Why are buffer solutions not neutral

Question

I am very confused about buffer solutions and I have lots of ideas about them which don’t integrate together so I really can’t tell which are correct and which are wrong. That being the case it’s likely the title question is the wrong one to ask. Here is my understanding of buffer solutions and the issues that arise from it.

To clarify: I am only considering buffers made from weak acids and their conjugate bases, I know that basic buffers exist but to simplify matters I am only considering acidic ones in the rest of this question

The acid has the following dissociation equation:

$$\ce{HA <=> H+ + A-}$$

I believe that the position of equilibrium would lie very far to the left as it is a weak acid and so will scarcely dissociate.

This would mean that the associated equilibrium constant, $\ce{K_a}$, would be very small as the numerator would be very small compared to the denominator.

I hope that is correct so far, if not please do tell me what I have misunderstood!

Now if the conjugate base, $\ce{A-}$, is added, I would say that that is an increase in the concentration of $\ce{A-}$ so the reaction quotient would also increase and would no longer be equal to $\ce{K_a}$. I would then apply Le Chatelier’s principle for increasing the concentration of a species in a system at equilibrium to find how the position of equilibrium would change to make the reaction quotient once again equal $\ce{K_a}$. The increased concentration of the species on the right side of the equation, $\ce{H+ + A-}$, would cause the position of equilibrium to shift to counteract the change. The only way that the concentration of the species of left side of the equation could decrease would be for $\ce{H+}$ and $\ce{A-}$ to react to form the original acid, $\ce{HA}$ thus decreasing the left concentration and increasing the right concentration until the ratio of the two concentrations (the reaction quotient) returns to being equal to the equilibrium constant, $\ce{K_a}$.

Is that correct so far?

The result of this new equilibrium position would be a decrease in $\ce{[H+]}$ but an increase in $\ce{[A-]}$.

Now to make buffer solutions you add rather large concentrations of the conjugate base. I would assume that the number of conjugate base ions added would by far outnumber the $\ce{H+}$ already in the system.

Therefore, I would predict that the conjugate base ions would react with all the $\ce{H+}$ that are present (as I have explained above). I would expect that there would be no hydrogen ions left. This brings two problems that (at least!).

Firstly: zero hydrogen ions would mean that all the non-water species in the buffer solution have a pH of 14, and so the pH of the buffer would not be acidic, (would it not be basic?) In my textbook it says that these buffer solutions are slightly acidic, due to the slight dissociation of the acid. That is a contradiction!

Secondly: If you add equal molar quantities of the acid and the conjugate base and the acid only slightly dissociates, surely it will never be possible for the quotient to equal $\ce{K_a}$, because you have increased the left hand concentrations so much and there aren’t enough hydrogen ions to react to shift it back. Would that mean that equilibrium is never established because the reaction quotient can never equal $\ce{K_a}$?

Or perhaps for some reason the position of equilibrium wouldn’t change as I have described above. Perhaps it would move to counteract the change a little but not so much that all the hydrogen ions would be ‘removed’ (as part of the acid). Why would this be the case and how could one possibly predict that Le Chatelier’s principle would apply only partially in this situation?

Perhaps there is another explaination. I hope so!

I hope I have made my confusion clear. As you can see I have come across many stumbling blocks and any help would be very much appreciated. I am studying A level chemistry so I can only understand chemistry at that level. I am aware that there are lots of simplifications made at A Level level and perhaps that is why I am confused. Please do tell me if it is just something that I have to accept rather than understand, like Le Chatelier’s Principle, that would be most helpful!

Thank you so much for reading my question!

Answer 1

Therefore, I would predict that the conjugate base ions would react with all the H+ that are present (as I have explained above). I would expect that there would be no hydrogen ions left. This brings two problems that (at least!).

Be aware that a small relative mutual conversion within a conjugate pair of a buffer leads to a big relative change of H+ concentration, leading to easy managing of dissociation equilibrium.

Imagine

$$K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{(c_\ce{0,H+} - x)(c_\ce{0,A-} - x)}{c_\ce{0,HA} + x}$$

with $$c_\ce{0,H+} \ll c_\ce{0,A-} \approx c_\ce{0,HA}$$

Mutual conversion HA <=> A- restablishes the equailibrium by a figurative finger snap.

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The alpha and omega is the equation of dissociation equilibrium for weak acids

$$\ce{HA(aq) <=> H+(aq) + A-(aq)} \tag{1}$$

that is for diluted solutions approximately (thermodynamic activities approaximated by concentrations)

$$K_\text{a} \approx \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}, \tag{2}$$

respectively in the rearranged logarithmic form, known as the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {\frac{[\ce{A-}]}{[\ce{HA}]}} \tag{3}$$

where $K_\mathrm{a}$ is the respective acidity constant and $[\ce{X}]$ denotes amount concentration of $\mathrm{X}$.

