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Comparing Concentrated and Dilute Solutions of a Weak Acid
Acetic acid (ethanoic acid), CH3COOH, is a weak acid, it only partially dissociates in water so undissociated (unionised) acetic acid molecules are in equilibrium with hydrogen ions (protons or hydrons), H+(aq), and acetate ions (ethanoate ions), CH3COO-(aq), as shown by the chemical equation given below:
| Dissociation of Acetic Acid (Ionisation of Ethanoic Acid) |
|---|
| Word equation: |
acetic acid
(ethanoic acid) |
⇋ |
hydrogen ions
(protons or hydrons) |
+ |
acetate ions
(ethanoate ions) |
|---|
| Balanced chemical equation: |
CH3COOH |
⇋ |
H+(aq) |
+ |
CH3COO-(aq) |
|---|
Imagine you have two reagent bottles, Bottle A and Bottle B, containing different concentrations of acetic acid, CH3COOH(aq), both at 25°C and 1 atm.
Bottle A contains 0.100 mol L-1 CH3COOH(aq) with a pH of 2.87
Bottle B contains 0.010 mol L-1 CH3COOH(aq) with a pH of 3.37
Let's compare these two bottles of acetic acid.
Bottle A contains a more concentrated solution of acetic acid than Bottle B.
| |
Bottle A |
|
Bottle A |
|---|
| [CH3COOH(aq)] (mol L-1) |
0.100 mol |
> |
0.010 |
|---|
| relative concentration |
more concentrated |
|
more dilute |
|---|
Notice that the pH of the more concentrated acetic acid solution in Bottle A is lower
| |
Bottle A |
|
Bottle A |
|---|
| [CH3COOH(aq)] (mol L-1) |
0.100 |
> |
0.010 |
|---|
| relative concentration |
more concentrated |
|
more dilute |
|---|
| pH |
2.87 |
< |
3.37 |
|---|
| relative pH |
lower |
|
higher |
|---|
Which tells us that the concentration of hydrogen ions, [H+(aq)], in Bottle A must be greater than it is in Bottle B:
| |
Bottle A |
|
Bottle A |
|---|
| [CH3COOH(aq)] (mol L-1) |
0.100 |
> |
0.010 |
|---|
| relative concentration |
more concentrated |
|
more dilute |
|---|
| pH |
2.87 |
< |
3.37 |
|---|
| relative pH |
lower |
|
higher |
|---|
| [H+(aq)] = 10-pH (mol L-1) |
1.35 × 10-3 |
> |
4.27 × 10-4 |
|---|
| relative [H+(aq)] |
higher |
|
lower |
|---|
We can also calculate the % dissociation (% ionisation) for the acetic acid solution in each bottle and compare the results:
| |
Bottle A |
|
Bottle A |
|---|
| [CH3COOH(aq)] (mol L-1) |
0.100 |
> |
0.010 |
|---|
| relative concentration |
more concentrated |
|
more dilute |
|---|
| pH |
2.87 |
< |
3.37 |
|---|
| relative pH |
lower |
|
higher |
|---|
| [H+(aq)] = 10-pH (mol L-1) |
1.35 × 10-3 |
> |
4.27 × 10-4 |
|---|
| relative [H+(aq)] |
higher |
|
lower |
|---|
| %dissociation = ([H+(aq)]/[CH3COOH(aq)]) × 100 |
1.35% |
< |
4.27% |
|---|
| relative %dissociation |
less |
|
greater |
|---|
and we find that the % dissociation of acetic acid in the more dilute solution (Bottle B) is greater than it is in the more concentrated solution (Bottle A).
This increased %dissocation of a weak acid on dilution is predicted using Le Chatelier's Principle.
Before dilution, molecules of undissociated acetic acid in solution were in equilibrium with hydrogen ions and acetate ions as shown below:
| solute |
+ |
solvent |
⇋ |
solution |
| acetic acid |
+ |
water |
⇋ |
hydrogen ions |
+ |
acetate ions |
| CH3COOH(aq) |
+ |
H2O(l) |
⇋ |
H3O+(aq) |
+ |
CH3COO-(aq) |
When we dilute the solution by adding more water, we perturb (or disturb) this equilibrium.
The equilibrium position shifts to the right, consuming more of the CH3COOH(aq) molecules to produce more H3O+(aq) (or H+(aq)) and more CH3COO-(aq).
Hence the number of CH3COOH(aq) molecules decreases and the number of H3O+(aq) (or H+(aq)) ions increases.
So the %dissociation (%ionisation) of the acid increases after dilution.
But remember, even though the number of hydrogen ions has increased, the volume of the solution has increased by an even greater amount after dilution.
