An acid and base react to form a salt. Therefore when an acid or a base is "neutralized" a salt is formed.

Depending on the acids and bases the salt that is formed can be neutral, acidic, or basic. This is all just a different language for what you have already learned. For example, if formic acid is combined with sodium hydroxide, it generates a salt, sodium formate and water

In this case, the salt is a basic salt since it contains the weak base, formate (HCOO^-) [and the spectator ion Na⁺].

In contrast, if a strong acid and a strong base are combined, like hydrochloric acid and potassium hydroxide you get a neutral salt, potassium chloride

This is because both the strong acid and the strong base result in ions that are merely spectators.

Finally, it is possible to make acidic salts by neutralizing a weak base such as ammonia, NH₃ with a strong acid like HCl

Here the neutralization of NH₃ forms the ammonium ion, NH₄⁺ which is a weak acid. Thus the ammonium chloride salt is acidic.

These salts can be isolated from solution by removing the water. This will leave behind the solid ionic compound.

What are some examples of basic salts? KCN, potassium cyanide. CN^- is the conjugate base of HCN. Na(HCOO), sodium formate. The formate ion, HCOO^- is the conjugate base of formic acid. What are some acidic salts? CH₃NH₃Cl, methylammonium chloride. Methylammonium is the conjugate acid of methylamine, CH₃NH_2.

Concept Question

The salt formed by neutralizing HCN with NaOH will be
(mouse over choices to get answer)

acidic
basic
neutral

Identifying Acids, Bases, and Salts.

You can't get anywhere in an acid/base problems if you can't identify what any of the compounds are. Which are weak acids, which are basic salts, etc...

Here some helpful guidelines to work through in identifying compounds. Rather than simply memorize them, use these ideas to help you think through the identification process.

First, you need to memorize the strong acids and bases. This is key. It will not only allow you to identify the strong acids and bases, but it will help you figure out which ions are simply spectator ions.

Second, you should realize that history has played a role in naming compounds. Therefore there are some compounds we typically think of as acids and some compounds we think of bases. However, every acid has a conjugate base and every base a conjugate acid. So we really have two forms of each compound: one protonated (the acid) and one deprontonated (the base).

Names of acids are typically "something"-acid. As such they are readily identified. Weaks acids are the acids that are not strong. Memorize the strong acids and you'll know everything else is weak. For example, the following are all weak acids: lactic acid, benzoic acid, oxalic acid, trichloroacetic acid, ...The chemical formula for the acids depends on the type of acid (hydroacid, oxoacid, or carboxylic acid). In particular you want to be able to recognize the carboxylic acid group that is found in many organic molecules. It is an oxygen double-bonded to a carbon with an OH group on the same carbon. This is denoted by RCOOH, where the R is a generic representation of the rest of the molecule. For example, formic acid is HCOOH (R = H), acetic acid is CH₃COOH (R = CH₃).

These weak acids are all compounds that are "uncharged" in their protonated state. If we neutralize them with an strong base we will end up with a salt composed of a cation and an anion. The anion is the conjugate base of the acid. This is an important clue to the identity of the species. If the acidic compound is not charged once you remove H⁺ from the compound you will form an anion. Where there is an anion there is always a cation to balance the charge. The cation will be a spectator. Thus the basic salts will be composed of a spectator ion like sodium, Na⁺, potassium, K⁺ or something like that and an anion that is the conjugate base of the acid. They can be identified as they are the same compound as the weak acid but they are "missing" a H⁺. Thus F^- is the conjugate base of HF. NaF is a basic salt. K(CH₃COO), potassium acetate, is a basic salt that contains the acetate anion that is the deprotantated form of acetic acid. Their names are also helpful in identification. The names of the anions often end in "ate". The benzoate ion is deprotonated benzoic acid. The formate ion is the conjugate base of formic acid.

