CHM 1046
General Chemistry II
Dr. Michael Blaber
Acid-Base Equilibria
Weak Acids
Most acids are weak acids - they only partially ionize in aqueous solution
- If weak acids only partially ionize, then the [H+] that is produced is less than the total concentration of acid added to the solution (note: a strong acid dissociates completely and therefore the [H+] is equal to the concentration of acid added to a solution).
- Since the [H+] produced by the ionization of weak acids is less than the concentration of acid added to the solution, you cannot calculate the pH directly from the concentration of acid (again, this would only be possible with strong acids that dissociate completely).
- The extent of their ionization can be expressed by using the equilibrium constant and equilibrium equation:
HA(aq) ó H+(aq) + A-(aq)
- Ka is the equilibrium constant for the ionization of an acid, often referred to as the acid dissociation constant
- What happens if the hydrated proton is represented by H3O+(aq) instead of H+(aq)?
HA(aq) + H2O(l) ó H3O+(aq) + A-(aq)
- Here the H2O is in liquid form, and therefore, its concentration never changes (and we can consider it to be a constant in the expression
- And Ka is the same expression in both cases
Weak acids are often composed entirely of C, H and O atoms
- H atoms are often bonded to either C or O atoms
- C electronegativity = 2.5
- H electronegativity = 2.1
- O electronegativity = 3.5
- The H atoms bonded to C do not ionize (not a polar bond)
- It is one or more H atoms bonded to an O
atom that can ionize in water (due to difference in electronegativity)
- Weak acids are often carboxylic acids:
- The magnitude of Ka is an indication of the tendency of the H atom to ionize
- If Ka is large, it means that the concentration of H+ ion is large, and this means the acid likes to ionize (the more it ionizes, the stronger the acid)
- Thus, the larger the value of Ka, the stronger the acid
Phenol is a very weak acid, its Ka = 1.3 x 10-10
Hydrofluoric acid is a weak acid, but is considered to be one of the strongest weak acids (if that makes sense, sort of like "Jumbo Shrimp"), its Ka = 6.8 x 10-4
Values for Ka for weak acids is typically less than 1 x 10-3
The pH of a 0.25M solution of acetic acid (HC2H3O2) is found to be 2.68. What is Ka for this solution and what % of the acid ionized?
pH = -log[H+]
2.68 = -log[H+]
-2.68 = log[H+]
10-2.68 = [H+]
2.09 x 10-3M = [H+]
Balanced equation and equilibrium equation:
HC2H3O2(aq)
ó H+(aq) + C2H3O2-(aq)
Ka = [H+][C2H3O2-] / [HC2H3O2]
- From the stoichiometry, 2.09 x 10-3M of H+ ions would be produced if 2.09 x 10-3M of acetic acid decomposed.
- Likewise, 2.09 x 10-3M of conjugate base (C2H3O2-) would be produced.
- If 2.09 x 10-3M of acetic acid decomposed, then the amount of acetic acid at equilibrium would be equal to (what we started with - amount that decomposed) = (0.25M - 2.09 x 10-3M) = 0.248M
Ka = (2.09 x 10-3) * (2.09 x 10-3) / (0.248)
Ka = 1.76 x 10-5
- The % of the acid that ionized is = (the concentration of [H+] at equilibrium) / (the initial concentration of the acid) * 100
(2.09 x 10-3M / 0.25M) * 100 = 0.84% (i.e. less than 1% of the acid ionized - truly a weak acid)
Calculating pH for Solutions of Weak Acids
Knowing the starting concentration of a weak acid and the value of the acid dissociation constant, Ka , we can calculate the resulting pH (i.e. [H+]) of the acid solution.
The Ka for the dissociation of hypochlorous acid is 3.0 x 10-8. What is the resulting pH of a 0.15M solution of hypochlorous acid?
- Begin by writing the balanced equation for the dissociation of the acid:
HClO(aq)
ó H+(aq) + ClO-(aq)
- Now you can write the equilibrium equation for the acid dissociation constant, Ka:
Ka = [H+][ClO-] / [HClO]
We know that we started with 0.15M of the acid, but we don't know exactly how much dissociates as it attains equilibrium. So, we will assign the variable
X as the amount that dissociates:
- From the stoichiometry, if
X moles of acid dissociate, then X moles of H+ are produced, and X moles of conjugate base (i.e. ClO-) are produced
We now have a description of the concentrations of the different components at equilibrium:
X
[ClO-] = X
[HClO] = (0.15 - X)
These equilibrium values can be plugged into the equilibrium equation:
Ka = 3.0 x 10-8 = [H+][ClO-] / [HClO] =
(X * X) / (0.15 - X)
Solving for X:
(3.0 x 10-8) * (0.15 - X) = X2
X2 + 3.0 x 10-8X - 4.50 x 10-9 = 0
- X can be solved using the quadratic equation
- This yields two possible solutions for X:
X = 6.71 x 10-5, or
X = -6.71 x 10-5
- Which is correct? Well, X is supposed to be the [H+], this number cannot be a negative value, so 6.71 x 10-5 is correct. The pH can be calculated from this value:
pH = -log[H+]
pH = -log(6.71 x 10-5)
pH = -(-4.17)
pH = 4.17
A potential shortcut to avoid having to use the quadratic equation for this type of problem:
If the initial concentration of weak acid is at least one-hundred fold greater than the value of Ka, then you can assume that the amount that dissociates in going to equilibrium is negligibly small compared to the initial concentration.
