Using the McAfoos Rules for Oxidation Numers

Here are the rules in use. Remember that Rule 1 is not only the most important, it is also the LAST thing you do each time.

Rule #1 – The sum of the oxidation numbers = the total charge. 

This rule assigns oxidation numbers to pure elements, and single element ions. Here are some examples:

Rule 1 is also used to assign the "last" oxidation number. That does NOT mean the oxidation of the element on the right (or the left for that matter). It means that once we have assigned oxidation numbers for all but one of the element in the compound or ion, we use math to assign the last.

Let's look at the compound \(CoF_2\). If we know that the F has an oxidation number of -1 (Rule 2 would allow us to know that), we can determine the the Co must have a charge of +2 according to the math below:

\(charge=sum~of~oxidation~numbers\)

The charge on this (or any) compound =  0, and each F is -1. Let's set the oxidation number of Co = x. That gives us:

\(0=x + 2(-1)\)

Rearranging gives us \( x=+2\), so the oxidation number of Co is +2.

For another example, let's look at the compound \(K_2O_2\). We can know (again from rule 2) that the potassium has an oxidation number of +1. We know that the charge is 0, so let's set the oxidation number of oxygen to "y". That gives us:

\(0 = 2(+1) + 2(y)\)

which, when rearranged, gives us \(y=-1\). This is an odd charge for oxygen, but it's common enough that we gave it a name ⎼ this is a peroxide.

Rule #2 – Single charge elements

This rule assigns oxidation numbers to those elements that only have one non-zero charge. The list is short: alkali metals are always +1 (this doesn't include H), alkaline earth metals are always +2, F is always -1. There are some other elements that have only one (non-zero) charge, but you can probably get away without thinking about them. 

Immediately after applying this rule, you should apply Rule #1 (if possible).

Here are some examples:

Rule #3 - Hydrogen is always +1

Remember that this rule is a LOWER priority than the ones above, so we would not apply it for the following:

\(H_2\):    Hydrogen's oxidation # = 0 (Rule 1)

KH:    Hydrogen's oxidation # = -1 (Rule 2)

Now let's look at some examples where Rule 3 does matter:

Rule #4 - Oxygen is always -2

Just like Rule 3, this seems to contradict some of the things we've already done. However, as long as you remember that these rules are in RANKED order, you'll be fine. 

For instance we would NOT use Rule 4 in the following cases:

\(O_2\): Oxygen's oxidation # = 0

\(CaO_2\):    Oxygen's oxidation # = -1

Now, let's look at some examples for which we would use Rule #4:

Rule #5 - The most electronegative element gets its most logical negative charge

This rule is only used occasionally. In fact, you may not even realize that you are using it.

Here's a simple example:

\(NiCl_3\)

None of the previous rules can be applied. We have more than one element (so no Rule 1), Both of these elements have more than one possible charge (Rule 2), there is no Hydrogen (Rule 3) and no Oxygen (Rule 4).

We do know that Chlorine is more electronegative than Ni. The logical charge on Chlorine is -1 (since that makes it isoelectronic with Argon.

That makes our math easy: \(0 = x + 3(-1)\) which gives \(x=+3\). So this is nickel III chloride.

A word of caution: Complex ionic compounds (those with polyatomic ions and especially with polyatomic ions that contain the same element) have to be treated carefully - both in terms of assigning oxidation numbers and physically. This should explain.

The Henderson/Hasselbalch Equation and the Mathematics of Buffers

In 1908, Lawrence Joseph Henderson derived an equation to calculate the \([H_3O^{+}]\) in a buffer solution. In 1909, Søren Peter Lauritz Sørenson introduced the idea of pH, which allowed Karl Albert Hasselbalch to rework Henderson’s equation resulting in what is known today as the Henderson-Hasselbalch equation:

\(pH = pK_a - log(\frac{[HA]}{[A^{-1}]})\)

Where \(pK_a\) is the -log of the acid dissociation constant (\(K_a\)), \([HA]\) is teh molar concentration of a weak acid, and \([A^{-1}]\) is the molar concentration of the conjugate base.

The equation is also written like this:

\(pH = pK_a + log(\frac{[A^{-1}]}{[HA]})\)

This is mathematically identical.

