# The Equilibria of Acid-Base Solutions

## Strong Acid and Base Accounting 101

In this section we will tackle the difficult challenge of calculating equilibrium concentrations of acid/base reactions. Fear not. We'll begin with the strong acids and bases. They almost fully dissociate into conjugate acids or bases and H_{3}O

^{+}or OH

^{-}. This makes calculation of the final equilibrium concentrations of H

_{3}O

^{+}or OH

^{-}a little easier for all of us.

Let's start with the strong acid HNO

_{3}(nitric acid) as an example. HNO

_{3}has one proton that it can transfer. Acids that have a single dissociable proton are called

**. It's kind of like how train tracks with only one rail are called**

*mono*protic*mono*rails. Makes us want to plan a vacation to Disney World. Other examples of monoprotic acids are HCl and HClO

_{4}. Because HNO

_{3}is a strong monoprotic acid, the amount of H

_{3}O

^{+}produced when HNO

_{3}is added to water will be essentially equal to the amount of HNO

_{3}added. The chemical equation is drawn this way:

The equation above tells us that if we were to add 0.1 M of HNO

_{3}to water, we would be left with basically zero HNO

_{3}and have 0.1 M H

_{3}O

^{+}and 0.1 M NO

_{3}

^{-}. Notice that we are ignoring the dissociation of water itself (remember,

*K*

_{w}= 1 × 10

^{-14}at 25 °C). We ignore water in this particular case because its dissociation into H

_{3}O

^{+}and OH

^{-}is negligible compared to the ~100% dissociation of 0.1 M HNO

_{3}. While 0.1 M HNO

_{3}gives us 0.1 M H

_{3}O

^{+}, the dissociation of water would only give us a tiny fraction of that concentration of H

_{3}O

^{+}. We'll be ignoring water in related scenarios in the next sections, just a heads up.

The pH of a solution of a strong monoprotic acid, like the HNO

_{3}example above, is calculated by taking the negative log of the H

_{3}O

^{+}concentration that results from the approximately 100% dissociation of the acid. In the example above the pH of the resulting solution would be 1 [=-log(0.1 M)].

Unlike HNO

_{3}, some strong acids can be

**diprotic**. Diprotic acids have two dissociable protons. One example is H

_{2}SO

_{4}. Unfortunately, calculating the pH of a diprotic acid is more complicated because usually only one proton dissociates fully and the other proton is much less than ~100% dissociated. We'll deal with these guys later.

Strong bases can be dealt with much like the strong acid example above. The most common strong bases are alkali metal hydroxides, like NaOH and KOH that follow the general equilibrium equation:

Just like a strong acid, a strong base will nearly fully dissociate in water and the concentration of strong base added will give the same concentration of OH

^{-}and M

^{+}as products. Therefore, a 0.1 M solution of NaOH will contain zero NaOH and 0.1 M OH

^{-}and 0.1 M Na

^{+}. The pH would be the pOH subtracted from 14 at 25 °C. The pH value in the NaOH example above would be 13.

Some alkaline earth metal hydroxides can release two OH

^{-}groups into solution. Many strong bases that have two hydroxide groups, like Mg(OH)

_{2}, will release both hydroxide groups when dissolved in water, like so:

If we add 0.001 M Mg(OH)

_{2}to water, we'll get out 0.002 M of OH

^{-}because both hydroxides dissociate. Many of these types of alkaline earth metal hydroxides are relatively insoluble so it's hard to achieve high concentrations of them in solution. For example, only up to about 0.01 g of Mg(OH)

_{2}will dissolve in 1 L of water. If anymore Mg(OH)

_{2}were added it would stay as solid Mg(OH)

_{2}and not release OH

^{-}.

Strong bases known as alkaline earth metal oxides are also capable of releasing not one but two hydroxides into solution. These heavy-duty bases first release O

^{2-}, which then goes on to react with water to form two hydroxides. Here's the visual:

An example of a strong base alkaline earth metal oxide is CaO (calcium oxide), which dissociates fully into Ca

^{2+}and O

^{2-}. As shown above, O

^{2-}will then go on to form two hydroxides.

### Brain Snack

CaO (or quicklime) is a component of Greek fire, which was used by the Greeks to pen its enemies.### That's So Weak

Solutions of weak acids present a bigger challenge for us in terms of calculating equilibrium concentrations. Remember that a strong acid or base will always be ~100% dissociated, whereas a weak acid or base is only partially dissociated. If a strong acid (or base) solution has the same equilibrium concentrations of H_{3}O

^{+}or OH

^{-}as a weak acid (or base) solution then it's purely coincidental. To calculate the equilibrium concentrations of a weak acid or base solution we'll have use equilibrium constants and some math.