Weak bases are typically compounds that are like substituted ammonia, NH₃. In these compounds the H's have been replaced with something else (like a carbon chain). For example, methyl amine, is CH₃NH₂. One H has been replaced with a CH₃. There is a near endless list of such compounds. As a class, they are called "amines". Thus "something something" amine is identifiable as a weak base. What about their conjugate acids? Since the amines have no charge, if they accept a proton they will add H⁺. This means their conjugated acids will be positively charged and will form salts with spectator anions. Therefore something like methylammonium chloride (CH₃NH₃)Cl can be identified as an acidic salt.

What about salts that combine a basic anion and an acidic cation? Consider for example, ammoniumfluoride (NH₄F). Here the ammonium ion (NH₄⁺) is an acid and the fluoride ion (F^-) is a base. These "mixed" salts are best thought of simply as either acids or bases.

There are also compounds with more than one acidic proton (polyprotic species). These will form anions that can be both acids and bases. These species are called, amphiprotic. They can be identified as anions that still have acidic protons. For example, HCO₃^-. This is carbonic acid, H₂CO₃ that has lost one proton. It is both an acid and base. This will be covered more in detail with polyprotic acids.

Finally, there are a few exceptions to these ideas. First there are neutral salts, like NaCl. In this case, both the cation and the anion are the conjugate pair of a strong acid and strong base. As such, a solution of these ions will be neutral. If you have memorized the strong acids and strong bases, these can quickly be identified. Second, there are a number of insoluble hydroxides, like Al(OH)₃ that are weak simply since they don't dissolve in water. They don't really have a conjugate acid. They are weak simply as a result of solubility.

What is the best way to learn to identify compounds? Practice. There are two worksheets on identifying acids and bases and salts on the worksheet page.

Concept Question

The salt NaCN is
(mouse over choices to get answer)

acidic
basic
neutral

Neutralization Reactions

It is critical in acid/base chemistry to first determine the majority of the chemical species that are in the solution. This is particularly true when mixing two solutions together. Once you know the dominate species, you can then worry about solving the equilibrium problem to determine any small concentrations of interest (such as the pH).

As such, when mixing two solutions together, you need to first look at any neutralization reaction to figure out what will (for the most part) remain in solution. A neutralization reaction is the reaction of an acid and base. In particular strong acids will always react in the presence of any base. Similarly strong bases will always react ion the presence of any acid.

Which side does this equilibrium favor? Note: This is the reverse reaction for the reaction of putting acetate (as weak base) into water. Therefore, this reaction strongly favors the righthand side of the reaction. We can assume this reaction goes 100% to the right. This reaction forms the salt sodium formate, Na(HCOO). We will see later that this salt is basic (since it forms a basic solution when placed in water). If we wanted to know the concentrations in a solution formed by mixing equal parts of formic acid and sodium hydroxide it would be the same as solving for the concentrations in a solution of sodium formate. This is because neutralizing formic acid with sodium hydroxide creates a solution of sodium formate.

To determine what is present after mixing any two acid/base solutions, we must realize that it is not possible to simultaneously have high concentrations of certain species.

The simplest case is the "neutralization" reaction when you have exactly the same amount of acid and base. That is neither the acid nor the base is in excess. They will react until one or the other of them is gone from the solution. In the case of perfect "neutralization" they will both be gone and you'll end up with 100% products. Then you can work the equilibrium problem. Note: for weak acids and weak bases neutralization does not end up forming a solution with a neutral pH

This is the procedure you want to use for all neutralization reactions. First react the H₃O⁺ and any base (weak or strong). Alternatively you would react OH^- and any acid (weak or strong). Next use the limiting reagent to determine what reactants (if any) will remain in solution. When you are finished, you should have either no remaining H₃O⁺ or no remaining base . Alternatively you should have no remaining OH^- or no remaining acid (or neither of either one). Then you can look at the solution and decide what type of solution you have. The remaining solution will fit into one of five categories:

You already know how to solve for the equilibrium concentrations of the first four types of solution. We will soon cover the buffer situation.

Let's look at the neutralization reactions for a generic weak acid HA (BH⁺). This would occur by mixing a weak acid solution with that of a strong base. This is the reaction we can assume will go 100% until either all of the HA is reacted or all of the OH^- is reacted.

The neutralization of a weak base, B (A^-), with H₃O⁺ can also be assumed to go 100%.

