Note that because we are discussing stoichiometry here, we are interested in quantities (moles) of reactants, not concentrations of reactants.

If we add 0.2 mol of sodium hydroxide to a solution containing 1.0 mol of a weak acid HA, then an equivalent number of moles of that acid will be converted into its base form A^–. The resulting solution will contain 0.2 mol of A^– and 0.8 mol of HA.

The important point to understand here is that we will end up with a “partly neutralized” solution in which both the acid and its conjugate base are present in significant amounts. Solutions of this kind are far more common than those of a pure acid or a pure base, and it is very important that you have a thorough understanding of them.

Thus starting with a solution of a pure weak acid or weak base in water, we can add sufficient strong base or strong acid, respectively, to adjust the ratio of the conjugate species — that is, the ratios [HA]/[A^–] in the case of an acid, or [B]/[BH⁺] for a base, to any value we want.

Ionization fractions

In order to express the relative concentrations of the protonated and deprotonated forms of an acid-base system present in a solution, we could use the simple ratio [HA]/[A] (or its inverse), but this suffers from the drawback of yielding an indeterminate result when the concentration in the denominator is zero. For many purposes it is more convenient to use the ionization fractions

(1-7)

The fraction₁ is also known as the degree of dissociation of the acid. By making appropriate substitutions using the relation

(1-8)

we can express the ionization fractions as functions of the pH:

(1-9)

Notice (left) that the the values for both of these functions are close to zero or unity except within the pH range pK_a ± 1.

Gallery of ionization fraction plots

Plots of thefunctions vs pH for several systems are shown below. Notice the crossing points where [HA] = [A] when [H⁺] = K_a. This corresponds to unit value of the quotient in the Henderson-Hasselbalch equation.

 

What is a buffer?

The word buffer has several meanings in English, most of them referring (in its verb form) to cushion, shield, protect, or counteract an adverse effect. In chemistry, it refers specifically to a solution that resists a change in pH when acid or base is added.

A buffer (or buffered) solution is one that resists a change in its pH when H⁺ or OH^– ions are added or removed owing to some other reaction taking place in the same solution. Buffer solutions are essential components of all living organisms.

• Our blood is buffered to maintain a pH of 7.4 that must remain unchanged as metabolically-generated CO₂ (carbonic acid) is added and then removed by our lungs. 
• Buffers in the oceans, in natural waters such as lakes and streams, and within soils help maintain their environmental stability against acid rain and increases in atmospheric CO₂. 
• Many industrial processes, such as brewing, require buffer control, as do research studies in biochemistry and physiology that involve enzymes, are active only within certain pH ranges.

How buffers work

The essential component of a buffer system is a conjugate acid-base pair whose concentration is fairly high in relation to the concentrations of added H⁺ or OH^– it is expected to buffer against. A simple buffer system might be a 0.2 M solution of sodium acetate; the conjugate pair here is acetic acid HAc and its conjugate base, the acetate ion Ac^–.

The idea is that this conjugate pair "pool" will be available to gobble up any small (≤ 10^–3 M) addition of H+ or OH^– that may result from other processes going on in the solution.

Exercise:  sketch out a similar diagram showing what happens when you remove H⁺ or OH^–.

When this happens, the ratio [HAc]/[Ac^–],
will remain substantially unchanged, as will the pH, as you will see below.

You can also think of the process depicted above in terms of the LeChâtelier principle: addition of H⁺ to the solution suppresses the dissociation of HAc,  partially counteracting the effect of the added acid, as illustrated by the equation at the bottom left of the above diagram.

 

The Henderson-Hasselbalch equation

The development given here was also briefly shown in an earlier lesson.

In order to develop this more quantitatively, we will consider the general case of a weak acid HA to which a quantity C_b of strong base has been added; think, for example, of acetic acid which has been partially neutralized by sodium hydroxide, yielding the same conjugate pair described above, although not necessarily identical concentrations.

The mass balance for such a system would be

[HA] + [A^–] = C_a + C_b = C_T(2-1)

in which C_t denotes the total concentration of all species in the solution. Because we added a strong (completely dissociated) base NaOH to the acid, we also note that

C_b = [Na⁺] (2-2)

Recalling the equilibrium expression for a weak acid

(2-3)

We can solve this for [H⁺]:

(2-4)

Re-writing this in terms of negative logarithms, this becomes

(2-5)

or, since pK_a = – log K_a, we invert the ratio to preserve the positive sign:

(2-6)

This extremely important relation is known as the Henderson-Hasselbalch equation. It tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system.

Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its pK_a. According to the above equation, when pH = pK_a, the log term becomes zero, so that the ratio
[AB^–] / [HA] = 10⁰ = 1, meaning that [HA] = [AB^–].  In other words,

When the pH of a solution is set to the value of the pK_a of an acid-base pair, the concentrations of the acid- and base forms will be identical.

This condition can be represented schematically on a proton-free energy diagram:

This says, in effect, that at when the pH of a solution containing both the acid and it conjugate baseis made identical to the acid pK_a, the forms HA and A^– possess identical free energies, and will therefore be present in equal concentrations.

Making and using buffer solutions

Selecting the buffer system

Buffers are generally most effective when the buffer system pK_a is not too far from the target pH. Under these conditions, both [HA] and [A^–] are large enough to compensate for the withdrawal or addition, respectively, of hydrogen ions.

Some common buffer systems

acid	formulas	pKa	pH range
phosphoric	H2PO4 / HPO4–	2.16	1-3
carbonic	H2CO3* /  HCO3–	3.75	3-5
acetic	HCOOH / HCOO–	4.75	4-6
dihydrogen phosphate	H2PO4– / HPO42–	7.21	6-8
boric	H3BO3 / H2BO3–	9.24	8-10
ammonium	NH4+ / NH3	9.25	8-10
bicarbonate	HCO3– / CO32–	10.3	9-11
monohydrogen phosphate	HPO42– / PO43–	12.3	11-13

Practical buffer calculations

Eq (2-4) above and its logarithmic equivalent (2-6) are of limited use in calculations because the exact values [A^–] and [HA] are known only for the special case when the pH of the solution is identical to the pK_a. Most buffer solutions will be adjusted to other pH's, and of course, once the buffered solution begins doing it work by counteracting the effect of additions of H⁺ or OH^–, both [A^–] and [HA] will have changed.

For these reasons, it is useful to rewrite the above two equations by replacing the equilibrium concentrations [A^–] and [HA] with the known initial values C_a and C_b:

(2-7)

(2-8)

Although these are approximations, they are usually justified because useful buffer systems are always significantly more concentrated than those responsible for adding or removing hydrogen ions.  Also, at these higher ionic concentrations, the K_a of the buffer system will seldom be precisely known anyway. Don't expect actual buffer pH's to match calculations to better than 5%.

 

 

 

Pitfalls of the Henderson-Hasselbalch equation

The H-H equation will be found in virtually all chemistry and biochemistry textbooks and is widely used in practical calculations. What most books do not tell you is that Eq 2-4 is no more than an “approximation of an approximation” which can give incorrect and misleading results when applied to situations in which the simplifying assumptions are not valid.

An exact treatment of conjugate acid-base pairs, including a correct derivation of the Henderson-Hasselbalch equation, is given in the chapter on Exact Calculations.

The H-H equation is only valid for fairly high concentrations

The approximations that lead to the H-H equation limit its reliable use to values of C_a and C_b that are within an order of magnitude of each other, and are fairly high. Also, the pK_a of the acid should be moderate.

The shaded portion of this set of plots indicates the values of C_a and K_a that yield useful results. Clearly, the smaller the buffer concentration, the narrower the range of useable acid pK_as.

Most buffer solutions tend to be fairly concentrated, with C_a and C_b typically around .01 - 0.1 M.

Thus a buffer based on a .01 M solution of an acid such as chloric (HClO₃)  with pK_a of 1.9 will fall just outside the "safe" boundary near the upper left part of the diagram.

 

Be careful about confusing the two relations

(2-9)

The one on the left is simply a re-writing of the equilibrium constant expression, and is therefore always true. Of course, without knowing the actual equilibrium values of [HA] and [A^–], this relation is of little direct use in pH calculations. The equation on the right is never true, but will yield good results if the acid is sufficiently weak in relation to its concentration to keep the [H⁺] from being too high. Otherwise, the high [H⁺] will convert a significant fraction of the A^– into the acid form HA, so that the ratio [HA]/[A^–] will differ from C_a /C_b in the above two equations. Consumption of H⁺ by the base will also raise the pH above the predicted value as we saw in the preceding problem example.