The Derivation

Although it is unlikely that you would need to know the derivation of the equation, seeing the derivation may help you understand when the equation can be used and its limitations.

We begin with the acid dissociation reaction and the equilibrium expression for that reaction:

\(HA + H_2O \rightleftharpoons A^{-1} + H_3O^{+}~~~~~~~~~~~K_a = \frac{[A^{-1}][H_3O^{+}]}{[HA]}\)

We can algebraically rearrange the equilibrium expression to solve for \([H_3O^{+}\) giving this equation:

\([H_3O^{+}] = K_a \cdot \frac{[HA]}{[A^{-1}]}\)

Taking the negative log of everything, and remembering that \(log(x \cdot y) = log(x) + log(y) gives us:

\(-log[H_3O^{+}] = -log(K_a) - log(\frac{[HA]}{[A^{-1}]})\)

which translates into:

\(pH = pK_a - log(\frac{[HA]}{[A^{-1}]})\)

Understanding and using the Henderson-Hasselbalch Equation

Like any good equation, the Henderson-Hasselbalch equation is only useful if you KNOW all but one of the variables. In other words, you can only use this equation to solve for pH if you KNOW the \(K_a\) and both the \([H_3O^+]\) and the \([OH^{-1}]\).

The good news for us, is that weak acids and bases have small \(K_a\) and \(K_b\) values which leads to “little x” problems. Take a look at this ice table:

Assuming that both “some” and “a bit” are non-zero amounts, then if this is a “little x” problem, those concentrations won’t (appreciably) change. So, in that situation, we already “know” the \([HA]\) and the \([A^{-1}]\).

The upside is that when you have a non-zero amount of both weak acid and conjugate base present in the solution, the Henderson-Hasselbalch equation can be used to find the pH.

Buffer solutions

We can use the Henderson-Hasselbalch equation to find the pH of a buffer solution when given an amount of both a weak acid and its conjugate base. For instance:

What is the pH of a buffer solution that has a \([HC_2H_3O_2]\) = 0.25 M and a \([C_2H_3O_2^{-1}]\) = 0.20 M. The \(K_a\) of acetic acid is \(1.8x10^{-5}\).

The solution:

\(pH = -log(1.8x10^{-5}) - log(\frac{0.25 M}{0.20 M}) = 4.65\)

Finding the pH during a weak titration

During a weak titration (for instance, a weak acid being titrated by a strong base), there is a region within which the Henderson-Hasselbalch equation can be used. This region falls between roughly 10% of the equivalence point volume and 90% of the equivalence point. This region is called the buffer region. These restrictions ensure that the “some” and “a bit” in the ICE table above are large enough so that the “little x” can be ignored. Let’s look at the 10.00 mL point in the problem below. This problem is FULLY worked out here.

50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

The 10.00 mL point corresponds to this:

So, the pH can be found by doing this:

\(pH = -log(2.56x10^{-5}) - log(\frac{(\frac{0.09 mols}{0.060 L})}{(\frac{0.010 moles}{0.060 L})}) = 3.64\)

Of course, since the acid and base are in the same solution and, therefore, have the same volume, you could also just have done this:

\(pH = -log(2.56x10^{-5}) - log(\frac{(0.09 mols)}{(0.010 moles)}) = 3.64\)

Finding \(K_a\) by weak titration

At the beginning of a weak titration the solution contains “only” the weak acid, HA. At the equivalence point, all of those acid molecules have lost their \(H^+\) and are now \(A^{-1}\). So, logically, when you are halfway to the equivalence point, then half of the acid molecules will remain and half will have been converted to \(A^{-1}\).

That means that halfway to the equivalence point the ratio \(\frac{[HA]}{[A^{-1}]}\) = 1. Since the -log(1) = 0, then \(pH = pK_a\). That point is called the maximum buffer point and we can use that idea to solve a problem like this one:

A student titrates 100.0 mL of 0.100 M HA with 0.200 M NaOH. After 25.00 mL of the base have been added, the pH of the solution is 6.12. What is the value of \(K_a\)?

