The goal of this section is to illustrate what a buffer is and how one is produced. The first concept we have to go over is the common ion effect.
Imagine that you had a solution consisting of hydrofluoric acid (`HF`) and a fluorine salt (`MgF_2`). How does the pH of this solution compare to the pH of a solution solely composed of HF?
It turns out that this new solution consisting of `HF` and `MgF_2` will be less acidic than the pure `HF` solution! To understand why this is, recall that the dissociation of `HF` follows that of a standard acid dissociation:
`HF ↔ H^+ + F^-`
Additionally, recall that the a salt in solution dissolves into its consistuent ions:
`MgF_2 ↔ Mg^(2+) + 2F^-`
The net reaction going on in the `HF` and `MgF_2` solution is therefore:
`HF + MgF_2 ↔ H^+ + Mg^(2+) + 3F^-`
Since `F^-` is in both dissociations, `F^-` is called the common ion.
From Le Chatelier's Principle, we know that the addition of `F^-` into the reaction will shift the reaction to the left. This means that less `H^+` will be produced, and thus the pH will be reduced. Since `Mg^(2+)` is a non-reactive ion in this reaction, it doesn't do anything. This shift in equilibrium from the addition of a common ion is called the common ion effect.
We can demonstrate the extent to this effect through a problem:
#1. Calculate the pH of the a solution containing `2M HF` and a solution containing `2M HF` and `1M MgF_2` . The `K_a` of `HF` is `7.2x10^(-4)` .
To calculate the pH of the `2M HF` solution, we have to determine the concentration of `H^+` in the solution at equilibrium. This is best accomplished through an ICE table:
| `HF` | `H^+` | `F^+` |
`2M` |
`0` |
`0` |
`-x` |
`+x` |
`+x` |
`2-x` |
`x` |
`x` |
The `K_a` for this reaction is therefore:
`K_a=([H^+][F^-])/[HF]=x^2/(2-x)`
We know the value of `K_a`, so we can just solve for the value of `x` using methods outlined in earlier sections. In this case, I used the small-number approximation and got `x=0.038` .
Since `x=[H^+]` , `[H^+]=0.038M` . The pH is therefore `-"log"[H^+]=-"log"[0.038]=1.42`
Now we're going to do this with a solution already containing `1M MgF_2` . Since `1M MgF_2` corresponds to `2M F^-` , we're beginning the reaction with `2M F^-` ion already present in solution. The `"I"` column in the ICE table should reflect this change:
| `HF` | `H^+` | `F^+` |
`2M` |
`0` |
`2M` |
`-x` |
`+x` |
`+x` |
`2-x` |
`x` |
`2+x` |
From this point, the solving process is essentially identical as above.
`K_a=([H^+][F^-])/[HF]=(x(2+x))/(2-x)`
Using the small number approximation once again, I get `x=7.2x10^(-4)` .
The pH of this solution is therefore 3.14.
Answer: pH = 1.42 and 3.14
The inclusion of the salt is enough to change the pH of nearly 2 orders of magnitude! As we'll see in the rest of this section, the common ion effect is what allows for buffer systems to exist.
A buffer is a solution that is able to resist changes in pH. A buffer is created through mixing a weak acid and its salt or a weak base and its salt.
The solution we played with above is an example of a buffer. Let's examine what happens to the solution when acids or bases are added to it.
The reaction is `HF ↔ H^+ + F^-` . The added `H^+` will react with `F^-` to drive the reaction backwards to form more `HF` . Since the concentrations of `HF` and `F^-` are greater than that of `H^+` due to the addition of the salt, the changes in `H^+` don't affect the equilibrium too much. The result is that the equilibrium budges slightly and the pH therefore remains relatively constant.
Similarly, what would happen if we added a base? We expect the pH of the solution to increase. The added `OH^-` ions will react with the `H^+` ions. According to Le Chatelier's Principle, this will drive the reaction to the left just slightly. Just like with the acid, the result is that the pH is changed only slightly.