Buffer index

The terms buffer intensity and buffer capacity are commonly employed as synonyms for buffer index, but in some contexts, buffer capacity denotes the quantity of strong acid or strong base which alters the buffer's pH by 1 unit. (see below)

How effective is a given buffer system in resisting changes in the pH? The most direct expression of this is the rate of change of the pH as small quantities of strong acid or base are added to the system: ΔC/Δ(pH). Expressed in calculus notation, this is the buffer index, defined as

   (2-10)

C here refers to the concentration of strong acid or base added to the solution.  Because added acid or base affects the pH in opposite ways, we take the absolute value of this function in order to ensure that β is always positive.

The value of β can be calculated analytically from C_a, C_b, K_a, K_b and [H⁺]. By taking the second derivative of β, it can be shown that the buffer index has a maximum value when the pH = pK_a.

 

 

 

Pure water is buffered by the H₃O⁺/H₂O conjugate pair at very low pH, and by
H₂O/OH- at high pH.  This is easily understood if you think about adding some strong acid acid to pure water; even one drop of HCl will send the pH shooting down toward 0.  If you continue adding acid, the pH will not drop significantly below 0, because there won't be enough free water molecules remaining to hydrate the HCl to produce H₃O⁺ ions — thus the solution is strongly buffered.

Note that if we were to subtract the effect of the NaAc buffering from the above plot, the remaining plot for water itself would exhibit a minimum at pH 7, where both [H₃O⁺] and [OH^–] share a common minimum value.

 

 

 

This alternative view shows how the distribution fractions of HAc and Ac^– relate to the effective buffering range of this conjugate pair, which is conventionally defined as ±1 pH unit of the pK_a.

 

Buffer capacity

As we mentioned in the sidebar above, the term buffer capacity is an alternative means of expressing the ability of a buffer system to absorb the addition of strong acid or base without causing the pH to deviate by more than one unit from that of the pure buffer.

In this example, the buffer capacities for addition of acid and base will differ because the buffer pH has been adjusted to a value that differs from its pK_a.

The importance of buffer concentration

Buffer systems must be appreciably more concentrated than the concentrations of strong acid or base they are required to absorb while still remaining within the desired pH range. Once the added acid or base has consumed most of one or other of the conjugate species comprising the buffer, the pH will no longer be stabilized.  And of course, very small buffer concentrations will approach the pH of pure water.

 

Because the concentrations of conjugate species can vary over many orders of magnitude, it is far more useful to express them on a logarithmic scale.  Since pH is already logarithmic, one can obtain a "bird's-eye view" of an acid-base system in a compact log-log plot.

Even better, these plots are easily constructed without any calculations or arithmetic — or even any graph paper — any scrap paper, even the back of an envelope, will be sufficient.  A ruler or other straightedge will, however, give more accurate results.

In addition, you can use these plots to estimate the pH of a solution of a monoprotic or polyprotic acid or base without contending with quadratic or higher-order equations —  or doing any arithmetic at all!  The results will not be as precise as you would get from a proper numerical solution, but given the uncertainties of how equilibrium constants are affected by the presence of other ions in the solution, this is rarely a problem. 

But aside from these advantages, the use of log-C vs pH plots will afford you an insight into the chemistry of acid-base systems that cannot be obtained simply by doing numerical calculations.

The basic form of the plot, and the starting point for any use of such plots, looks like this:

If you examine this plot carefully, you will see that it is nothing more than a definition of pH and pOH, as well as a definition of a neutral solution at pH 7. Notice, for example, that when the pH is 4.0, 
[H⁺] = 10^–4 M and
[OH^–] = 10^–10 M.

 

You will have to admit that estimating a pH in this way is far more painless than an ordinary numerical calculation would be! 

 

 

Ammonia - a weak base

Ammonia, unlike most gases, is extremely soluble in water. Solutions of NH₃ in water are properly known as aqueous ammonia, NH₃(aq); the name "ammonium hydroxide" is still often  used, even though there is no evidence to suggest that a species NH₄OH exists.

Ammonia is the conjugate base of the ammonium ion NH₄⁺. The fact that ammonia is a base has no special significance insofar as the construction and interpretation of the log-C vs pH diagram is concerned; such diagrams always refer to conjugate acid-base systems, rather than to individual acids and bases.

Similarly, at the crossing point , [NH₃] ≈ [H⁺].  This comes from the proton condition [NH₃] + [H⁺] – [OH^–] = 0 by dropping the [OH^–] term on the assumption that its value is negligible compared to [H⁺].

