Equilibria for acids and bases
Acids and bases
Dissociation of acids
When acids release H+, we talk about dissociation. Dissociation is not restricted to acids, it just means that a part of the chemical compound is separated from the chemical compound.
Consider a monovalent (monoprotic) model acid, which we will call HB. When mixed with water, it will dissociate an give off an H+:
HB(aq)
H+(aq) + B−(aq)
The part left behind, which is B−, is called the conjugate base or corresponding base. The reaction is always an equilibrium, as written here. However, when it comes to the acids called strong acids, the equilibrium point is so far to the right, that we normally write this as an irreversible reaction. But, it is a reversibel process. A strong acid like that could be hydrochloric acid, which is written like this:
HCl(aq)
H+(aq) + Cl−(aq)
When it comes to polyvalent acids (polyprotic acids), dissociation becomes a bit more complicated. Consider a divalent model acid, which we will call H2B. When mixed with water, it will in theory dissociate and release 2 H+:
H2B(aq) 2 H+(aq) + B2−(aq)
This, however, is not always the case. H+ is released stepwise, so the 2 H+ are removed one at a time, like this:
To obtain the total reaction, with full dissociation, both H2B and HB− have to be strong acids. Otherwise you will only have partial dissociation. If the dissociation is only partial, you get a system that is a bit more messy to work with, as we shall see further down. Had it been a trivalent acid, we would have had a reaction step 3, and a step 4, had it been a tetravalent acid, and so forth.
Sulfuric acid is one of the strong acids, where you cannot see the stepwise dissociation, so this is the reaction you see:
H2SO4(aq) 2 H+(aq) + SO42−(aq)
Because the equilibrium is shifted entirely to the right, like HCl, it also looks as if the reaction is irreversible. BUT: The dissociation occurs in two steps and it is an equilibrium.
Moving away from the strong acids, what happens is, that you only get a partial dissociation. Now we have an equilibrium like the ones we known from other equilibria. This could be acetic acid:
CH3COOH(aq) H+(aq) + CH3COO−(aq)
For this we have the equilibrium constant, Ka. This is the acid ionization constant. So the acid ionization constant is the same as the equilibrium constant, we just use a different name when working with acids.
Instead of Ka, we traditionally use the value pKa, which is calculated in the same manner as pH, i.e.:
pKa = -log(Ka)
pKa is an experimental value, which we will see, when we get to titration curves, so it has a practical application, and is not just some math we add to acid/base equilibria.
Looking at how Ka is calculated from the equilibrium, we also see why we do not have Ka values for strong acids. As the concentration of acid on the left side approaches zero, we get a mathematical expression where we are dividing by zero, which is impossible. In practice this is solved by defining the pKa value for strong acids as zero.
The Ka and pKa values are used for dividing the acids according the strength:
In some cultures and educational systems you may encounter acids only being divided into strong and weak acids.
When we get to the polyvalent acids that are not strong, it becomes a bit complicated. Dissociation happens stepwise, as always with polyvalent acids, but what will happen here is that not all the H2B og HB− dissociates, so in the solution you have a mixture of H2B, HB−, H+ and B2−.
In principle it looks like this:
2 H2B(aq) HB−(aq) + 3 H+(aq) + B2−(aq)
To make it more understandable and reflect the actual conditions, which is the whole point with the reaction equations, you use this representation:
H2B(aq) H+(aq) + HB−(aq) 2 H+(aq) + B2−(aq)
You have now shown that the equilibrium is actually two equilibria.
Because we are working with multiple equilibria, we also have multiple equilibrium constants for polyvalent acids. These are numbered 1, 2, etc., so if we have two steps
this will result in two equilibrium constants:
The two Ka values may be the same, but they do not have to be. It is not a matter of symmetry that determines whether the two Ka values are the same, as could be expected. An acid like oxalic acid, which has the formula HOOC-COOH, have pKa1 = 1.23 and pKa2 = 4.19.