We've already dealt with the equilibrium of one weak acid (or base)—our old friend water. Like we did for water, we can characterize the equilibrium of a given weak acid (or base) using an equilibrium constant:

For the chemical equation above, the formal equilibrium constant is:

Here, the concentration of H

_{2}O is much larger relative to all the other concentrations and will not change much during the equilibration. Therefore, we can lump it in with

*K*

_{c}to give us the definition of the

**acid-dissociation constant**,

*K*

_{a}:

The value of the acid-dissociation constant tells us exactly how much a weak acid will dissociate. It is a direct indication of the strength of any acid. As an aside, chemists don't typically quote

*K*

_{a}values to communicate the strength of an acid. Instead, they take the negative log of the

*K*

_{a}value to calculate what's called the p

*K*

_{a}of an acid. The lower the value of the p

*K*

_{a}of an acid, the stronger the acid is. Think of it like golf scores; the lower the score of a golfer, the better at golf the player is. Check out the table below for a list of acids, their acid-dissociation constants, and their corresponding p

*K*

_{a}values.

With the acid-dissociation constant now in our arsenal, we'll forge ahead with an example of how to calculate equilibrium concentrations of a weak acid solution. Let's think about a 0.25 M solution of CH

_{3}COOH. If we put 0.25 M of CH

_{3}COOH into water we'll get four possible species at different concentrations. These concentrations will depend on the

*K*

_{a}value of CH

_{3}COOH. The possible species are CH

_{3}COOH, CH

_{3}COO

^{-}, H

_{3}O

^{+}, and OH

^{-}(we'll still have H

_{2}O around, too).

To begin this problem, we'll list out the possible acid-base reactions that could be happening. First, since CH

_{3}COOH is a weak acid, one reaction will be the transfer of the CH

_{3}COOH proton to water. Second, H

_{2}O can dissociate so we have to keep track of that reaction as well.

Now that we have the equations in front of us, we need to compare dissociation constants to determine which reaction will make significant contributions to concentrations of each species. The acid-dissociation constant for CH

_{3}COOH is more than a billion times larger than the dissociation constant for water. Therefore, the H

_{3}O

^{+}released from water will be practically zero and all the H

_{3}O

^{+}will come from CH

_{3}COOH. The CH

_{3}COOH reaction is called the

**principle reaction.**The water reaction in this case is called the

**subsidiary reaction.**To calculate the equilibrium concentration of H

_{3}O

^{+},we'll ignore the subsidiary reaction and only focus on the principle reaction.

To start, we'll construct a table that lists the initial concentrations, the change in those concentrations as the solution equilibrates, and the final equilibrium concentrations:

Notice that we're not including water here. No offense to water, but it's so concentrated relative to the other molecules that we know its concentration isn't going to change significantly.

Now we can solve for

*x*using the relationship between the acid-dissociation constant for CH

_{3}COOH (1.8 × 10

^{-5}M) and the equilibrium concentrations:

We could launch into solving this equation using the quadratic formula. However, making an assumption can make our life simpler: that

*x*will be much smaller than 0.25 M.

In this particular problem, notice that the

*K*

_{a}of the principle reaction is really small and the reaction will not proceed that far to the right. Therefore, the equilibrium concentration of products will be very small relative to the starting concentrations. For now we'll assume (0.25 M –

*x*) ≈ 0.25 M. Now the equation is much easier to solve for

*x*:

**Check Those Assumptions—You'll Thank Yourself Later**

We can do a quick check at this point to make sure our assumption above is okay. The value for

*x*of 0.0020 M, when subtracted from 0.25 M, does not give us a difference that is significant. Yep, our assumption is okay. It's always good to do this check because for some problems the assumption will not be okay and we'll have to use the quadratic formula to solve for

*x*.

With the value of

*x*solved, we now know the concentrations of three of the species involved in the equilibrium. Since both [CH

_{3}COO

^{-}] and [H

_{3}O

^{+}] equal

*x*, we have established their concentrations at equilibrium to be 0.0020 M. We also know that the concentration of CH

_{3}COOH doesn't significantly change—at equilibrium it's still 0.25 M. Also, since the CH

_{3}COOH dissociation reaction is the primary reaction the concentration of H

_{3}O

^{+}is fully determined by this reaction alone. The subsidiary reaction of water dissociation does not significantly contribute to the H

_{3}O

^{+}concentration.