You will have noticed that we referred to Eq ?? as the "proton condition".  Sorry to sneak it in without warning, but we hope that having piqued your curiosity, you might take the time to look at the following discussion of this important tool for estimating the pH of a solution of a pure acid, base, or a salt.

Mass- and charge balance, and the proton condition

In any aqueous solution of an acid or base, certain conservation conditions are strictly observed. Together with concentration and the K_a or K_b , these put certain constraints on the system that determine the state of the system.

The most fundamental of these are conservation of mass and of electric charge, which of course apply to chemical changes of all kinds. We commonly express these as mass balance and charge balance, respectively.  (The latter of these is sometimes referred to as the electroneutrality condition.)

Using the above example of the ammonium system as an illustration of this, we are interested in two particular instances of practical importance: what conditions apply to solutions made by dissolving a) pure ammonia,
or b) ammonium chloride, in water?

Mass balances

C_{T, NH3} = [NH₃] + [NH₄⁺] = 10^–2 M 

C_{T, Ac} = [HAc] + [Ac^–] = 10^–2 M

Charge balances

NH₃: [H⁺] + [NH₄⁺] = [OH^–]

NaAc: [Na⁺] + [H⁺] = [Ac^–] + [OH^–]

OK so far?  That's really all we need, but it's usually more convenient to combine these with mass balances on the protons alone, so that we have a single equation for each of the two solutions. The resulting equations are known as the proton conditions for the two solutions.

In order to avoid a lot of algebra, there is a simple short-cut for writing a proton condition equation:

1. Identify the substance you are starting with — the pure acid, base, or salt, and also H₂O itself, that make up the solution whose pH you wish to know at the concentration for which the log C-pH chart is drawn. we call this the proton reference level (if that appellation scares you, it is sometimes referred to as the "basis substance".) 
2. On the left side of the proton condition, write concentration expressions for all species that possess protons in excess of the reference level.
3. On the right side, write concentrations of all species that possess fewer protons the reference level.

Let's try it for our ammonia system. 

Solution of NH₃ in water:

The proton reference level (PRL) is defined by NH₃(aq) and H₂O.

Proton condition: [H₃O⁺] + [NH₄⁺] = [OH^–]

(Notice that the substances that define the PRL (in case, H₂O and NH₃) never appear in the proton condition equation.)

Solution of NH₄Cl (or of any other strong-anion ammonium salt) in water:

The PRL is defined by NH₄⁺ and H₂O.

Proton condition: [H₃O⁺] = [NH₃] + [OH^–]

(The chloride ion has nothing to do with protons, so it does not appear here.)

This proton condition stuff seems awfully complicated: why do we bother?

Good question: the answer is that what seems complicated at first sometimes turns out to be simpler and easier to understand than any alternative.

Think of it this way: acid-base reactions involve the transfer of protons: some protons jump up to higher proton-free energy levels
(e.g., NH₃  + H₂O → NH₄⁺ + OH^–), others drop down to proton-vacant levels (H₃O⁺ → NH₃ → H₂O + NH₄⁺).  But no matter what happens, the total number of "available" protons (that is, all "dissociable" protons, including those in
H₃O⁺) must be conserved — thus the "mass balance on protons".

So let's look again at the log C - pH plot for the ammonium system, which we reproduce here for your convenience:

Consider first the solution of ammonium chloride, whose proton condition is given by [H₃O⁺] = [NH₃] + [OH^–].  We know that NH₄Cl, being the salt of a weak base and a strong acid, will give a solution that is slightly acidic. In such a solution, [OH^–] will be quite small so that we can neglect it without too much error.  We can then write the proton condition as [H⁺] ≈ [NH₃]. On the plot above, this corresponds to point where the lines representing these quantities cross. At this pH of around 6.2, [OH^–] is about two orders of magnitude smaller than [NH₃], so we are justified in dropping it.

Similarly, for a solution of ammonia in water, the proton condition
[H₃O⁺] + [NH₄⁺] = [OH^–] that we worked out above can be simplified to
[NH₄⁺] ≈ [OH^–] (point ) with hardly any error at all, since [H⁺] is here about six orders of magnitude smaller than [NH₄⁺].

