Wednesday, November 25, 2009

PH OF ACID AND BASE

pH Calculations and Problem-Solving

In order to understand acid-base chemistry, we will need to familiarize ourselves with the appropriate jargon. This section covers the skills necessary to process information on acid and base solutions and solve non- buffered acid-base problems. There are few new chemical ideas presented, but the focus is on problem-solving strategies. Our goal is to become comfortable with such acid-base tools as pH-value conversions and calculations of dissociation in aqueous solutions. The section starts with simple calculations of the pH of strong acid or strong base solutions. We will then move into more difficult situations that deal with mixtures of weak acids and weak bases and hydrolysis reactions. Finally we discuss how to calculate the pH of a polyprotic acid solution.

Terms

Acid - A substance that has the potential to donate a proton or accept an electron pair.
Acidic - Having a pH less than 7 or a pOH greater than 7.
Base - A substance that can accept a proton, release OH-, or donate an electron pair.
Basic - Having a pH greater than 7 or a pOH less than 7.
Buffer - A solution composed of an acid and its conjugate base that serves to moderate the pH of the solution.
Conjugate Acid - A molecule that can be described as a base that has gained one proton.
Conjugate Base - A molecule that can be described as an acid that has lost one proton.
Dissociate - Separate into its ion constituents.
Hydrolysis - A reaction that modifies a water molecule. In acid-base chemistry this usually refers to the reaction of a solute which changes the pH of an aqueous solution.
pH - A measure of the hydrogen ion concentration, it is equal to - log [H+].
pK a - A measure of the strength of an acid, it is equal to - log K a, where K a is the acid dissociation constant in water.
pK b - A measure of the strength of a base, it is equal to log K b, where K b is the base dissociation constant in water.
pOH - A measure of the hydroxide ion concentration, it is equal to - log [OH- ].
Polyprotic Acid - An acid that can donate more than one proton.
Strong Acid - An acid with a pK a less than zero. Strong acids completely dissociate in water.
Strong Base - A base with a pK b less than zero. Strong bases completely dissociate in water.
Weak Acid - An acid with a pK a greater than zero. Weak acids do not completely dissociate in water.
Weak Base - A basewith a pK b greater than zero. Weak bases do not completely dissociate in water.

The pH of Non-Buffered Solutions

Calculating the pH of Strong Acid and Strong Base Solutions

When a strong acid or a strong base is added to water, it nearly completely dissociates into its ion constituents because it has a pK a or pK b less than zero. For example, a solution of H2SO4 in water contains mostly H+ and SO4 2-, and almost no H2SO4 is left undissolved. This makes calculating the pH of a strong acid or strong base solution exceedingly simple--the concentration of acid equals the concentration of H+. Recall that pH is computed by taking the negative log of of [H+]. Common strong acids that should be memorized include HCl (hydrochloric), HNO3 (nitric), HClO4(perchloric), and H2SO4 (sulfuric). Strong bases include Group I hydroxides (LiOH, NaOH, KOH, etc.) and Group II hydroxides except for Be(OH)2 and Ba(OH)2.


Calculating the pH of Weak Acid and Weak Base Solutions


Calculating the pH of weak acid and weak base solutions is much more complicated than the above case--weak acids and bases do not completely dissociate in aqueous solution but are in equilibrium with their dissociated forms. Therefore, we must apply what we know about equilibria to solve these types of problems. For example, let's calculate the pH of a 0.10 M solution of acetic acid in water. To do this, we first write down the equilibrium involved and the expression for the equilibrium constant:

Figure %: Note that H+ represents H3O+ in the equilibrium constant expression.

Next, you should compile a table of values for the concentration of all species involved in the equilibrium. We already know that the initial concentration, [ ]o, of acetic acid is 0.10 M and that the initial concentration of H+ is 10-7 (since the solvent is neutral water). Even though there is an initial concentration of H+ in solution, it is so small compared to the amount produced by the acid that it is usually ignored. From the stoichiometry of the reaction, one mole of H+ and one mole of acetate (Ac-) are produced for each mole of acetic acid dissociated. Therefore, if we denote the amount of acetic acid dissociated as x, the final concentrations of H+ and Ac- are both x, and the final concentration of HAc in solution is 0.10 M - x. This data is summarized in :

Figure %: Table of equilibrium values for the dissociation of acetic acid

After you have compiled the table of values, you can substitute the equilibrium concentration values for each species into your expression forK a as shown below:

You should note that the above equation is a quadratic equation in x and therefore requires use of the quadratic equation to solve for x. If, however, we can make the assumption that [HAc]o - x = [HAc]o, the equation becomes much easier to solve. We can do this if HAc is a weak enough acid that it dissociates very little, and the change in [HAc] is negligible.

Is this assumption valid? Solving the quadratic equation using the quadratic formula gives a pH of 2.88. The difference of 0.01 pH unit is small enough to be insignificant, so the assumption is valid in this case and will certainly save you some time on a test. We can make the approximation [HA]o - x = [HA]o so long as x is less than 5% of the initial concentration of HA. X will be greater than 5% of [HA]o with stronger weak acids at low concentrations. Consider these guidelines when you decide whether or not to make the approximation: you can simplify the quadratic equation if the solution is more concentrated than 0.01 M and the pK a of the acid is greater than 3.

