Titrations I - Monoprotic/Monoprotic

When we're dealing with acid-base chemistry, we have to be able to determine the concentration of the acids and bases that we're using. This most often means performing a titration where one of the components is known. The absolutely 100% most important thing about titrations is to remember that titrations are stoichiometry problems, and they should be treated as such. There's nothing new and magical about titrations, they're just stoichiometry problems applied to a specific situation. The 4 steps to solve every stoichiometry problem are:

1) Write a balanced chemical equation

2) Calculate moles of one component (“moles of known”)

3) Convert moles of known to moles of interest using the ratios in the balanced chemical equation

4) Convert moles of interest to whatever you want to find

With acid-base titrations, it is most often convenient to monitor pH of the solution. To explore the pH-dependence of a titration, we can look at a system where we know all the concentrations. Let's say we are titrating 25.00mL of 0.85M nitrous acid, HNO₂(aq), with 0.85M KOH(aq). At the beginning, we have a solution of a weak acid, so we can calculate the initial pH using a K_a expression. Looking up the K_a of nitrous acid online gives a couple different values, let's say that it's around 5x10^-4 . K_a is just like every other equilibrium, so let's set up a table...

	HNO2(aq) +	H2O(l) ↔	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.85M	XXXX	10-7 M	0 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.85 – x) M	XXXX	(10-7 + x) M	x M

Plugging in values:

That expression is solvable by the quadratic formula (and I encourage you to solve it for practice and to convince yourself that the things we're about to do are valid), but we might be able to simplify it if we make some assumptions about the size of “x”. Since this is a weak acid (K_a < 1), it's probably reasonable to assume that most of the HNO₂ molecules will still be intact when this system reaches equilibrium, so although we will lose some (Δ [ ] = -x), the value of “x” will probably be quite small compared to 0.85, so we can assume that (0.85 – x) ≈ 0.85. At the same time, although “x” is probably small, 10^-7 is really small, so there's a pretty good chance that (10^-7 + x) ≈ x. With these two assumptions (which we have to check later) in place, the K_a expression simplifies to:

x = 0.0206

Before we do anything else, we need to pause to check that the assumptions we made are indeed reasonable. 0.0206 is truly massive compared to 10^-7 so the second assumption is great. For the first assumption, it's a little closer; 0.0206 is certainly smaller than 0.85, but is it smaller enough?

(0.0206 / 0.85 ) *100 = 2.4%

Usually, the limit is around 5%, so we should be OK here as well. The pH of this solution at the beginning of the experiment should be:

pH = -log[H₃O⁺] = -log(0.0206) = 1.686

What happens to the pH when we start adding KOH(aq)? Well, KOH is a base, so the pH will go up, but how will it go up? We can calculate that. Starting with 25.00mL of 0.85M nitrous acid, let's add 5.00mL of 0.85M KOH(aq). To make it easier to keep track of everything, let's start by writing out a chemical equation and converting to moles:

HNO₂(aq) + KOH(aq) ↔ H₂O(l) + KNO₂(aq)

Mols HNO₂: (0.02500L)(0.85M) = 0.02125mols HNO₂

Mols KOH(aq): (0.00500L)(0.85M) = 0.00425mols KOH

We can assume that all of the OH^-1(aq) that is added will react with the H⁺/HNO₂ that is present in solution, so after this 5.00mL addition, we should have a solution that contains:

0.02125 – 0.00425 = 0.017mols HNO₂(aq)

This nitrous acid is now in (25.00 + 5.00 = 30.00mL) of solution {if we assume that the volumes are additive}, so the concentration of nitrous acid is:

0.017mols / 0.03000L = 0.5667M

And the concentration of nitrite ions is:

0.00425mols / 0.03000L = 0.1417M

We can calculate the pH by treating this like an equilibrium problem, just like above, setting up a table. The pH of the solution should be the same as if we had made a new solution by adding 0.017mols of HNO₂ and 0.00425mols of NO₂^-1 to enough water to make 30.00mL of solution:

	HNO2(aq) +	H2O(l) ↔	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.5667M	XXXX	10-7 M	0.1417 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.5667 – x) M	XXXX	(10-7 + x) M	(0.1417 + x) M

Making similar assumptions, the expression simplifies to:

x = 0.002000

Checking assumptions again, they are all valid, so:

pH = -log[H₃O⁺] = -log(0.002000) = 2.699

We can continue adding 5.00mL portions of the KOH(aq) and calculating to get the values shown in the table:

mL KOH(aq) added	pH
0.00	1.686
5.00	2.699
10.00	3.125
15.00	3.477
20.00	3.903

But what happens when 25.00mL of the KOH(aq) solution is added? At that point, the mols of OH^-1 that have been added is exactly equal to the mols of acid that were present in the original solution. This is an equivalence point in the titration. If we look at the balanced chemical equation for the process,

HNO₂(aq) + KOH(aq) ↔ H₂O(l) + KNO₂(aq)

The solution we have produced by titration is the exact same solution that we would have produced if we had simply dissolved 0.02125mols of KNO₂ in enough water to make 50.00mL of solution. Since NO₂^-1(aq) is the conjugate base of nitrous acid, we can think about it being involved in a K_b equilibrium with water, K_b(NO₂^-1) = 2x10^-11. Set up a table...