In our discussion of the plot for the acetic acid system, we arrived at the proton conditions [H⁺] ≈ [Ac^–] (for HAc in water) and [H⁺] ≈ [OH^–] (for NaAc) by simply using the stoichiometries of the reactions. This can work in the simplest systems, but not in the more complicated ones involving polyprotic systems. In general, it is far safer to write out the complete proton condition in order to judge what concentrations, if any, can be dropped.

 

Ammonium acetate solution: solution of a salt

 

Each system retains its own system point,  for the acetate system and for the ammonium system.

Note that the two plots previously shown were for 10^–3 M solutions, so the non-system points for this more concentrated solution occur at different pH values. The pH's of 10^–2 M solutions of HAc , NaAc , and NH₃ are located by using the same proton conditions as before.

The only new thing here is the point , which corresponds to the ammonium acetate solution in which we are interested. The proton condition for this solution is [H⁺] + [HAc] = [NH₃(aq)] + [OH^–].  Inspection of the plot reveals that [H⁺] << [HAc] and [OH^–] << [NH₃(aq)], so the proton condition can be simplified to [HAc] ≈ [NH₃(aq)], corresponding to the intersection of these two lines at .

Polyprotic systems

Now that you are able to find your way around log C-pH plots that encompass two acid-base systems, the polyprotic systems that we describe below should be no trouble at all.

The carbonate system

A thorough understanding of the carbonic acid-carbonate system is essential for understanding the chemistry of natural waters and the biochemistry of respiration in humans and other animals.

There are two special things about this system that you need to know:

• A solution of carbon dioxide in water is mostly in the form of hydrated CO₂, CO₂(aq). But a small fraction of dissolved CO₂ reacts with water to form carbonic acid, H₂CO₃.  Because this acid cannot be isolated, common practice is to add an asterisk to the formula to designate the "total" CO₂ concentration
  [CO₂ + H₂CO₃]: thus H₂CO₃*.
• Because CO₂ is a gas, solutions containing carbonate species of all kinds can equilibrate with the atmosphere, which normally contains small quantities of CO₂. This means that a solution of sodium bicarbonate that is open to the atmosphere can lose CO₂ to the atmosphere at low pH, and absorb it at high pH. A system of this kind is said to be open. Alternatively, a system can be closed to exchange with the atmosphere. Sillén plots for open and closed carbonate systems have very different appearances.

 

This Sillén plot for the
H₂CO₃* closed system is typical of other diprotic acid systems in that there are three conjugate species and two system points. The 10^–3 M concentration plotted here is typical of what is found in groundwaters that are in contact with carbonate-containing sediments.

The pH of a 10^–3 M  solution of CO₂, NaHCO₃ and Na₂CO₃ are indicted by the blue annotations.

solute, (reference level)	proton condition	approximation
carbon dioxide (H2CO3*)	[H+] = [HCO3–] + 2[CO32–] + [OH–]	[H+] ≈ [HCO3–]
sodium bicarbonate (NaHCO3)	[H+] + [H2CO3] = [CO32–] + [OH–]	[H2CO3] ≈ [OH–]
sodium carbonate (Na2CO3)	[H+] + 2[H2CO3*] + [HCO3–] = [OH–]	[HCO3–] ≈ [OH–]

You should take a few moments to verify the approximations in the rightmost column.

Notice the factors of 2 that multiply some of the concentrations.  For example, the "2[CO₃^2–]" term in the first line of the table means that the CO₃^2– species is two steps away from the proton reference level H₂CO₃*, so it would take two hydrogen ions to balance a single carbonate ion.

Another new feature introduced here is the doubling of the slopes of the lines representing [H₂CO₃*] and [CO₃^2–] where they cross the pH values corresponding to the K_as that are one step removed from their own system points and near –log C = 7.

The phosphate system

This triprotic system is widely used to prepare buffer solutions in biochemical applications. In its fundamental form, there is nothing new here, other than an additional system point corresponding to the pK_a of H₂PO₄^–.