The equivalence point of the titration is 50.00 mL of NaOH, so 25.00 mL is halfway. At that point \(pH = pK_a\). So,

\(10^{pH} = 10^{pK_a} ~~~~~~~~~ K_a = 7.59x10^{-7}\)

Buffer Solutions

 Buffer solutions are an important, but potentially confusing idea involving weak acids and bases. We’ll start with a simple definition, then look at what buffers can do that makes them so interesting. 

The Definition 

A buffer is a solution that contains a weak conjugate pair.

Let’s pull that apart. First, a buffer solution must contain BOTH an acid and a base. Secondly, both must be weak, since if one or both was strong, they would immediately react ~100%. Lastly, the acid and base need to be conjugates.

A simple example of a buffer is a solution that contains some acetic acid (\(HC_2H_3O_2\)) and some acetate (\(C_2H_3O_2^{-1}\)). Another example would be a solution that contains both ammonia (\(NH_3\)) and ammonium (\(NH_4^{+1}\)).

What makes a buffer so interesting? 

A buffer solution is one that resists changes in pH when an acid or base is added to the solution.

To understand that, let’s compare what happens when you add HCl (a strong acid) to water and when you add it to a buffer solution (we’ll use the \(HC_2H_3O_2 ~and~ C_2H_3O_2^{-1}\) for our example.

When you add HCl to water, the following reaction occurs:

\(HCl + H_2O \rightleftharpoons H_3O^{+1} + Cl^{-1}\)

Because this reaction produces hydronium (\(H_3O^{+1}\))the pH will go down.

However, if you add HCl to the buffer solution, the HCl will react with the conjugate base (\(C_2H_3O_2^{-1}\)) in the following reaction:

\(HCl + C_2H_3O_2^{-1} \rightleftharpoons HC_2H_3O_2 + Cl^{-1}\)

Because this reaction does NOT produce hydronium, the pH is unchanged.

We see the same sort of thing when we add a base, like NaOH.

When we add NaOH to water, it dissolves, creating a solution of hydroxide:

\(NaOH_{(s)} \rightleftharpoons Na^{+1}_{(aq)} + OH^{-1}_{(aq)}\)

The addition of hydroxide ions to the solution means that the pH will rise.

However, if you add NaOH to the buffer solution, the NaOH will react with the acid present (\(HC_2H_3O_2\)). You can think of this as a double displacement reaction:

\(NaOH + HC_2H_3O_2 \rightleftharpoons NaC_2H_3O_2 + H_2O\)

Although it is more accurate to think about the net-ionic equation that occurs with the hydroxide ions:

\(OH^{-1}_{(aq)} + HC_2H_3O_{2 (aq)} \rightleftharpoons H_2O_{(l)} + C_2H_3O_{2 (aq)}^{-1}\)

No matter which version of the reaction you look at, no hydroxide is added to the solution, so the pH is unchanged.

Buffer Math 

It is, of course, not quite correct to suggest that the pH doesn’t change at all. What we really mean is that the pH does not change dramatically.

How it changes, and by how much, can be worked out using the Henderson/Hasselbalch equation. This math is beyond the scope of most first year chemistry classes, but you can find some information here.

Buffers in Your Life 

It turns out that buffer solutions are present in lots of situations. Here are just a few:

Swimming Pools

As we all know, children make regular acidic “donations” to swimming pools. In addition, because an in-ground pool is walled in cement which contains lime, the walls themselves can make the pool  more basic. In addition, rain not only adds water to the pool, but also washes in debris. Insects, and perhaps frogs, fall in the pool and die. All in all, there are multiple factors that can change the pH of the water in your pool. Since you cannot control, or predict these in any meaningful way, the pH of your pool is controlled with a buffer.

The Ocean

The ocean is a vast solution containing an incredible number of different compounds and ions. Two of those ions are carbonate (\(CO_3^{-2}\)) and bicarbonate (\(HCO_3^{-1}\)). These two ions, along with the carbonic acid formed from the dissolution of carbon dioxide, create a complex buffer solution that holds the ocean’s pH steady.