Here is a Question

If a 100mL of a 1M solution of NaOH is combined with 200 mL of a 0.5M HF solution, which of the following will have the highest concentration?

F^-
HF
OH^-

Buffers

A buffer is a solution that contains substantial amounts of a compound in both its protonated and deprotonated forms. As such, it is "resistant" to pH change upon the addition of strong acid or strong base. This is because the protonated form can neutralize any strong base and the deprotonated form can neutralize any strong acid. When this happens the ratio of protonated to deprotonated changes, but if their initial amounts are significant, then these changes are small. The total amount of acid or base that can be added to the buffer before neutralizing all of one of the forms of the compound is its buffer capacity. The pH of a buffer is generally in a range around the pKa of the acidic form of the compound (or you can think of this as the pOH is around the pKb of the basic form). So if you want to maintain the pH in a specific region, you should use a buffer where the conjugate pair has a pKa near that value.

For example, let's imagine we wanted to make a buffer with a pH around 3.2, you would then like a compound with a pKa near this value. Benzoic acid has a pKa of 3.18 so this would be a good choice. Now you need to make a solution that contains both benzoic acid (the protonated form) and the ion benzoate (the deprotonated form).

There are three ways that we can accomplish this. The first is to mix a solution of benzoic acid with a solution of a salt that contains the benzoate ions (such as sodium benzoate). If we put in comparable numbers of moles of benzoic acid and the benzoate ion we'll end up with a buffer solution. For the pH to exactly match the pKa the concentrations of these two need to be equal. However, the buffer does not need to have exactly equal concentrations of these two. We can make a slightly more acidic buffer with more acid (benzoic acid). We can make a slight more basic buffer with more base (benzoate). Alternatively we can make the buffer by starting with a solution that is benzoic acid and partial neutralizing it to generate the benzoate. So if I have a 1 mole of benzoic acid in solution and I add 0.5 moles of OH^-, the resulting solution will have 0.5 moles of benzoic acid and 0.5 moles of the benzoate ion. The third way is the opposite of this, I can start with a solution that essentially has all of the compound in the deprotonated from of the benzoate ion and I can partially neutralize this with a strong acid.

Note: In neutralizing the solutions to form the buffer, I cannot neutralize all of the acid or all of the base or I won't make a buffer.

Finally, I can make buffers of different capacity by using different amounts of the compounds. Let's take the example above of a solution with 1 mole of benzoic acid that is neutralized with 0.5 moles of OH^-. The resulting buffer contains 0.5 moles of both the protonated and deprotonated species. It has a buffer capacity of 0.5 moles for both acid and base. Conversely, if I started with a solution with 0.1 moles of benzoic acid and neutralized it with 0.05 moles of OH^-, the resulting buffer would have 0.05 moles of each benzoic acid and the benzoate ion. This buffer would had the same pH as the previous example since there are equal concentrations of protonated (acid) and deprotonated (base) forms. However, the capacity of this buffer is 10 times less, since there are 10 times fewer moles of each. The addition of 0.05 moles of strong acid to the first buffer would only make a small change as it would neutralize only ten percent of the base. The addition of 0.05 moles to the second buffers would neutralize all of the base.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the concentrations of a protonated form of a weak acid (base) to the deprotonated form. It can be used to easily calculate (predict) the pH of a buffer where we have substantial concentrations of both the protonated and deprotonated forms of the compound. For simplicity, we'll write it here for the acid, but it could just as easily be written for the base.

where the pKa = -log(K_a). Thus we can see that in a situation in which we have equal amounts of the protonated and deprotonated forms of a conjugate pair, the pH = pKa. Alternatively, you could say that at a pH that is equal to the pKa, these concentrations are equal. This means that for any compound, you can determine if the majority will be in the protonated or deprotonated form at a given pH by comparing the pH of the solution to the pKa. If the pH is less than the pKa, the solution is "too acidic" and the majority of the compound will be in the protonated (acidic) form. If the pH is greater than the pKa, the majority will be in the deprotonated (basic) form. The exact ratio can be obtained from the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation can also be approximated to determine the pH of a buffer from the initial concentrations that I use to make the solution. If one makes a solution with both the protonated and deprotonated forms of a compound, the differences in concentration at equilibrium compared to the initial concentrations will be very small. Thus one can approximate the concentrations at equilibrium as the initial concentrations, in this case you can predict the pH is

where the concentrations of [HA]₀ and [A^-]₀ are the initial concentrations which we have placed into the equilibrium expression (assuming the change is small). Sometimes we show the initial or mixed concentrations as \(C_{\rm HA}\) and \(C_{\rm A-}\) which would give the following H-H equation:

