b of NH_3 from Table
We can determine the concentration of conjugate acid NH_4^+
K_a = K_w / K_b = 10^(–14) / 1.77 × 10^(–5)
= 5.64 × 10^(–10).

### Di- and Polybasic Acids and Di- and Polyacidic Bases

=> Some of the acids like oxalic acid, sulphuric acid and phosphoric acids have more than one ionizable proton per molecule of the acid. Such acids are known as polybasic or polyprotic acids.

The ionization reactions for example for a dibasic acid color{red}(H_2X) are represented by the equations :

color{red}(H_2X (aq) ⇌ H^+(aq) + HX^–(aq))

color{red}(HX^–(aq) ⇌ H^+(aq) + X^(2–)(aq))
And the corresponding equilibrium constants are given below:
color{red}(K_(a_1) = {[H^+][HX^–]} // [H_2X]) and color{red}(K_(a_2) = {[H^+][X^(2-)]} // [HX^-])

Here,  color{red}(K_(a_1)) and color{red}(K_(a_2)) are called the first and second ionization constants respectively of the acid color{red}(H_2X). Similarly, for tribasic acids like color{red}(H_3PO_4) we have three ionization constants.

=> The values of the ionization constants for some common polyprotic acids are given in Table 7.8.

=> It can be seen that higher order ionization constants color{red}(( K_(a_2) ,K_(a_3)) are smaller than the lower order ionization constant color{red}( K_(a_1) ) of a polyprotic acid. The reason for this is that it is more difficult to remove a positively charged proton from a negative ion due to electrostatic forces. This can be seen in the case of removing a proton from the uncharged color{red}(H_2CO_3) as compared from a negatively charged color{red}(HCO_3^–). Similarly, it is more difficult to remove a proton from a doubly charged color{red}(HPO_4^(2–)) anion as compared to color{red}(H_2PO_4^–).

=> Polyprotic acid solutions contain a mixture of acids like color{red}(H_2A, HA^–) and color{red}(A^(2–)) in case of a diprotic acid. color{red}(H_2A) being a strong acid, the primary reaction involves the dissociation of color{red}(H_2 A), and color{red}(H_3O^+) in the solution comes mainly from the first dissociation step.

### Factors Affecting Acid Strength

=> Having discussed quantitatively the strengths of acids and bases, we come to a stage where we can calculate the color{red}(pH) of a given acid solution. But, the curiosity rises about why should some acids be stronger than others? What factors are responsible for making them stronger? The answer lies in its being a complex phenomenon. But, broadly speaking we can say that the extent of dissociation of an acid depends on the strength and polarity of the color{red}(H-A) bond.

=> In general, when strength of color{red}(H-A) bond decreases, that is, the energy required to break the bond decreases, color{red}(HA) becomes a stronger acid. Also, when the color{red}(H-A) bond becomes more polar i.e., the electronegativity difference between the atoms color{red}(H )and color{red}(A) increases and there is marked charge separation, cleavage of the bond becomes easier thereby increasing the acidity.

=> But it should be noted that while comparing elements in the same group of the periodic table, color{red}(H-A) bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, color{red}(H-A) bond strength decreases and so the acid strength increases. For example,

color{green}("Size increases")

color{red}(underset(→) overset(→)( HF < < HCl < < HBr < < HI ))

color{green}("Acid strength increases")
Similarly, color{red}(H_2S) is stronger acid than color{red}(H_2O).

But, when we discuss elements in the same row of the periodic table, color{red}(H-A) bond polarity becomes the deciding factor for determining the acid strength. As the electronegativity of A increases, the strength of the acid also increases. For example,

color{green}("Electronegativity of A increases ")

color{red}(underset(→) overset(→) (CH_4 < NH_3 < H_2O < HF))

color{green}("Acid strength increases")