The Bloodstream

As your body burns sugars for energy, carbon dioxide is produced. As we know, \(CO_2\) forms carbonic acid when it reacts with water. If that process was unmodulated, the pH of your bloodstream would drop dramatically when you exercised (burning lots of sugars) and would rise as you exhaled. Such an unstable situation would be unworkable for the complex chemistry that keeps us alive. Fortunately, the same complex buffer solution found in the ocean (\(H_2CO_3, ~HCO_3^{-1}, ~and~ CO_3^{-2}\)) is also present in your body. In fact, the tiny variations in pH (generally less than 0.02 points on the pH scale) are how your body regulates breathing. The urge to breathe when you are holding your breath is determined, NOT by the lack of oxygen, but instead by the slight drop in blood pH as the concentration of \(CO_2\) increases.

Country Time Lemonade\(^®\)

Country Time Lemonade mix is advertised as being “not too tart, and not too sweet”. Of course making sure that your lemonade mix is not too sweet is simple. Just add too much sugar. However “tart” is a little different. Tart is just a nice word for sour. That means that Country Time is claiming that their mix will not be too acidic. However, they have no control over the pH of the water in my sink that I use to make their lemonade. To control the pH of my drink mix, they have used a buffer. The first ingredient here is citric acid. The second and third ingredients are potassium and sodium citrate. Citrate ions are, of course, the conjugate base of citric acid.

PolyProtic Titration Graphs

 

Although the mathematics of polyprotic titrations are beyond the scope of this text, we can glean some information from their graphs. 

Let’s start with what it means for an acid to be polyprotic.

By the Bronsted/Lowry definition, an acid is a compound that can donate an \(H^{+1}\) ion to water or to a base. Any compound that can donate more than one \(H^{+1}\) ion is polyprotic. That means that \(H_2SO_4, ~H_2CO_3, ~and ~H_3PO_4\) are all polyprotic acids. This can be seen in the formula where more than on H is found at the beginning.

This is, in fact, why inorganic chemists write the formula of acetic acid as \(HC_2H_3O_2\) with the “H’s” split. Acetic acid is monoprotic, that is, it can only donate ONE H ion, so we write the formula with only ONE H at the beginning.

Just because an acid CAN donate 2 (or more) \(H^{+1}\) ions, doesn’t mean that those ions will be donated at the same time. Polyprotic acids can be thought of as two (or more) consecutive acids. So, sulfurous acid (\(H_2SO_3\)) reacts with water in a two step process:

Step 1: \(H_2SO_3 + H_2O \rightleftharpoons H_3O^+ + HSO_3^{-1}\)

Step 2: \(HSO_3^{-1} + H_2O \rightleftharpoons H_3O^+ + SO_3^{-2}\)

The same two step idea applies during a titration. When a stong base,like NaOH, is added to a solution of sulfurous acid, the hydroxide (\(OH^{-1}\)) first reacts with (and removes) ONE \(H^+\) ion from each acid:

\(H_2SO_3 + OH^{-1} \rightarrow H_2O + HSO_3^{-1}\)

Then, any additional NaOH added will remove the second \(H^+\) from the acid:

\(HSO_3^{-1} + OH^{-1} \rightarrow H_2O + SO_3^{-2}\)

This “two-step” process is visible on a titration graph. The graph will have two equivalence points: the first when the moles of base = the moles of acid. That portion of the process will involve removing ONE \(H^+\) ion from each acid molecule. After that point, the base will begin removing the second \(H^+\) ion from the acid, leading to a second equivalence point.

The graph below was created to show the general shape of a diprotic acid titration graph.

In addition to the two equivalence points, there are two maximum buffer points, since there are (essentially) two different acids (\(H_2SO_3 ~and~ HSO_3^{-1}\))

Notice that the volume required to reach the second equivalence point is exactly twice the volume needed to reach the first equivalence point. That should make sense, since the number of \(H^+\) ions removed in each case is the same.

The graph rises from left to right. Although that seems obvious, it points out several things:

• The first acid (in this case \(H_2SO_3\)) is stronger, and therefore more acidic, than the second acid (\(HSO_3^{-1}\)).
• \(K_{a1}\) (the \(K_a\) for \(H_2SO_3\)) > \(K_{a2}\) (the \(K_a\) for \(HSO_3^{-1}\))

Strong/Weak Titration Graph

 

Let's look at how the pH changes during a weak/strong titration from a graphical perspective. (A strong titration graph is found here) 

We will be looking at this for the problem here: 

50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

We did all of the calculations here, so if you haven't looked through those yet, it would be worth checking out.