Titrations

A titration is an analytical chemistry technique that is often used to characterize an acid/base solution. In a titration, a strong acid/base of accurate concentration is added stepwise in small amounts (aliquots) to incrementally neutralize the solution. Titration are performed to either measure the concentration of an unknown solution and/or to determine the K_a (K_b) of an unknown acid (base).

During a titration you have two solutions: the analyte and the titrant. The analyte is the "unknown" solution for which you would like to know either the concentration or the equilibrium constant. The titrant is the "known" solution which has a precise and accurate concentration. The analyte can either be an acid or base and it can be either weak or strong. The titrant is generally a strong acid or base. Since the titration is a neutralization, acid analytes are titrated with strong bases. Basic analytes are titrated with strong acids. The titration is performed by slowly adding the titrant to the analyte solution in small amounts called aliquots. After each addition of an aliquot the pH of the solution is measured. This is performed until the solution has essentially experinced the entire range of pH conditions from acidic to basic (or basic to acidic).

As an example, take a strong acid solution as an analyte that is titrated with a strong base. As the equilibrium constant of a strong acid is not of interest, the key to this titration is to accurately measure the concentration of the analyte solution. In this case, the pH of the analyte starts out very low (as the anlyte solution is a strong acid). As strong base is titrated into the solution, the pH increases slightly but in general will not change much. However, at some point the number of moles of base that have been added to the solution will be equal to the number of moles of acid in the original analyte. At this point the pH will change very dramatically as the solution will now be neutral. After this adding any more of the strong base will rapidly make the solution basic. It is the very large change in pH over a small range of the added titrate volume that is the reason to perform a titration. The point at which the number of moles of added base are equal to the number of moles of acid in the analyte solution is called the equivalence point. It is easy to identify this point in a titration because it is the volume at which the pH is rapidly changing. Technically, the equivalence point is where the titration curve exhibits an inflection point. At this point the curve has the steepest slope. The volume at the equivalence point can be used with the known concentration of the titrant to determine how many moles have been added to the solution. At the equivalence point the moles of added base will be equal to the moles of original acid, this allows the determination of the number of moles of original acid. This can then be combined with the original volume of the analyte solution to determine its concentration. In practice it is very important to use small aliquots to accurately determine the exact volume at the equivalence point.

The graph above shows the titration of a 50mL of a strong acid, HBr, of unknown concentration vs a volume of NaOH added. The concentration of the NaOH solution is known to be 0.1M. The equivalence point is at 150 mL. At the equivalence point 0.015 moles of OH^- have been added to the solution (.15L x 0.1 M). This means that the original acid solution contained 0.015 moles of HBr. The concentration of the original solution must have be 0.3M (0.015 moles/0.05 L). (Note: for a strong acid and strong base titration the equivalence point is at a pH=7. This is because at this point you have equal moles of added base as acid in the original solution. Therefore at the equivalence point the solution has formed a neutral salt and the pH is 7).

