# Buffer Solution in Action – Acid-Base Reactions

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00:00 Buffer Solutions. A common buffer solution, which is relatively easy to explain than it is to understand because it consists only of 1 deproteination step would beat ethanoic acid and sodium ethanoate solution otherwise known as acetic acid in old terminology and sodium acetate. The buffer is prepared, in this particular case, that's 1.0 moles per liter acetic acid, ethanoic acid and sodium acetate or sodium ethanoate. The equilibrium that I showed you before, that structural form, is shown here. Well we have the ethanoic acid plus an equivalent of water giving us the hydronium ion and CH3CO2Na. On one side, we have the H+, on the other side we have the sodium acetate. If we rearrange this equation, so instead of using our model examples of H+ and A- we now have a real world structure to consider. The Ka, as we said before, for acetic acid is 1.8 x 10-5. If we rearrange this equation to determine the concentration of H+ in our solution, we get the following equation on the board which is shown right at the bottom where Ka multiplied by the concentration of acetic acid divided by the concentration of the sodium acetate results in a ___ calculation for the concentration of H+ in the form of the hydronium one. So, let's go back to our equation. Let's assume that the equilibrium concentrations do not differ much from the initial concentrations. So in this particular case if we were to have a 1 molar equivalent of acetic acid and sodium acetate, this would give us a concentration of H+ of 1.8 x 10-5. What that means in pH is if we do the negative log to the base 10 of this value we get a value of 4.74. This is shown on the board there. Good. So far so good. We have a solution, we know that the pH is 4.74. Now let us add a strong base to neutralize 2% of the acid. So, we'll add a strong base so we know it's going to be completely dissociated into, let's say sodium hydroxide and a + and OH-. This amount of base change the pure water, as we saw right here on the scale, from pH 7.0 to 12.3. The OH- signs will react with the ethanoic acid reducing its concentration. However, the acid molecules that react will become ethanoate ions. In buffers, the neutralization of 1 buffer component converts it to the other. So, let us apply that. We've neutralize 2%. This results now in a concentration of acetic acid of 0.98 molar and a concentration of sodium acetate of 1.02 molar. Now, let's use our knowledge of those concentrations to then determine what the pH is going to be now we've made that change. As you can see, we now have a situation where the very very smooth change in concentration of sodium acetate and acetic acid results in a minute change, if that is barely observable, in the concentration of H+ which is available.

03:41 1.7 x 10-5. This equates to a pH of 4.77. So whereas before in the case of water where we saw a substantial change in pH with a minute amount of a strong acid or strong base. Here, where we add a small amount of a very strong base, we see a fundamentally unchanged buffer solution. The pH has hardly been affected. That is an example of a buffer scenario.

04:09 However, let's repeat with the same concentration of acid. So now what we did beforehand we're actually adding some more acid to the scenario so we're converting more of the sodium ethanoate into the ethanoic acid. So we are moving the equilibrium back towards the reactants. This is represented in the equation here. 1.02/0.98. This results in a marginally increased concentration of H+ of 1.9 x 10-5. Now, if we do the negative log to the base 10 of that concentration of H+, we see that we end up with a pH value of 4.72. Again, not very different to the original pH that we started off with when we used 1 mole equivalent of either the sodium acetate and the acetic acid. The equilibrium in the buffer between acid and conjugate base has prevented the pH from changing much and this is effectively the same.

05:08 Right, okay. So let's bring this into play of how does it work and what are the assumptions that we make when we do these reactions when we set these buffers up. We can adjust the equation to actually work out what the concentration of H+ and therefore what the pH will be by virtue of assuming a number of things. One of the principal assumptions and this is why you need either a weak acid or a weak base is that the amount of sodium acetate that you add pretty much stays as sodium acetate and the amount of acetic acid you add pretty much stays as acetic acid. That's pretty much your assumption, your principal assumption. So you can control precisely how much H+ is in that solution. If we take the negative log to the base 10 of each of these particular terms as you can see on the board, we can determine pH, which of course is the negative log to the base 10 of the H+ component and by doing the log to the base 10 of the others we can see that pH equals PKA, so this is the log of the acidity constant with association constant for the acid plus the log of the concentration of sodium acetate over acetic acid. In its general form, the Henderson-Hasselbalch equation is given thus. So where we have a concentration of a conjugate base (in this case our acetate) over our concentration of weak acid (in this case our acetic acid), we add the log of that relationship or ratio to our log of PKA to give us the pH or log the negative base end of ___ H+. This equation is very useful for buffer calculations. So if you are ever unsure and you're using, let's say a monoprotic acid, weak acid, maybe for example dichloro or monochloroacetic acid, as long as you know the Ka and as long as you know approximately what's being added in terms of the conjugate base (in this case our sodium acetate) and the amount of free acid you are adding, you can calculate approximately what the buffer pH will be. The biggest assumption is that the equilibrium concentrations are the same as the concentrations of the components which we mix to make the buffer. So in other words they remain principally unchanged. In the previous example of a 1.00 molar solution of ethanoic acid and sodium ethanoate are mixed to make the buffer. The assumption in this case is that the concentrations will be the same in dynamic equilibrium. To ensure that the effects of this assumption are minimized, the following should be observed. When the concentration of conjugate base is divided by the concentration of the weak acid, the resulting value should be between 0.10 and 10. Concentrations of the components should exceed Ka by at least 100 times.

The lecture Buffer Solution in Action – Acid-Base Reactions by Adam Le Gresley, PhD is from the course Ionic Chemistry.

### Included Quiz Questions

1. 5.69
2. 4.34
3. 4.74
4. 5.04
5. 7.40
1. Because during buffer neutralization, one component of the buffer system gets converted into the other.
2. It prevents pH changes via the breakdown of the acid component of the buffer system until its last molecule.
3. It prevents pH changes via the breakdown of the conjugate base component of the buffer system until its last molecule.
4. It prevents pH changes via the breakdown of the base component of the buffer system until its last molecule.
5. It prevents pH changes via the breakdown of the conjugate acid component of the buffer system until its last molecule.
1. It is a complex derivation dealing with redox reactions.
2. It assumes that the equilibrium concentrations are equal to the concentrations of the components mixed in the buffer preparation.
3. The ratio of the concentration of conjugate base to that of weak acid should be between 0.10 and 10.00.
4. The concentrations of components should exceed Ka by at least 100 times.
5. pH = pKa + log {(conjugate base concentration) / (weak acid concentration )}
1. pH = pKa + log ([CH3COO-] / [CH3COOH])
2. pH = pKa + log ([CH3COOH] / [CH3COO-])
3. pKa = pH + log ([CH3COO-] / [CH3COOH])
4. pKa = pH + log ([CH3COOH] / [CH3COO-])
5. pH = pKa - log ([CH3COO-] / [CH3COOH])