If we graph the points from that problem, we get this:

Adding in a few more points to "smooth" things out a bit, we get this:

In simplest terms, at the beginning, when we have excess acid, the solution is acidic (with a low pH). When we to equivalence, the pH becomes 7. Once we pass the equivalence point, the solution contains excess base and the pH is high. The equivalence point is the middle of the steep vertical part of the graph. Stated differently, the equivalence point is the point in the graph where the slope is the most vertical.

We can find that point graphically, by making a graph of the slope vs. volume of base added. For those of you with some level of higher math, this is simply the derivative of the previous graph.

If it helps to visualize what this graph is showing, here are the two graphs on the same axes:

Strong graphs vs. weak graphs

It's worth pointing out the differences between this graph (for a weak titration) and the graph of a strong titration. Here they are side by side:

The first difference to note is that in the strong titration, the pH at the equivalence point is 7, while in the weak titration it is above 7. (For the record, if we had used a weak base with a strong acid, the pH at equivalence would have been below 7, so the key issue here is that it is NOT 7.)

Secondly, the pH starts much lower in the strong titration than it does in the weak titration. This, of course, depends to some extent on the concentration of the acids, but in this comparison, both acids were of equal concentration.

It's also important to note that in the strong titration graph, both the beginning and end of the titration graph are nearly flat, while in the weak titration graph, the beginning of the graph is a more gradual slope. This gently sloped area is the buffer region.

The Math of Weak Titration

 

Let’s work through our weak titration problem. Here it is again: 

50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

Before the Titration

The first part of the question is just an equilibrium problem. Here is our ICE table:

That leads to the following math:

\(2.56x10^{-5} = \frac{(x) \cdot (x)}{(2.00 - x)}\)

Solving for x, gives us: 

\(x = [H_3O^+] = 0.00714 M\). So, pH = -log(0.00714) = 2.15

It is worth noting that this is notably higher than the starting point of the strong titration.

Between the beginning and equivalence

Now we’ll start the real work of the problem. For each step of the problem, we’ll start with the same table set-up we used before to find the moles of the acid (HA) and the conjugate base \((A^{-1})\). All of the problems leading up to the equivalence point have the same structure.

First we work through the reaction of the acid with the strong base:

then, we set up the ICE table

and solve it

\(2.56x10^{-5} = \frac{(x) \cdot (0.0196 + x)}{(1.94 - x)}\)

Solving for x, gives us:

\(x = [H_3O^+] = 0.00227 M\). So, pH = -log(0.00714) = 2.64

Here's the second row:

\(2.56x10^{-5} = \frac{(x) \cdot (0.167 + x)}{(1.50 - x)}\)

\(x = [H_3O^+] = 2.30x10^{-4} M\). So, pH = -log(0.00714) = 3.64

and the third row:

\(2.56x10^{-5} = \frac{(x) \cdot (0.50 + x)}{(0.50 - x)}\)

\(x = [H_3O^+] = 2.56x10^{-5} M\). So, pH = -log(0.00714) = 4.59

and the fourth row:

\(2.56x10^{-5} = \frac{(x) \cdot (0.66 + x)}{(0.0135 - x)}\)

\(x = [H_3O^+] = 5.24x10^{-7} M\). So, pH = -log(0.00714) = 6.28

The equivalence point

Our next point is the equivalence point, but that presents a problem. We cannot do the same math (using \(K_a\)) because there is NO HA present after the acid base reaction. In fact the only thing present (other than water) after the acid/base reaction is the conjugate base \((A^{-1})\)

When that base is present in water, the following reaction can occur:

\(A^{-1} + H_2O \rightleftharpoons HA + OH^{-1}\)

In order to use this reaction, we need a K value. We know that \(K_w = K_a \cdot K_b\), so solving for \(K_b\) gives us the following:

\(K_b = \frac{K_w}{K_a} = \frac{1.00x10^{-14}}{2.56x10^{-5}} = 3.91x10^{-10}\)

We can create a new ICE table with our new reaction:       

We can now solve this for x:

\(3.91x10^{-10}=\frac{(x) \cdot (x)}{(.667-x)}\) which gives \(1.61x10^{-5}\)

We need to remember that \(x=[OH^{-1}]\) not \([H_3O^+]\), so -log(x) will give pOH, rather than pH.