	NO2-1(aq) +	H2O(l) ↔	OH-1(aq) +	HNO2(aq)
[ ]initial	0.425M	XXXX	10-7 M	0 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.425 – x) M	XXXX	(10-7 + x) M	x M

Again, we can make assumptions similar to the above examples to get x = 2.92x10^-6 = [OH^-1].

pOH = -log[OH^-1] = -log(2.92x10^-6) = 5.535

pH = 14 - pOH = 8.465

If we keep adding KOH(aq), the mixture will continue to get more basic, eventually leveling off as the concentration of excess hydroxide reaches a limit. {Sounds like a fascinating calculus problem, the concentration should asymptotically approach 0.85...} If we plot this pH data, we get a titration curve like the one shown below.

Measuring the Acidity of a Solution

It is often useful and necessary to measure the acidity or basicity of a solution. There are a number of ways this could be described or reported, but in the context of the Bronsted-Lowry definitions of acids and bases, it is probably most convenient to monitor the concentration of H⁺(aq). The [H⁺] in a solution can be a very small number. Although modern pocket calculators can readily handle very small numbers, it's a little easier to describe these small concentrations using pH. pH expresses very small concentrations without having to use scientific notation and will be useful for a number of quantities.

pH = -log[H₃O⁺]

Reversing and re-solving that expression:

[H₃O⁺] = 10^-pH

Why do we follow [H⁺] or [H₃O⁺]? We know from the K_w expression that [H₃O⁺] and [OH^-] are related. Depending upon the type of problem, sometimes it's easier to think in terms of [OH^-], but [OH^-] is also usually a very small number, so it's useful to define an analogous quantity, pOH:

pOH = -log[OH^-]

Reversing and re-solving that expression:

[OH^-] = 10^-pOH

pH and pOH are related by K_w and we can derive that expression from K_w:

K_w = [H₃O⁺][OH^-]

-log(K_w) = -log([H₃O⁺][OH^-])

If we generalize that “-log” can be replaced by “p”, and remember that log AB = log A + log B, then:

pK_w = -log [H₃O⁺] + (-log[OH^-]) = pH + pOH

At 25°C, K_w = 10^-14, so pK_w = 14.

Strengths of Acids and Bases

If an acid is an H⁺-donor, then the stronger an acid is, the more effectively it will be able to donate H⁺. If that acid is in aqueous solution, we can think of “strength” by the following equilibrium for the generic acid, HA(aq):

HA(aq) + H₂O(l) ↔ H₃O⁺(aq) + A^-(aq)

In equilibrium terms, the stronger an acid is, the more product-favored the equilibrium will be. Since this equilibrium expression can be applied to any acid, and acids are an important and diverse class of compounds, we define this equilibrium as an acid dissociation equilibrium and call its corresponding equilibrium constant the acid dissociation constant, or K_a.:

Similarly, we can think of the strength of a base using the equilibrium for the generic base, B(aq):

B(aq) + H₂O(l) ↔ OH^-(aq) + BH⁺(aq)

With its corresponding base dissociation constant, Kb:

Looking at the K_a equilibrium equation, the water that appears on the reactant side is accepting H⁺ to become H₃O⁺(aq). If water is accepting a proton, it is acting as a Bronsted-Lowry base. Considering the reverse reaction, H₃O⁺(aq) is a proton-donor so it is acting as an acid, while A^-(aq) is accepting a proton as B-L base. These acids and bases are not independent of each other, they are conjugate acid-base pairs. A^-(aq) is the conjugate base of HA(aq), and HA(aq) is the conjugate acid of A^-(aq); H₂O(aq) is the conjugate base of H₃O⁺(aq), and H₃O⁺(aq) is the conjugate acid of H₂O(aq). A conjugate acid-base pair are related by the addition or removal of a single H⁺. The same relationships can be described for the K_b equilibrium expression.