For the acid we get an overall Ka by multiplying the Ka values for the two reaction steps:
Consider a monovalent (monoprotic) model acid, which we will call HB. When mixed with water, it will dissociate an give off an H+:
HB(aq)
H+(aq) + B−(aq)The part left behind, which is B−, is called the conjugate base or corresponding base. The reaction is always an equilibrium, as written here. However, when it comes to the acids called strong acids, the equilibrium point is so far to the right, that we normally write this as an irreversible reaction. But, it is a reversibel process. A strong acid like that could be hydrochloric acid, which is written like this:
HCl(aq)
H+(aq) + Cl−(aq)When it comes to polyvalent acids (polyprotic acids), dissociation becomes a bit more complicated. Consider a divalent model acid, which we will call H2B. When mixed with water, it will in theory dissociate and release 2 H+:
H2B(aq)
2 H+(aq) + B2−(aq)This, however, is not always the case. H+ is released stepwise, so the 2 H+ are removed one at a time, like this:
| Reaction step 1: | H2B(aq) H+(aq) + HB−(aq) |
| Reaction step 2: | HB−(aq) H+(aq) + B2−(aq) |
| Total: | H2B(aq) 2 H+(aq) + B2−(aq) |
To obtain the total reaction, with full dissociation, both H2B and HB− have to be strong acids. Otherwise you will only have partial dissociation. If the dissociation is only partial, you get a system that is a bit more messy to work with, as we shall see further down. Had it been a trivalent acid, we would have had a reaction step 3, and a step 4, had it been a tetravalent acid, and so forth.
Sulfuric acid is one of the strong acids, where you cannot see the stepwise dissociation, so this is the reaction you see:
H2SO4(aq)
2 H+(aq) + SO42−(aq)Because the equilibrium is shifted entirely to the right, like HCl, it also looks as if the reaction is irreversible. BUT: The dissociation occurs in two steps and it is an equilibrium.
Moving away from the strong acids, what happens is, that you only get a partial dissociation. Now we have an equilibrium like the ones we known from other equilibria. This could be acetic acid:
CH3COOH(aq)
H+(aq) + CH3COO−(aq)For this we have the equilibrium constant, Ka. This is the acid ionization constant. So the acid ionization constant is the same as the equilibrium constant, we just use a different name when working with acids.
| Ka = | [H+] · [CH3COO−] |
| [CH3COOH] |
Instead of Ka, we traditionally use the value pKa, which is calculated in the same manner as pH, i.e.:
pKa = -log(Ka)
pKa is an experimental value, which we will see, when we get to titration curves, so it has a practical application, and is not just some math we add to acid/base equilibria.
Looking at how Ka is calculated from the equilibrium, we also see why we do not have Ka values for strong acids. As the concentration of acid on the left side approaches zero, we get a mathematical expression where we are dividing by zero, which is impossible. In practice this is solved by defining the pKa value for strong acids as zero.
The Ka and pKa values are used for dividing the acids according the strength:
| pKa ≤ 0 | Strong acid |
| 0 < pKa ≤ 4 | Medium strength acid |
| 4 < pKa ≤ 10 | Weak acid |
| pKa > 10 | Very weak acid |
In some cultures and educational systems you may encounter acids only being divided into strong and weak acids.
When we get to the polyvalent acids that are not strong, it becomes a bit complicated. Dissociation happens stepwise, as always with polyvalent acids, but what will happen here is that not all the H2B og HB− dissociates, so in the solution you have a mixture of H2B, HB−, H+ and B2−.
In principle it looks like this:
2 H2B(aq)
HB−(aq) + 3 H+(aq) + B2−(aq)To make it more understandable and reflect the actual conditions, which is the whole point with the reaction equations, you use this representation:
H2B(aq)
H+(aq) + HB−(aq) 2 H+(aq) + B2−(aq)You have now shown that the equilibrium is actually two equilibria.
Because we are working with multiple equilibria, we also have multiple equilibrium constants for polyvalent acids. These are numbered 1, 2, etc., so if we have two steps
| Reaction step 1: | H2B(aq) H+(aq) + HB−(aq) |
| Reaction step 2: | HB−(aq) H+(aq) + B2−(aq) |
this will result in two equilibrium constants:
| Ka1 = | [H+] · [HB−] |
| [H2B] |
| Ka2 = | [H+] · [B2−] |
| [HB−] |
The two Ka values may be the same, but they do not have to be. It is not a matter of symmetry that determines whether the two Ka values are the same, as could be expected. An acid like oxalic acid, which has the formula HOOC-COOH, have pKa1 = 1.23 and pKa2 = 4.19.