\(-log(1.61x10^{-10}) = 4.79\)

Then, since pH + pOH = 14, pH = 14 - pOH. So pH = 9.21. This is distinctly ABOVE 7, which means that the equivalence point is NOT neutral.

Beyond equivalence

Past the equivalence point, the problem changes again. Since we have run out of acid (at the equivalence point), any additional amount of base added after that will remain in solution. Since this base is strong, the pH will depend ONLY on this excess base.

\(pOH = -log(\frac{moles~excess~strong~base}{total~volume}) = -log(\frac{0.020 moles}{0.170 L}) = 0.93\)

Therefore the pH (14 - pOH) = 13.07

If this seems like a bit of a “cheat” at the end of a difficult problem, it is. The “truth” is that you should really do another ICE table, using \(K_b\) and the new values.

However, because \(K_b\) is very small (\(10^{-10}\)) and because there is already an amount of product present, the “x” will be very small. Since the \([OH^{-1}]\) = 0.118 + x, if that “x” value is very small, it will not have a mathematically significant effect on the final calculated pOH. 

Solve it yourself if you aren’t sure!

A note about buffers and weak titration

During a weak titration, a buffer solution is present for part of the process. Specifically, between the time you have added about 10% of the base needed to reach equivalence until about  90% of the equivalence volume, the solution behaves as a buffer. That means for those points, you could use the Henderson/Hasselbalch equation. In the problem above, since 100 mL of base were required to reach equivalence, the H/H equation to find the pH between 10 mL and 90 mL. 

How Weak Titration Differs from Strong Titration

Here, we are assuming that you have already looked at the mathematical discussion for a strong titration (that is, a titration with a strong acid and a strong base). As on that page, we will be looking at the titration of an acid with a base, but in this case, we will be using a weak acid (with a known \K_a value) and a strong base. It is, of course, also possible to use a weak base and a strong acid. The math is very similar in that case. Although, in the lab, you can also add a weak base to a weak acid, the mathematics of that situation is significantly more complex and beyond the scope of this text. 

For this discussion, we’ll focus our thoughts on this problem:

50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

Again, let’s consider what is happening in the flask. At the beginning, there is only acid in the flask, so the pH is relatively low, although to figure out what it actually is, we will need to solve an equilibrium problem.

When we start to add the base, a reaction will occur between the acid and base. Since the base is strong, that reaction will (effectively) go 100%. This makes for a simple limiting reagent problem. However, after that simple process, we will be left with some weak acid and some conjugate base. To find the pH, we will again need to do an equilibrium problem.

For example, say we have 10 moles of acid in the flask and add 2 moles of base. The following reaction will occur:

\(HA + XOH \rightarrow HOH + XA\)

Since the acid and base react in a 1:1 ratio, we will use up 2 moles of the acid, leaving 8.

\(2~moles~XOH \cdot \frac{1~HA}{1~XOH} = 2~moles~HA~used\)

\(10~moles~HA~originally~present~-2~moles~HA~used = 8~moles~HA~remaining\)

In this case, however, we need to ALSO keep track of the amount of conjugate base produced. Fortunately, this is just equal to the amount of acid used, so in our example, 2 moles \(A^{-1}\)

We will then need to solve for the \([H_3O^+]\) using the equilibrium reaction:

\(HA + H_2O \rightarrow H_3O^+ + A^{-1}\)

In our ICE table, the “I” row would contain the following:

\([HA] = \frac{8 ~moles~ HA}{total ~volume ~in ~liters}\)

\([A^{-1}] = \frac{2 ~moles~ A^{-1}}{total ~volume~ in~ liters}\)

Let’s take a hard look at the math.

Buffers

There is one other important difference between strong and weak titration. Weak titrations involve the production of a buffer solution in in the region between the beginning of the titration and the equivalence point. 

It is possible to do all of the titration calculations without ever thinking about this, but it also means that you can do some of the calculations using the Henderson/Hasselbalch equation.