There are two other important equilibrii

$$\ce{A-(aq) + H2O <=> HA(aq) + OH-(aq)}$$ and $$\ce{H2O <=> H+(aq) + OH-(aq)}$$

All three equilibrii are related by the equation

$$K_\mathrm{w} = K_\mathrm{a} \cdot K_\mathrm{b}$$

The same is valid for weak bases, if we consider their conjugate acids and their acidity constants.

For pure solutions of just the acids, there are possible conditional simplifications:

$$K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \overset{\text{All H+ from acid}}{\approx}\frac{[\ce{H+}]^2}{c_0 - [\ce{H+}]} \overset{\text{low dissociation}}{\approx}\frac{[\ce{H+}]^2}{c_0} \implies \\ \implies [\ce{H+}] \approx \sqrt{K_\text{a} \cdot c_0} \implies \text{pH} \approx \frac 12 (\text{p}K_\text{a} - \log{c_0}) \tag{4}$$

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If there are mixed both a weak acid and its conjugate base to form solution of significant concentrations of both then changes of their concentration due equilibrium reactions are negligible.

the equations (2), (3) above become:

$$K_\text{a} \approx [\ce{H+}] \cdot C,\tag{5}$$

respectively

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log {(C)}\tag{6}$$

where $C$ is the concentration ratio $\dfrac{[\ce{A-}]}{[\ce{HA}]}$ .

Answer 2

I believe that the position of equilibrium would lie very far to the left as it is a weak acid and so will scarcely dissociate.

Only if you start with weak acid in pure water. If you start with the weak base in pure water, you get a basic pH.

Therefore, I would predict that the conjugate base ions would react with all the H+ that are present (as I have explained above).

Water can always make more H+ by dissociating. Alternatively, you can think of the weak base ($\ce{A-}$) react with water instead of with $\ce{H+}$:

$$\ce{A- + H2O <=> AH + OH-}$$

This would be the combination of

$$\ce{A- + H+ <=> AH}$$

and

$$\ce{H2O <=> H+ + OH-}$$

[…] would have a pH of 14.

That’s impossible because it would be more basic than a weak base. The weak acid alone is a bit acidic, the weak base alone is a bit basic, and the mixture (the buffer) somewhere in between.

[OP in comments] Oh so are you saying that the addition of the conjugate base would increase the dissociation of water? Are the hydrogen ions common to both equilibria, that of water dissociation and acid dissociation? I didn’t consider that perhaps it is the source of all my confusion! Would there not be an increase in pH if water dissociates as although the ‘lost’ hydrogen ions are replaced hydroxide ions are also formed? Wouldn’t there be a net increase in alkalinity?

Yes, the hydrogen ions are common to both equilibria, which makes it a bit difficult to think through these situations in our head. A math engine simply solves the system of coupled equations and spits out the answer. This does not help you understand what is going on conceptually, though. Again, yes, if you take away most of the hydrogen ions, they would be replaced by water dissociating, also producing hydroxide. This is how a weak base like ammonia can get a pH higher than 7. The hydroxide ions come from water, and most of the hydrogen ions that ammonia accepts to become ammonium also come from water (as a neutral solution contains just $\pu{1E-7 mol/L}$ of $\ce{H+}$ to begin with).

Answer 3

In aqueous solutions the acidity is mostly governed by the concentration of hydronium ions[H3O+ aka H+] and usually denoted by its negative logarithm, pH. Although H3O+ is most important, weak acids and weak bases present have their own properties. Some reactions happen at an optimum pH and pH control is advantageous. This can be accomplished in several ways. The easiest is to simply use a stoichiometric excess of a strong acid or base or even a weak acid or base if that gives the appropriate acidity. This method is used in permanganate or dichromate oxidations and many reactions. When an intermediate pH is necessary the direct method fails because any change in acid or base gives a large change in H3O+. The answer is to use an excess of the weak acid or base and add the right amount of the conjugate base or conjugate acid to give the proper pH. When the proper amount of buffer is used the reaction pH will stay near the optimum pH. If acid is consumed the weak acid ionizes. If acid is formed the weak base accepts it. The equilibrium conditions satisfy both the ionization quotient of the acid and Kw.

LeChatlier's principle starts with a system at active equilibrium. The buffer is perturbed by a slight change in acidity; either water or the weak acid will change to restore equilibrium. Agreed the rates are not equal. Remember the properties of H3O+ and weak acids differ. Instead of an extreme shock treatment use small equilibrium steps.

example: Reaction uses some acid, solution needs acid to reestablish equilibrium so water and weak acid both ionize to give more H3O+ each to its ability[acid probably ionizes faster that is why it is an acid]. Since H3O+ was consumed the overall solution is now at a higher pH. Since the solution is buffered the change in pH is less than if the acidity were a trace amount of strong acid. In a nonbuffered solution, when a reaction requires acid, as acid is consumed the reaction slows, all the acid is not consumed. Concentrations do not reach zero [pH infinity not 14].