Titrations of weak acids or bases can also be used to determine the K_a (K_b) of the analyte in addition to the concentration. For example, examine the the titration of an unknown acid with strong base. Initially the analyte has essentially all of the acid in its protonated form. It is simply a solution of a weak acid. As weak acid dissociates only slightly the majority of the compound is in the protonated form. As strong base is titrated into the solution this compound is converted from its protonated state to its deprotonated state. The titration takes the analtye from essentially fully protonated to fully depronated. Now at the equivalence point all of the anlayte will have been neutralized to form a basic salt. Therefore at the equivalence point the solution is essentially that of a weak base and the pH will be greater than 7. This point can again be utilized to calculate the concentration of the analyte solution in the exact same way as was done in the strong acid example above. However now additional information can be deduced from the half-equivalence point. The half equivalence point is when exactly half of the original analyte has been neutralized. This volume is easy to determine form the titration as it is half of the volume at the equivalence point. At this volume the concentration of the protonated form and deprotonated form of the acid are equal. This is a buffer solution. It is also a very special buffer. Since the two species have equal concentration this is the point at which for a weak acid the pH = pK_a. Thus in the titration of a weak acid with a strong base the pK_a of the acid can simply be read off the graph as the pH at the half-equivalence point (to the value of the Henderson-Hasselbach approximation).

The graph above shows the titration of 50 mL of a weak acid solution with 0.1 M NaOH. The equivalence point is clearly at 100 mL of added NaOH. Note that at the equivalence point the pH is basic (~8.8). Since the concentration of the NaOH is 0.1 M we can deduce that the concentration of the original acid solution must have been 0.2 M (equal moles of acid in the original 50 mL solution as in the 100 mL of added 0.1 M strong base). The K_a of the acid can also be determined. The half-equivalence point is the volume that is half the volume at the equivalence point. In this example that would be 50 mL. The pH at this point is 4.75. At this point the system should be a buffer where the pH = pK_a. Thus the pK_a of this acid is 4.75. The K_a is then 1.8 x 10^-5 (10^-4.75).

Titrations of weak bases with strong acids are very similar. Except that at the equivalence point all of the base will have been converted to its conjugate acid. The half-equivalence point will be a basic buffer with the pOH = pK_b.

Here are two help sheets on Titration Curves

Approximate pH formulas

You will begin to notice that we keep solving the same equilibrium problem over and over again with the same approximations.

It is easy at some point to remember these simple results (along with the approximations we made to get here).

We general make two approximations when working acid/base problems. First, we assume the amount of H₃O⁺ and OH^- in pure water are significantly small that we can ignore them. This works well as long as we don't have extremely dilute solutions. Second, we assume that the changes in concentrations for weak acids and bases are small compared to the concentrations. This is the step where we say at equilibrium we have some concentration minus a change that is approximately equal to the initial concentration. This works best when the change is small. That is the equilibrium constant is small. And when the concentration is high. That is the solution is not dilute.

That is the H₃O⁺ concentration is simply the concentration of the initial acid, C_a. To see how this might break down think about the pH of a super dilute solution (10^-35 M HCl is not pH = 35).

When do these go wrong? They are technically always approximate so the real question is how accurate an answer do you want. In general the approximation are better for concentrated solutions (since we are generally ignoring small changes in concentration) and for "weak" equilibria (since we are assuming the changes are small).

Concept Question

For which of the following will an approximate pH calculation be the most accurate?
(touch choices to toggle feedback on/off)

0.1 M HCOOH Ka = 1.8 x 10^-4
2x10^-5 M HCOOH Ka = 1.8 x 10^-4
0.1 M CHCl₂COOH Ka = 3.3 x 10^-2
2x10^-5 M CHCl₂COOH Ka = 3.3 x 10^-2

Thinking About it

With all neutralization problems, it is important to think about the problems systematically. This is what is meant by "thinking like a chemist". Before leaping to a formula, you need to know what you have in solution and what reactions are taking place. Which concentrations are dominant and which ones are very small.

1. Identify all the compounds (acids, bases, strong, weak, spectator ions, ...).

3. Neutralize any strong acids or bases (if there are other bases/acids in solution). This will require looking for the limiting reagent, reacting the compounds to completion, and identifying what remains in solution.

4. After figuring out what is left in the solution, solve the equilibrium. The remaining solution will either be a strong acid, weak acid, buffer, weak base, or strong base solution.

Remember, if you have any H₃O⁺ after neutralization you have a strong acid solution. If you have any OH^- after neutralization you have a strong base solution. If you have substantial amounts of both the protonated and deprotonated forms of a conjugate pair then you have a buffer.