For the acid we get an overall Ka by multiplying the Ka values for the two reaction steps:
| Ka = Ka1 · Ka2 = | [H+] · [HB−] | · | [H+] · [B2−] | = | [H+]2 · [B2−] |
| [H2B] | [HB−] | [H2B] |
Bases uptake of H+
Bases are the opposite of acids, i.e. they bind H+ instead of releasing them. Most people associate bases with OH− and the reaction
OH−(aq) + H+(aq) H2O(l)
because that is what they learned in school.
It is not wrong, per se, but it is not entirely correct either. First of all, the reaction really looks like this:
OH−(aq) + H+(aq) H2O(l)
It is an equilibrium. It is ALWAYS an equilibrium! The equilibrium is shifted so far to the right that we think of it as an irreversible reaction as written in the previous reaction equation.
That it is an equilibrium, and that it does no have to be OH− can be seen with ammonia, NH3, which is a weak base. Here the reaction looks like this:
NH3(aq) + H+(aq) NH4+(aq)
Since a base does not have to be OH−, how do you determine pOH and thereby pH for a solution? The answer to that lies in water's ability to act as an ampholyte. Ammonia reacts with water, giving you this equilibrium:
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
Therefore, a base in aqueous solution can provide OH− ions for measuring pOH, though the base itself does not contain any OH−.
Similar to acids, we have an equilibrium and an equilibrium constant that goes with it, which we call Kb for bases. This is the base ionization constant. The base ionization constant is thus, just like we saw for the acids, the same as an equilibrium constant, we just use a different name whan talking about bases.
For ammonia in water, the formula for the base ionization constant looks like this:
Notice that the concentration of water is not included in the calculation. The concentration of water in an aqueous solution is defined as 1, as it is normally done for equilibria.
Instead of Kb, it is common practice to use the value called pKb, which calculated in the same manner as pH and pKa, i.e.:
pKb = -log(Kb)
pKb is an experimental value, which we will see, when we get to titration curves, so it has a practical application, and is not just some math we add to acid/base equilibria.
The Kb and pKb values are used for dividing the bases according the strength:
In some cultures and educational systems you may encounter bases only being divided into strong and weak bases.
Like the acids, bases can be polyvalent, and like the acids, we are dealing with equilibria and the uptake of H+ happens stepwise just like the release of H+ for acids. If we are working with strong bases like Ca(OH)2, you cannot see a difference between the pKb values for the two hydroxide ions, whereas ethane-1,2-diamine, which has the formula H2N-CH2-CH2-NH2, has pKb1=3.288 and pKb2=6.436. Just like we saw with the acids, there is no relation between pKb values and whether the bases are symmetrical or not.
A strong divalent base will thus take up 2 H+, in a manner that, at first sight, looks like this:
B(aq) + 2 H+(aq) H2B2+(aq)
In reality, it goes through the same procedure as the other bases. H+ is taken up stepwise, so the 2 H+ are taken up one at a time, like this:
A more accurate description of the two equilibria is this:
B(aq) + 2 H+ H+ + HB+(aq) H2B2+(aq)
The two steps in the reaction gives us two equilibria constants:
For the base we get an overall Kb by multiplying the Kb values for the two steps:
OH−(aq) + H+(aq)
H2O(l)because that is what they learned in school.
It is not wrong, per se, but it is not entirely correct either. First of all, the reaction really looks like this:
OH−(aq) + H+(aq)
H2O(l)It is an equilibrium. It is ALWAYS an equilibrium! The equilibrium is shifted so far to the right that we think of it as an irreversible reaction as written in the previous reaction equation.
That it is an equilibrium, and that it does no have to be OH− can be seen with ammonia, NH3, which is a weak base. Here the reaction looks like this:
NH3(aq) + H+(aq)
NH4+(aq)Since a base does not have to be OH−, how do you determine pOH and thereby pH for a solution? The answer to that lies in water's ability to act as an ampholyte. Ammonia reacts with water, giving you this equilibrium:
NH3(aq) + H2O(l)
NH4+(aq) + OH−(aq)Therefore, a base in aqueous solution can provide OH− ions for measuring pOH, though the base itself does not contain any OH−.
Similar to acids, we have an equilibrium and an equilibrium constant that goes with it, which we call Kb for bases. This is the base ionization constant. The base ionization constant is thus, just like we saw for the acids, the same as an equilibrium constant, we just use a different name whan talking about bases.
