# 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 and easy to understand because it consists only of one deprotonation step, would be ethanoic acid and sodium ethanoic solution, otherwise known as acetic acid, in old terminology - sodium acetate.

00:21 The buffer is prepared, in this particular case, as 1.0 M, or moles per litre, acetic acid or ethanoic acid and sodium acetate or sodium ethanoate. The equilibrium that I showed you before in structural form is shown here where 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.

00:56 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 × 10 to the -5.

01:10 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 ion.

01:36 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 M equivalent of acetic acid and sodium acetate, this would give us a concentration of [H+] of 1.8 × 10 to the -5.

02:00 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.

02:13 We have a solution. We know that the pH is 4.74. Now, let us add a strong base to neutralise 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 NA+ and OH-. This amount of base changed the pure water, as we saw earlier on the scale, from pH 7.0 to 12.3. The OH- ions will react with the ethanoic acid reducing its concentration. However, the acid molecules that react will become ethanoate ions. In buffers, the neutralisation of one buffer component converts it to the other.

02:59 So, let us apply that. We’ve neutralised 2%. This results now in a concentration of acetic acid of 0.98 M and a concentration of sodium acetate of 1.02 M. 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.

03:23 As you can see, we now have a situation where a very, very small change in concentration of our sodium acetate and our acetic acid results in a minute change, in fact is barely observable, change in the concentration of [H+] which is available 1.7 × 10 to the -5. This equates to a pH of 4.77.

03:46 So, whereas before in the case of the 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 very strong base, we see a fundamentally unchanged buffer solution. The pH has hardly been affected.

04:06 That is an example of a buffer scenario. However, let’s repeat with the same concentration of acid. So, know 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.

04:21 So, we’re moving the equilibrium back towards the reactants. This is represented in the equation here - 1.02/0.98.

04:30 This results in a marginally increased concentration of [H+] of 1.9 × 10 to the -5.

04:37 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.

04:47 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.

04:58 The equilibrium in the buffer between acid and conjugate base has prevented the pH from changing much and this is effectively the same.

05:05 Right. Okay. So, let’s bring this into play. 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+], therefore, the pH would be, by virtue of assuming a number of things.

05:27 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.

05:41 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.

05:55 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.

06:06 And, by doing the log at the base 10 of the others, we can see that pH = pKa. So, this is the log of the acidity constant or dissociation constant for the acid plus the log of the concentration of sodium acetate over acetic acid.

06:24 In its general form, the Henderson–Hasselbach equation is given thus.

06:28 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, a log to the negative base 10 of hydrogen or [H+].

06:51 This equation is very useful for buffer calculations. So, if you’re ever unsure and you’re using, let’s say a monoprotic acid, a weak acid, maybe it was, 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’re adding, you can calculate approximately what the buffer pH will be.

07:23 The biggest assumption is that the equilibrium concentrations are the same as the concentrations of the components which we mixed to make the buffer. So, in other words, they remain principally unchanged.

07:34 In the previous example of a 1.00 M solution of ethanoic acid and ethanoic... ethanoate...

07:41 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 minimised, the following should be observed.

07:56 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.

08:07 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. 5.70
2. 4.34
3. 4.74
4. 5.04
5. 7.40
1. …during buffer neutralization one component gets converted into the other.
2. …it prevents the pH changes via breakdown of acid component of the buffer system till its last molecule.
3. …it prevents the pH changes via breakdown of conjugate base component of the buffer system till its last molecule.
4. …it prevents the pH changes via breakdown of base component of the buffer system till its last molecule.
5. …it prevents the pH changes via breakdown of conjugate acid component of the buffer system till its last molecule.
1. It is a complex derivation dealing with redox reactions.
2. It assumes that the equilibrium concentrations are same as 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 of conjugate base concentration)/ (log of 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])