In this case, the term [INITIAL CONC OF ACID - X] is ~ [INITIAL CONC OF ACID]
Let's see if we can apply this shortcut to the above problem:
(initial conc)/Ka = (0.15) / (3.0 x 10-8) = 5.00 x 106
This number (5 million) is > 100
- therefore we can assume [initial conc of acid - amount that dissociates] ~ [initial conc]
Ka = 3.0 x 10-8 = [H+][ClO-] / [HClO] = (X * X) / (0.15 - X)
Using the assumption above:
(X * X) / (0.15 - X) ~ (X * X) / (0.15)
thus
3.0 x 10-8 = (X * X) / (0.15)
X2 = 4.5 x 10-9
X = 6.71 x 10-5
- Check above and you will see that this number agrees with the answer using the quadratic.
The quadratic answer is always accurate, the answer using the shortcut is approximate, but will be reasonably accurate if (initial conc of acid)/Ka > 100
For weak acids, the concentration of [H+] is typically only a small percentage of the total concentration of acid in solution.
- How does the percentage of acid molecules that ionize vary with the starting concentration of acid?
What is the % of Nitrous acid (HNO2) molecules that ionize in a 0.1M and 0.01M solution? (The Ka of nitrous acid is 4.5 x 10-4)
- The balanced equation for acid dissociation is:
HNO2(aq)
ó H+(aq) + NO2-(aq)
- The acid dissociation constant is defined as:
Ka = [H+][NO2-] / [HNO2]
For a given starting concentration of acid we can set X = the amount of acid that dissociates in attaining equilibrium. From the stoichiometry, X also equals the concentration of [H+] and [NO2-] at equilibrium. Thus:
Ka = 4.5 x 10-4 = [H+][NO2-] / [HNO2] = X2/(starting concentration - X)
4.5 x 10-4 = X2/(0.1 - X)
Solve for X using the quadratic equation yields:
X = 6.49 x 10-3 = amount of acid dissociated
% = (amount dissociated/initial conc) * 100
% = (6.49 x 10-3/0.1) * 100 = 6.49%
4.5 x 10-4 = X2/(0.01 - X)
Solve for X using the quadratic equation yields:
X = 1.91 x 10-3 = amount of acid dissociated
% = (amount dissociated/initial conc) * 100
% = (1.91 x 10-3/0.01) * 100 = 19.1%
There is a 3-fold increase in the number of acid particles that dissociate in response to a 10 fold dilution of the sample. Why is this?
A mathematical answer…
- The percent dissociation is proportional to [H+]/[HA]. From the equilibrium expression we have:
Ka = [H+][A-]/[HA]
Rearranging:
Ka/[A-] = [H+]/[HA]
Thus, if [A-] decreases (e.g. as a consequence of dilution) then a proportionally larger concentration of [H+]/[HA] should result
An answer based on Le Chatelier's principle…
- We have seen how Le Chatelier's principle can be applied to considerations of concentration, heat, and pressure (i.e. number of molecules). The number of molecules situation might be applied here.
- The dissociation equilibrium is an equilibrium that involves different numbers of particles for reactant versus product. A shift to the right increases the overall number of particles, a shift to the left decreases the overall number of particles.
- With regard to the number of particles, dilution would represent a perturbation of the equilibrium due to a net removal of particles (acid, proton and conjugate base). In this case we would predict that diluting the acid would cause the equilibrium to shift in favor of producing more particles (this would cause the reaction to be driven to the right and a greater percentage of acid molecules will dissociate at lower concentration)
Polyprotic Acids
Many acids have more than 1 ionizable H atom. Such atoms are known as polyprotic acids
- Sulfurous acid is a polyprotic acid:
H2SO3(aq) ó H+(aq) + HSO3-(aq) Ka1 = 1.7 x 10-2
HSO3(aq) ó H+(aq) + SO32-(aq) Ka2 = 6.4 x 10-8
- Ka1 is the equilibrium constant for ionization of the first proton
- Ka2 is the equilibrium constant for ionization of the second proton
Ka2 is much smaller than Ka1
Its more difficult to remove the second proton. This is because the second proton has to be removed from a negatively charged species (HSO3-) which will have an electrostatic attraction for this second proton
It is always easier to remove the first proton from a polyprotic acid (Ka values become successively smaller as protons are removed)
Usually Ka values for successive protons in polyprotic acids differ by a value of 103 or greater
- We can estimate pH values by considering only the Ka1 value for polyprotic acids
© 2000 Dr. Michael Blaber