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.