The Henderson-Hasselbalch Equation, often called the "HH-Equation", allows us to calculate the pH of a buffered system. To derive the equation, we start with the definition of `K_a` :
`K_a=([H^+][A^-])/[HA]`
Taking the `-"log"` of both sides, we get
`-"log"(K_a)=-"log"([H^+][A^-])/[HA]`
From the rule of logarithms, we can simplify the right side:
`-"log"(K_a)=-"log"[H^+]-log([A^-]/[HA])`
Recall that the `-"log"(K_a)` is just `pK_a` ! Similarly, the `-"log"[H^+]` is `pH` . Substituting these in, we get
`pK_a = pH - "log"([A^-]/[HA])`
Rearranging this to solve for pH and we have the Henderson-Hasselbalch Equation:
`pH=pK_a +"log"([A^-]/[HA])`
`pK_a` is the dissociation constant of `HA`
`[A^-]` is the concentration of the conjugate base
`[HA]` is the concentration of the acid
#2. Calculate the pH of a solution with starting concentrations of `2M HCH_2COOH ` and `1M NaCH_2COOH` . The `pK_a` of acetic acid is 4.76
We have to start off with an ICE table in order to determine the equilibrium concentrations.`
| `HCH_2COOH` | `H^+` | `CH_3COO^-` |
`2M` |
`0` |
`1M` |
`-x` |
`+x` |
`+x` |
`2-x` |
`x` |
`1+x` |
Since `pK_a=-"log"(K_a)` , we can solve for the `K_a` as `10^(-pK_a)=1.74x10^(-5)`
`K_a=([H^+][CH_3COO^-])/[HCH_2COOH]=(x(1+x))/(2-x)`
Using the small number approximation, we get `x=3.48x10^(-5)`
The pH is therefore `-"log"[3.48x10^(-5)]=4.46`
Alternatively, we can just use the Henderson-Hasselbalch Equation:
`pH= pK_a + "log"([A^-]/[HA])=4.76 + log([1M]/[2M])=4.46`
Either way, the answer will be the same. The Henderson-Hasselbalch Equation is clearly the simpler and quicker method, however.
Answer: `pH=4.46`
Notice that the `pH` of the solution only depends on the `[[A^-]]/[[HA]]` ratio and not the magnitudes of `[A^-]` and `[HA]` . It doesn't matter whether the concentrations are `[[5M]]/[[10M]]` or `[[0.1M]]/[[0.2M]]` ; as long as the ratio is the same, the pH will be the same.
The implicit assumption within solving problems using the HH-equation is that the initial concentrations and final concentrations will be the same. This is appropriate when dealing with weak acids, since weak acids don't dissociate much.
All buffering systems have a limit to how much acid and base they can buffer, called the buffering capacity of the buffer solution. The buffering capacity of a solution depends on the magnitudes of `[A^-]` and `[HA]` , as opposed to the ratio in the HH-equation.
A buffer "breaks," or stops being able to buffer properly, when the amount of acid or base added consumes the inherent base or acid. For example, if a solution has a base concentration of 0.5M and one added 10M strong acid to it, the base would all be consumed and nothing would be left to buffer the acid. The point at which the entirely of the acid/base and/or its conjugate is fully consumedis the buffering capacity of the buffer.
There are three main points about buffering capacity we should take note of:
1. The buffering capacity depends on the magnitudes of `[A^-]` and `[HA]` . The larger the magnitudes, the higher the buffering capacity.
2. The closer the `pH` is to the `pK_a` of the acid, the higher the buffering capacity.
3. The buffering capacity is the point at which either weak acid/base or its conjugate is fully consumed.
1. The common ion effect states that the addition of a common ion into a reaction will shift the equilibrium position of the reaction.
2. A buffer is a solution consisting of a weak acid/base and its conjugate salt. The purpose of buffers is to resist changes in pH in response to additions of acid/base.
3. The Henderson-Hasselbalch Equation allows us to simply calculate the `pH` of a buffered solution.
4. The `pH` of a buffer solution depends only on the ratio of `[A^-]` to `[HA]` , not the magnitudes of either.
5. The buffering capacity is the ability of the buffer to successfully buffer.
6. The buffering capacity depends on the magnitudes of `[A^-]` and `[HA]` , but not the ratio.
7. The closer the `pH` of the solution is to the `pK_a` of the acid, the higher the buffering capacity is.
#1. Buffers in our bodies
Our body naturally comes equipped with various buffer systems that are amazingly adept at resisting changes in `pH` . The most common and simplest one is in our blood, consisting of carbonic acid and its conjugate salt:
`H_2CO_3 ↔ H^+ + HCO_3^-`
Because of the various buffers present in our blood and elsewhere, our body successfully resists changes in `pH` which could damage our health. For example, certain sodas contain phosphoric acid which would undoubtedly decrease out bodies `pH` if it weren't for the buffers inherent in our bodies. `pH` is one of the most important metrics to maintain for a healthy human as a healthy human has a `pH` of approximately `7.4` ; it is in our best interest to maintain this resting `pH` .
Next time you're at the supermarket, check out the drinks section and see if you can find alkaline or basic water. If you read the labels, you'll find the argument that, since we consume so much acidic food and drinks, we need to consume more alkaline substances to neutralize the acid and prevent its effects. Do you think that the alkalinity in these drinks would have an effect on our bodies's `pH` , with consideration to the buffer systems in our body?