 

 

The pH of a 10^–3 M  solutions of H₃PO₄, NaH₂PO₄, Na₂HPO₄ and Na₃PO₄can be estimated by setting up the proton conditions as follows. Unfortunately, these estimates become a bit more complicated owing to the proximity of the three system points. But it's far easier than doing a numerical solution which requires solving a firth-degree polynomial!

solute, (reference level)	proton condition	approximation
phosphoric acid (H3PO4)	[H+] = [H2PO4–] + 2[HPO42–] + 3[PO43–] + [OH–]	[H+] ≈ [H2PO4–]
sodium dihydrogen phosphate (NaH2PO4)	[H+] + [H3PO4] = [HPO42–] + 2[PO43–] + [OH–]	[H+] + [H3PO4] ≈ [HPO42–]
disodium hydrogen phosphate (Na2HPO4)	[H+] + 2[H2PO4–] + [H2PO4–] = 3[PO43–] + [OH–]	[H3PO4-] = [PO43–] + [OH–]
trisodium phosphate (Na3PO4)	[H+] + 3[H3PO4] +2 [H2PO4–] + [HPO42–] = [OH–]	[HPO42–] ≈ [OH–]

Because the phosphate plot is rather crowded, we show here a modified one in which the details of the pH estimates for the solutions of H₃PO₄  and its salts are emphasized.

 

 

 

 

Estimation of the pH of 10^–3 M solutions of phosphoric acid and trisodium phosphate , based on the approximations in the above table are straightforward.

Those for NaH₂PO₄ and Na₂HPO₄ , however, are complicated by the presence of two concentration terms on the right side of the respective proton condition approximations.

Unless you are in an advanced course, you may wish to skip the following details.  The important thing to know is that graphical estimates of these more problematic systems are fundamentally possible.

This is more advanced material that can be skipped by first-year students.

The relation between the pH of a solution and the concentrations of weak acids and their conjugate species is algebraically rather complicated. But over most of the pH range, simplifying assumptions can be made that, when expressed in logarithmic form, plot as straight lines having integral slopes of 0, ±1, ±2, etc.

It is only within the narrow pH range near the pK_a that these simplifying assumptions break down, but on a logarithmic plot, one can draw a smooth curve that covers this range without introducing significant error.

Here is your basic stationery — a plot showing the two ions always present in any aqueous solution.  In practice, you can usually cut off the bottom part where concentrations smaller than 10–10 M.
For any given acid-base system, you need to know the pKa and nominal concentration CT. Find the point at which these two lines intersect.
Locate and mark the system point on the pKa line.  It will be 0.3 of a log-C unit below the concentration line. This is where [HA] = [A–] = ½ Ct . (Recall that log.5 = .3)
For pH < pKa, [HA] plots as a line through the system point with slope = 0. For pH > pKa, the line assumes a slope of –1. Use the [H+] and [OH–] lines as guides.
In the pH interval of ±pKa, the slope changes from 0 to +1.  You can approximate this with a sloped line as shown, or "prettify" it with a smooth curve.
Finally, draw the lines for the conjugate form [A–], using slopes +1 and 0.

What's all this good for?

Suppose you have a solution of NaA whose pH is 8. Estimate the concentration of the acid HA in the solution.

All you need do is look at the plot.  The [HA] line crosses the line marking this pH at – log C = 6, or [HA] = 10^–6. This quantity too small to significantly reduce [A^–], which remains at approximately CT = 10^–6 M.

 

Boundless outlines on buffer solutions. These summarize the main points and problem examples as presented in major textbooks. They are an excellent aid for pre-exam review. Chemistry: The Central Science (8 subtopics)

A collection of practice exercises (with solutions) on buffers and titrations from Bryn Mawr College.

The Henderson-Hasselbalch equation: its history and limitations - Henry N. Po and N.M. Senoza, J.Chem. Ed. 78 (11) 2001 (1499-1503)

A detailed treatment of log-C vs pH diagrams can be found in the book Acid-base diagrams by Kahlert and Schotz.  It is available online in some college libraries.

Make sure you thoroughly understand the following essential concepts that have been presented above.

• State the values of C_a and C_b after a weak monoprotic acid or base has been partially neutralized with strong base or acid.  (Problem Example 1)
• Define the ionization fraction, and calculate the percent ionization, of a C_a M solution of a weak monoprotic acid.
• Sketch out a rough plot showing the distribution of conjugate species in a solution of a weak acid having a given value of K_a.
• Define a buffer solution, and explain how it works.
• Calculate the approximate pH of a solution of a weak monoprotic acid having a given pK_a that has been partially neutralized by addition of a strong base.
• Specify how to make a given volume of a buffer solution having a certain pH, starting with a weak acid and sodium hydroxide.  (Problem Example 3)
• Define the buffer index, and explain its significance and the conditions under which its value is maximized.
• Sketch out a –log C vs pH (Sillén) plot for a weak monoprotic acid/base system having a given nominal concentration and pK_a, and use it to estimate the pH of the solution.