Acids and Bases together

So far, we only been looking at solutions of either acids or bases. When you mix acids and bases, however, you have to account for the effects of each acid on the $[H^+]$ levels.

As you know that acids and bases have opposite effects on $[H^+]$. This means that when you mix an acid and a base, the two can counteract one another’s effect on the $pH$. Let’s go through and see what happens in the most basic form, the combination of a strong acid and a strong base.

Reviewed hypothetical 25 °C final-volume concentration model: the added analytical NaOH and HCl amounts correspond to 5.00 M and 4.00 M in the same final volume. Treat activities as ideal. Find pH after neutralization.

This is a pretty high $pH$, but it’s actually the $pH$ of a solution that is $1M\text{ in }NaOH$. This is true for all solutions that have concentrations of strong acids and bases. Therefore a solution that is $4M \text{ in }HCl$ and $4M\text{ in }NaOH$ will have a $pH=7$. This is because if you combine the three involved equations we had in that solution, you get a neutralization reaction.

$NaOH\leftrightarrow Na^+ + OH^-$
$HCl\leftrightarrow H^++Cl^-$
$H_2O\leftrightarrow H^++OH^-$

$NaOH+HCl\leftrightarrow Na^++H_2O+Cl^-$

You might see a familiar face present after this neutralization reaction. Table Salt, $NaCl$, is the salt formed by this reaction. In fact, the definition of a salt is an ionic solid formed by and acid-base neutralization reaction.

Another great use of the neutralization reactions is its applications in titrations. A titration is a laboratory procedure in which you add a precise amount of some solution of known concentration to a precise amount of another solution of unknown concentration. The end goal of the procedure is to fully react the two solutions, reaching the equivalence point, the point where the two chemicals are stoichiometrically equal in solution according to their reaction equation.

Leaving it in such generic terms makes that hard to follow, doesn’t it? Let’s go ahead and look at one of the most common titrations: Acid-Base Titration.

As was pointed out previously, acids and bases react with one another in a neutralization reaction. For example:

$NH_3+HCl\leftrightarrow NH_4Cl$

Equivalence means that titrant and analyte have reacted in the stoichiometric ratio of the balanced equation. For NH₃+H⁺→NH₄⁺ this ratio is 1:1. It does not mean their remaining equilibrium concentrations are equal. The endpoint is the observed signal used to stop a titration and should be close to equivalence.

Typically you can’t perfectly identify the equivalence point, but you can get pretty close. This is why your end point will not quite be the same point as your equivalence point, but it’s pretty close.

So how do we determine the equivalence point of an Acid-Base titration?

If we want to identify the equivalence point, we have to first determine what is going on in the solution. Let’s keep with the Ammonia and Hydrochloric Acid titration. We need to determine which one is the known concentration and which is the unknown.

If you can choose what you are performing a titration with, it’s always easier if the known solution is a strong acid or base. This helps to more easily identify the equivalence point, as you will notice later.

As such, we will declare that in this titration, we know the concentration of the strong Hydrochloric acid and we are trying to determine the concentration of the weak base Ammonia. We will be adding the known to the unknown, therefore our titrant, the solution being added, is Hydrochloric acid. Our other player is Ammonia, which is acting as our analyte.

So what is going to happen as we add Hydrochloric acid to an Ammonia solution?

Well, we know that before we add any acid, a solution of ammonia is a weak base, so our $pH$ will be elevated. As it is a weak base, it means that the Ammonia will exist as both Ammonia and the Ammonium Ion ($NH_3\text{ and}NH_4^+$). The relevant reaction is:

$NH_3+H_2O\leftrightarrow NH_4^++OH^-$

Once we start adding a source of $[H^+]$ from $HCl$, this reaction changes to become a neutralization reaction:

$NH_3+HCl\leftrightarrow NH_4Cl$

The increased $[H^+]$ also reacts with the Hydroxide ion generated from the base to form water:

$H^++OH^-\leftrightarrow H_2O$

Now as we add the $HCl$, it will slowly decrease the $pH$ until there is no more base left to neutralize. From that point on, as $HCl$ is a strong acid, it will quickly drop the $pH$, as it is simply adding $[H^+]$ to the solution. Therefore, we can identify the equivalence point as the point where the $pH$ begins to drop considerably faster.

Now, if we keep going, eventually the $pH$ will start approaching the $pH$ of the titrant itself, but as it has been diluted, it won’t quite reach there. Therefore, if we draw up a graph of the $pH$ as a function of the volume of titrant added, it would have a shape something like the graph in Interactive 7.2.1.

This particular example curve is distinct from other titration curves, as it begins with a high $pH$ and tends towards a very low $pH$. This is because it is the curve of a titration of a weak base with a strong acid. If you look at the titration curves that result from other situations, they look a bit different. Those examples are shown in Gallery 7.2.1.

The general point of a titration is to identify the sharp drop-off or climb in $pH$ that occurs around the equivalence point. As you track $pH$ whilst adding your titrant, you should see a sharp change in $pH$ occur after a certain amount, that quickly slows. That jump means you have reached your equivalence point.

Read the equivalence volume from the titration curve and use the balanced reaction to calculate the original analyte amount. Equal moles apply to a 1:1 reaction; other coefficients require a corresponding mole ratio.

For a A+b B→products, equivalence satisfies n(A)/a=n(B)/b.

For solutions, C(A)V(A)/a=C(B)V(B)/b. The familiar C_titrantV_titrant=C_analyteV_analyte applies only to a 1:1 stoichiometric reaction.

On the titration curve, we identify the equivalence point as the middle of the jump that appears. The above relations only hold true at that point. From that point, you can read $V_{titrant}$, and you should already know $C_{titrant}$ and $V_{analyte}$. Using these values, you can find the concentration of the analyte.

Example Problem 7.2.1:
You are titrating 750mL of an acetic acid solution with 10M Potassium Hydroxide. The titration curve is shown below. What is the concentration of acetic acid?

Example Problem 7.2.2:
What volume of 4M $NaOH$ is needed to completely neutralize 800mL of 15M $HCl$ ? What is the $pH$ of the neutralized solution?