In the K_a equilibrium, water is acting as a base, while in the K_b equilibrium, water is acting as an acid. So what is water, an acid or a base? The answer is BOTH! The acid or base behavior of a substance is dependent upon its environment because acid and base are relative terms. In the case of water, if the water molecules are interacting with something that is more acidic than water, then water will act as a base. Likewise, if the water molecules are interacting with something that is more basic than water, the water will act as an acid. This brings up an interesting question: is water an acid or a base when it's not interacting with any other substance? Consider the following equilibrium:

H₂O(l) + H₂O(l) ↔ H₃O⁺(aq) + OH^-(aq)

Once again, water is acting as both an acid and a base. This process is called autoionization. Since water is such an important substance for life on Earth, this equilibrium also has a specific letter assigned to it, K_w, the autoionization constant for water:

Like almost all equilibria, K_w is dependent upon temperature. At 25°C, K_w = 10^-14. For pure water, this means that at 25°C, [H₃O⁺] = [OH^-] = 10^-7M. If this equilibrium constant only applied to pure water, it would be interesting but of limited use. Fortunately, it can be applied to any relatively dilute aqueous solution to understand the relative amounts of hydronium and hydroxide ions present in the solution. Consider an acid, HA(aq), and its conjugate base, A^-(aq), both interacting with water:

HA(aq) + H₂O(l) ↔ H₃O⁺(aq) + A^-(aq)

A^-(aq) + H₂O(l) ↔ OH^-(aq) + HA(aq)

If these equilibrium equations are added together, the result is the K_w equilibrium equation. When two (or more) sequential equilibria are added together, the equilibrium constant for the overall process is the product of the equilibrium constants for the individual steps. This means that for any conjugate acid – conjugate base pair, K_a x K_b = K_w. This also implies a general relationship, the stronger an acid is, the weaker its conjugate base, and vice versa.

This has been a lot of acid and base information wrapped up in a big discussion of equilibrium. K_a, K_b, and K_w all describe specific systems, but the most important thing to remember is that at their core, these are all equilibrium constants. They behave like every other equilibrium constant, they follow all the same rules as every other equilibrium constant, and they can be manipulated just like any other equilibrium constant. The only thing special about them is that they refer to a specific type of chemical equation.

Acids and Bases - Definitions

A lot of the behaviors and properties of acids and bases can be described by equilibrium.  Before we get into that, we need to establish some definitions.  In Gen Chem 1, we tried to recognize acids by looking for H⁺ and bases by looking for OH^-.  That's a good start, and is the basis for the Arrhenius definitions of acids and bases:

Acid = H⁺-donor

Base = OH^--donor

These very brief definitions work, but we quickly run into a problem.  Aqueous ammonia is a base.  Aqueous ammonia {NH₃(aq)} does not have a OH^- to donate.  Arg.  Fortunately, there's a qualifier there, this is aqueous ammonia, and if there's water around, we can use it:

NH₃(aq) + H₂O(l)  <=>  NH₄⁺(aq) + OH^-(aq)

So NH₃(aq) can be forced to fit the simple definition above by using water.  In fact, a more proper and complete version of the Arrhenius definitions of acids and bases is:

Acid = a substance that, when dissolved in water, increases the concentration of H⁺(aq)

Base = a substance that, when dissolved in water, increases the concentration of OH^-(aq)

These definitions are a little better, but they still seem a bit restrictive.  To make definitions that are a little more general, let's look at that ammonia equation again.  In that equilibrium, ammonia is "increasing the concentration of OH^-(aq)" by removing H⁺ from water.  If we're trying to understand the function of acids and bases, it might be nice to follow one consistent thing around, so maybe a "better" definition of acids and bases could be:

Acid = H⁺-donor

Base = H⁺-acceptor

These are the Bronsted-Lowry definitions of acids and bases, and they are the definitions we will use most often in Gen Chem 2.  It's important to note that the Bronsted-Lowry definitions and the Arrhenius definitions are (and must be) consistent with each other.  It wouldn't be very helpful if 1 definition called something a base while the other definition called that exact same substance an acid.

Since we're dealing with definitions here, what should we call "H⁺(aq)"?  "H plus" works, but we're also going to run across a few other descriptions.  If we think about the subatomic particles present in "H⁺(aq)", there's 1 proton (the atomic number of hydrogen, all hydrogens have to have 1 and only 1 proton), the "+1" charge means that the single electron of a hydrogen atom has been removed, and in the most common isotope of hydrogen there are zero neutrons.  This means that, in terms of subatomic particles, "H⁺(aq)" is just a proton, and that is what it is very often called in discussions of acids and bases.  Acids are "proton donors" and bases are "proton acceptors.
If we think in the other direction, a proton is a pretty concentrated little lump of positive charge.  If this little lump of positive charge is floating around in a polar solvent like water, it's probably going to attract (or be attracted to) the negative end of the water dipole.  We can write a chemical equation to describe this:

H⁺(aq) + H₂O(l)  <=>  H₃O⁺(aq)

"H₃O⁺(aq)" is called a "hydronium ion".  {For practice, draw Lewis structures of all the species in that equation.}
So when we're talking about acids and bases, we're likely to encounter a number of different descriptions of the same thing.  To keep them straight, remember that:

"proton" = H⁺(aq) = H₃O⁺(aq)

Now that we have a few definitions in place, what else can we do to set the table for acids and bases...