Small increase in moles of H+(aq), larger increase in volume of solution.
So, the concentration of hydrogens ion in solution, the moles of hydrogen ions per litre of solution, has decreased after dilution.
The important things to remember when we dilute an aqueous solution of a weak acid at 25°C and 1 atm are that AFTER dilution the resultant solution has:
- a lower concentration of acid
- a lower concentration of hydrogen ions
- a higher pH
- a greater % dissociation (% ionisation)
Now we are ready to perform calculations to determine the pH of an aqueous solution of weak acid after it is diluted by adding more water.
Calculating the pH of an Aqueous Solution of a Weak Acid After Dilution
In our example above, Bottle A contained a more concentrated acetic acid solution than Bottle B.
Bottle A : 0.100 mol L-1 solution of acetic acid, CH3COOH(aq), with a pH of 2.87 (25°C, 1atm).
Bottle A : 0.010 mol L-1 solution of acetic acid, CH3COOH(aq), with a pH of 3.37 (25°C, 1atm).
If we wanted to prepare the more dilute acetic acid solution in Bottle B we have a couple of options. For example, we could:
- dissolve 0.010 mol of acetic acid in enough water to make 1 L of aqueous solution, [CH3COOH(aq)] = 0.010 mol L-1
- dilute 10 mL of the 0.100 mol L-1 solution in Bottle A with enough water to make 100.0 mL of solution (a 1 in 10 dilution, or, dilution by a factor of 10), [CH3COOH(aq)] = 0.010 mol L-1
Although each solution of 0.010 mol L-1 CH3COOH(aq) will have been prepared in different ways, they will both have:
- same [CH3COOH(aq)]
- same [CH3COO-(aq)]
- same [H+(aq)]
- same Ka (acid dissociation constant)
- same %dissociation (%ionisation)
- same pH
However, when the amount of solute we need to dissolve becomes too small to accurately measure, or the volume of solution becomes too large to be able to measure accurately, we prefer to prepare a more concentrated solution and then dilute that down to the desired concentration.
So let's calculate the pH of an acetic acid solution after dilution.
First, we pipette 10.00 mL of the 0.100 mol L-1 acetic acid solution into a 100.00 mL volumetric flask, then we add enough distilled water to make the solution up to the mark.
We will need to find the concentration of acetic acid in the resultant solution (solution after dilution).
We can use the "dilution formula" to calculate this:
c1V1 = c2V2
c1 = [CH3COOH(aq)(transferred)] = 0.100 mol L-1
V1 = V(transferred) = 10.00 mL = 10.00 mL ÷ 1000 mL/L = 0.0100 L
c2 = [CH3COOH(aq)(resultant)] = ? mol L-1
V2 = V(resultant) = 100.00 mL = 100.00 mL ÷ 1000 mL/L = 0.100 L
Substitute the known values into the equation:
c1V1 = c2V2
0.100 × 0.0100 = c2 × 0.100
0.00100 = c2 × 0.100
Divide both sides of the equation by 0.100
0.00100 ÷ 0.100 = (c2 × 0.100 ) ÷ 0.100
0.0100 mol L-1 = c2
The concentration of acetic acid in the resultant solution (the solution after dilution) is 0.0100 mol L-1
But what is the concentration of hydrogen ions, c(H+(aq)) in this resultant solution?
Acetic acid is a weak acid, it only partially dissociates, as shown by chemical equation below:
CH3COOH(aq) ⇋ H+(aq) + CH3COO-(aq)
so the concentration of H+(aq) in solution will be less than the concentration of molecular acetic acid, CH3COOH(aq).
But how much less?
We should begin by writing the equilibrium law expression for this dissociation, the equation for the acid dissociation constant (Ka):
| Ka |
= |
[H+(aq)][CH3COO-(aq)]
[CH3COOH(aq)] |
If you look up a table of acid dissociation constants (Ka values), you will find Ka = 1.8 × 10-5 for the dissociation of acetic acid at 25°C.
| Ka |
= |
[H+(aq)][CH3COO-(aq)]
[CH3COOH(aq)] |
= |
1.8 × 10-5 |
Let x represent the amount of CH3COOH(aq) that does dissociate.
So, after dissociation, [CH3COOH(aq)] = 0.0100 - x
Since the stoichiometric ratio (mole ratio) of CH3COOH to H+ and CH3COO- is 1:1:1 (from the balanced chemical equation for the dissociation) we can say that the [H+(aq)] = [CH3COO-(aq)] = x
We can now substitute these values into the equation for Ka:
[H+(aq)][CH3COO-(aq)]
[CH3COOH(aq)] |
= |
1.8 × 10-5 |
[x][x]
[0.0100 - x] |
= |
1.8 × 10-5 |
[x]2
[0.0100 - x] |
= |
1.8 × 10-5 |
| [x]2 |
= |
1.8 × 10-5 [0.0100 - x] |
| [x]2 |
= |
1.8 × 10-7 - 1.8 × 10-5x |
| [x]2 + 1.8 × 10-5x - 1.8 × 10-7 |
= |
0 |
Unfortunaely we would have to solve this quadratic equation if we wanted to determine the value of x.