For ammonia in water, the formula for the base ionization constant looks like this:
| Kb = | [NH4+] · [OH−] |
| [NH3] |
Notice that the concentration of water is not included in the calculation. The concentration of water in an aqueous solution is defined as 1, as it is normally done for equilibria.
Instead of Kb, it is common practice to use the value called pKb, which calculated in the same manner as pH and pKa, i.e.:
pKb = -log(Kb)
pKb is an experimental value, which we will see, when we get to titration curves, so it has a practical application, and is not just some math we add to acid/base equilibria.
The Kb and pKb values are used for dividing the bases according the strength:
| pKb ≤ 0 | Strong base |
| 0 < pKb ≤ 4 | Medium strength base |
| 4 < pKb ≤ 10 | Weak base |
| pKb > 10 | Very weak base |
In some cultures and educational systems you may encounter bases only being divided into strong and weak bases.
Like the acids, bases can be polyvalent, and like the acids, we are dealing with equilibria and the uptake of H+ happens stepwise just like the release of H+ for acids. If we are working with strong bases like Ca(OH)2, you cannot see a difference between the pKb values for the two hydroxide ions, whereas ethane-1,2-diamine, which has the formula H2N-CH2-CH2-NH2, has pKb1=3.288 and pKb2=6.436. Just like we saw with the acids, there is no relation between pKb values and whether the bases are symmetrical or not.
A strong divalent base will thus take up 2 H+, in a manner that, at first sight, looks like this:
B(aq) + 2 H+(aq)
H2B2+(aq)In reality, it goes through the same procedure as the other bases. H+ is taken up stepwise, so the 2 H+ are taken up one at a time, like this:
| Reaktion step 1: | H+(aq) + B(aq) HB+(aq) |
| Reaktion step 2: | H+(aq) + HB+(aq) H2B2+(aq) |
| Total: | 2 H+(aq) + B(aq) H2B2+(aq) |
A more accurate description of the two equilibria is this:
B(aq) + 2 H+
H+ + HB+(aq) H2B2+(aq)The two steps in the reaction gives us two equilibria constants:
| Kb1 = | [HB+] |
| [H+] · [B] |
| Kb2 = | [H2B2+] |
| [H+] · [HB+] |
For the base we get an overall Kb by multiplying the Kb values for the two steps:
| Kb = Kb1 · Kb2 = | [HB+] | · | [H2B2+] | = | [H2B2+] |
| [H+] · [B] | [H+] · [HB+] | [H+]2 · [B] |
Fractional concentration, α
Because there is a difference in degrees of dissociation for the various acids and bases, you have the concept fractional concentration, α, which is the the fraction of the molecules that have dissociated. For an acid and a base at 100 % dissociation, α = 1, at 50 % dissociation α = 0,5 etc.
For the fictional acid HB, dissociating, we get:
HB(aq) + H2O(l) H3O+(aq) + B−(aq)
For the fictional base, B, taking up an H+, we get:
B(aq) + H2O(l) BH+(aq) + OH−(aq)
The fractional concentration is used in connection with Ostwald's dilution law, which is a rewriting of the equilibrium equation for medium strength and weak acids.
For the acid HB, dissociating in water:
HB(aq) + H2O(l) H3O+(aq) + B−(aq)
For the fictional acid HB, dissociating, we get:
HB(aq) + H2O(l)
H3O+(aq) + B−(aq)| α = | [H3O+] | = | [B−] |
| c(HB) | c(HB) |
For the fictional base, B, taking up an H+, we get:
B(aq) + H2O(l)
BH+(aq) + OH−(aq)| α = | [HB+] | = | [OH−] |
| c(B) | c(B) |
The fractional concentration is used in connection with Ostwald's dilution law, which is a rewriting of the equilibrium equation for medium strength and weak acids.
For the acid HB, dissociating in water:
HB(aq) + H2O(l)
H3O+(aq) + B−(aq)| Ka = | [H3O+] · [B−] | = | [H3O+]2 | , standard equilibrium equation |
| [HB] | c(HB) - [H3O+] |
⇕
| Ka = | c(HB)2 · α2 | = c(HB) · | α2 | , Ostwald's dilution law |
| c(HB) - c(HB) · α | 1 - α |