However, the subsidiary reaction is the only reaction that produces OH

^{-}molecules so it looks like we still have some work to do. To calculate the equilibrium concentration of OH

^{-}we'll call up one of our favorite equilibrium constants,

*K*

_{w}. For any solution the product of the equilibrium concentrations of H

_{3}O

^{+}and OH

^{-}must equal

*K*

_{w}.

*K*

_{w}= [H

_{3}O

^{+}][OH

^{-}]

The value of

*K*

_{w}(at 25 °C) is 1.0 × 10

^{-14}M and we solved for the concentration of H

_{3}O

^{+}above. Substituting these values into the

*K*

_{w}relationship and solving for [OH

^{-}] gives a value for [OH

^{-}] of 5.0 × 10

^{-12}M. Now we can check that the dissociation of water is indeed the subsidiary reaction when it comes to H

_{3}O

^{+}production.

For every one molecule of OH

^{-}that is produced when water dissociates, one molecule of H

_{3}O

^{+}is produced. As a result, the dissociation of water only gives 5.0 × 10

^{-12}M of OH

^{-}as well as H

_{3}O

^{+}. This H

_{3}O

^{+}concentration is really small compared to the concentration of H

_{3}O

^{+}produced from the dissociation of CH

_{3}COOH, which jives perfectly with our original expectation that the dissociation of water would not significantly influence the concentration of H

_{3}O

^{+}at equilibrium.

Lastly, we can calculate the pH of this 0.25 M CH

_{3}COOH solution by taking the negative log of the H

_{3}O

^{+}concentration that we calculated.

pH = -log([H

_{3}O

^{+}]) = -log(0.0020 M) = 2.7.

As a reality check, notice that the pH is well below 7. This means the solution is acidic. This is exactly what we'd would expect from putting a weak acid into water. Another handy metric chemists sometimes use to describe weak acid (or base) solutions is called

**percent dissociation**. The percent dissociation of a solution is defined as the concentration of weak acid (or base) dissociated at equilibrium divided by the initial concentration of weak acid (or base) added to the solution. Here's the percent dissociation for the CH

_{3}COOH example we worked through above:

The percent dissociation value reflects the strength of the acid solution. However, watch out because the percent dissociation value also depends on the concentration of acid added to the solution. Go back and resolve the CH

_{3}COOH example above but start with 1.00 M as your initial concentration of CH

_{3}COOH. We can wait.

Did you get 0.42%? This shows that the more concentrated the initial acid concentration is, the less percent dissociation we'll get.

The same approach we used to calculate equilibrium concentrations in a weak acid solution can be used to calculate equilibrium concentrations in a weak base solution. First, you'll want to identify which species will be present at equilibrium. Then we need to identify the principle reaction— that's the reaction that has the larger dissociation constant—and identify the subsidiary reaction. Finally, set up a table like we did above and calculate the equilibrium concentrations using the relationship between dissociation constants and equilibrium concentrations.

For weak bases, the dissociation constant describing the extent of proton uptake is defined as the

**base-dissociation**

**constant**(which is sometimes more accurately called the base-protonation constant):

Again, the concentration of H

_{2}O is left out of the relationship. In dilute solutions like the ones we've been dealing with, the concentration of water does not significantly change.

### It's All Related

There's an overarching relationship between the acid-dissociation constant and the base-dissociation constant that you'll want to imprint on your neural circuitry. The dissociation constants for an acid-base conjugate pair when multiplied must equal the dissociation constant of water. Let's look at an example using the weak base hydrazine to see this:The sum of the two reactions, after canceling species that are on both sides of the equilibrium arrows, results in a net equation representing the dissociation of water:

Keep in mind that when chemical equations are added, the equilibrium constant for the resulting net equation is equal to the product of the equilibrium constants for the individual reactions that were added. Let's prove this with the dissociation constants for hydrazine shown above:

Now we have it. The product of the acid and base dissociation constants equals the dissociation constant for the net reaction, which is simply the dissociation of water.

This relationship holds for any conjugate acid-base pair:

*K*

_{a}×

*K*

_{b}=

*K*

_{w}The acid/base dissociation constant relationship tells us that the stronger the acid, the weaker its conjugate base. Inversly, the stronger the base, the weaker the conjugate acid. Keep in mind that we can now calculate

*K*

_{b}for a particular base if we're given the

*K*

_{a}of it's conjugate acid. (Hint, hint.)