However, since the value of Ka is small we know that little of the molecular acetic acid dissociates, so x must be small compared to the concentration of the acetic acid, in which case:
0.0100 - x ≈ 0.0100
so we can use this approximation in our equation above:
[H+(aq)][CH3COO-(aq)]
[CH3COOH(aq)] |
= |
1.8 × 10-5 |
[x][x]
[0.0100 - x] |
= |
1.8 × 10-5 |
[x]2
[0.0100 - x] |
= |
1.8 × 10-5 |
| assume 0.0100 - x ≈ 0.0100 |
[x]2
[0.0100] |
≈ |
1.8 × 10-5 |
Multiply both sides of the equation by 0.0100
0.0100 × [x]2
[0.0100] |
≈ |
0.0100 × 1.8 × 10-5 |
| [x]2 |
≈ |
1.8 × 10-7 |
Take the square root of both sides of the equation to find the value of x:
| √[x]2 |
≈ |
√(1.8 × 10-7) |
| x |
≈ |
4.24 × 10-4 |
Since [H+(aq)(resultant solution)] = x
we now know the concentration of hydrogen ions in the resultant solution:
[H+(aq)(resultant solution)] = x = 4.24 × 10-4 mol L-1
And once we know the concentration of hydrogen ions in solution we can calculate the pH of the solution:
pH = -log10[H+(aq)(resultant solution)] = -log10[4.24 × 10-4] = 3.37
Just a word of caution before we move on.
If the acetic acid solution becomes extremely dilute and the concentration of acetic acid becomes very tiny, say 10-12 mol L-1, you will end up with silly results using this calculation method!
Let me show you...
[x]2
10-12 |
≈ |
1.8 × 10-5 |
| [x]2 |
≈ |
10-12 × 1.8 × 10-5 |
| [x]2 |
≈ |
1.8 × 10-17 |
| x |
≈ |
4.24 × 10-9 |
| pH |
≈ |
-log10[4.24 × 10-9] |
| pH |
≈ |
8.37 |
Clearly the resultant solution cannot have an excess of OH- that would produce a pH > 7 (at 25°C) so some of the assumptions we made must have broken down.
For example, we assumed that very little of the CH3COOH dissociated (x is negligible compared to the concentration of acetic acid).
But, as the solution becomes increasingly more dilute, the %dissocation increases.
At some point, we will not be able to ignore the dissociation of CH3COOH and the value of x (determined by solving the quadratic eqation) will approach the concentration of the acid (approaches 100 % dissociation)
When [CH3COOH(aq)] = 10-6 mol L-1, solving the quadratic equation gives [H+(aq)] = 9.5 × 10-7 mol l-1 (about 95% dissociation) and a pH = 6, but if you assume x can be ignored in calculating [CH3COOH(aq)] then [H+] = 4.4 × 10-6 and the pH = 5.4
When [CH3COOH(aq)] = 10-7 mol L-1, solving the quadratic equation gives [H+(aq)] = 9.95 × 10-8 mol l-1 (about 99.5% dissociation) and a pH = 7, but if you assume x can be ignored in calculating [CH3COOH(aq)] then [H+] = 1.34 × 10-6 and the pH = 5.9
But what happens if we continue to dilute the acetic acid past this point? What would be the pH of 10-12 mol L-1 acetic acid?
When [CH3COOH(aq)] = 10-12 mol L-1 we notice that the concentration of acetic acid is actually much, much less than the concentration of hydrogen ions in neutral water!
| [CH3COOH(aq)] |
<< |
[H+(aq)(water)] |
| 10-12 mol L-1 |
<< |
10-7 mol L-1 |
Even if all the acetic acid molecules dissociated (100% dissociation), the maximum concentration of hydrogen ions as a result would be 10-12 mol L-1, which is much, much, less than the contribution of the self-dissociation of water to the concentration of hydrogen ions in the solution.
So, as the dilution of acetic acid becomes extreme, the dissociation of acetic acid is not the main contributor to the concentration of hydrogen ions in the solution.
We ignore these H+ from the dissociation of the acid and assume all the hydrogen ions in solution come from the self-dissociation of water, so the concencentration of hydrogen ions in solution approaches 10-7 mol L-1 (at 25°C) and the pH of the solution approaches 7.