The pH of a weak base solution is calculated in the same manner as that of a weak acid solution, using K b instead of a K a.

To calculate the pH of a mixture of acids in aqueous solution, first decide which acid has the lowest pK a. Calculate the pH as if the strongest acid were the only one in solution. We can ignore the contributions to the pH of the weaker acids because they will be minor in comparison to that of the strongest acid in the group. The exact solution to such a problem requires more complex mathematics and will not be covered in this SparkNote.

Hydrolysis Reactions

A salt of a strong acid and a strong base (such as NaCl from HCl and NaOH) produces a neutral solution when dissolved in water. However, when a salt of a weak acid and a strong base (e.g. NaAc from acetic acid and NaOH), a strong acid and a weak base (e.g. NH4Cl from ammonia and HCl), or a weak acid and a weak base (NH4Ac) is dissolved in water, the solution does not have a neutral pH. These phenomena are explained by the reaction of the salts of weak acids and weak bases with water in hydrolysis reactions. As shown in , these hydrolysis reactions produce the parent weak acids and weak bases of the salts:

Figure %: Hydrolysis reactions

As you can see in the figure above, the salt of a weak acid, such as acetate ion, acts as a base in water, and the salt of a weak base, such as ammonium ion, acts as a weak acid. From our previous discussion on the reactions of acids and bases with water in Disassociations, you should know that the K b of acetate ion can be calculated from the K a of acetic acid, and that the K a of ammonium ion can be calculated from the K b of ammonia, as shown in .

Another type of hydrolysis reaction comes from the reaction of metal ions with high charges. Such ions act as Lewis acids to water molecules, as shown in . A metal ion can bond to water by accepting an lone pair from the oxygen of a water molecule, and this increases the acidity of water molecule. Like other acids, we can calculate the K a and calculate the pH of a solution containing such ions.

Figure %: Why metal ions decrease the pH of aqueous solutions

To calculate the pH of a solution containing the salt of a weak acid or a weak base, treat the problem exactly as you did when calculating the pH of weak acid and base solutions above in Calculating pH's, Heading . Mixtures of salts of weak acids and weak bases present a challenging mathematical problem that we will not cover in our treatment of acid-base chemistry.

Polyprotic Acids

So far we have dealt with acids that donate only one proton per molecule. However, this is not the case for polyprotic acids--acids that can donate more than one proton per molecule. Two key features of polyprotic acids are that they lose their protons in a stepwise manner and that each proton is characterized by a different pK a. The factors contributing to the pK a of each acidic proton in a polyprotic species are the same factors that determine the relative acidity of monoprotic acids--the dominant factor is strength of the acid-H bond. Consider, for example, the triprotic acid H3PO4shown in :

Figure %: Acidity of phosphoric acid

As each proton is lost from phosphoric acid, the phosphorous becomes more electron rich, and less electron withdrawing. Therefore, the loss of each proton strengthens the O-H bond and increases the pK a of the phosphate species. This trend is evident in the pK a data given in . In general, it is true that K a1, K a2, K a3, and so on, for polyprotic acids.

As you may have guessed, calculating the pH of a polyprotic acid solution is not as simple as it is for monoprotic acids. In fact, it is quite a messy problem. However, that mess can be quickly cleaned up by making the assumption, as we did for a mixture of acids, that only the strongest acid (i.e. only the first dissociation) has a significant effect on the pH. Making that assumption turns the problem into one you already know how to solve--calculating the pH of a weak acid solution.


Buffers

Buffered solutions are quite important in chemical and biological systems. A buffer allows for the maintenance of a fairly narrow range of pH even while another reaction is producing acids or bases. Because a buffer is a mixture of a weak acid and its conjugate base, it can react with either acid or base to remove the acid or base from solution. The functional pH range of a buffer is described in the Henderson-Hasselbalch equation.

In order to solve problems with buffered solutions, it is essential to have an intuitive grasp of which acid-base reactions are most likely to occur. In this section, we will learn how to determine which reactions are most likely to occur in buffer problems. The following material is focused on problem solving and will require close attention to details in problems and solutions presented.

Terms

Acid - A substance that has the potential to donate a proton or accept an electron pair.
Acidic - Having a pH less than 7.
Base - A substance that can accept a proton, release OH-, or donate an electron pair.
Basic - Having a pH greater than 7.
Buffer - A solution composed of an acid and its conjugate base that serves to moderate the pH of the solution.
Conjugate Acid - A molecule that can be described as a base that has gained one proton.
Conjugate Base - A molecule that can be described as an acid that has lost one proton.
Dissociate - Separate into its ion constituents.
Indicator - A molecule whose conjugate acid or conjugate base has a different color.
pH - A measure of the hydrogen ion concentration, it is equal to - log [H+].
pK a - A measure of the strength of an acid, it is equal to – log K a, where K a is the acid dissociation constant in water.
pK b - A measure of the strength of a base, it is equal to – log K b, where K b is the base dissociation constant in water.
Strong Acid - An acid with a pK a less than zero. Strong acids completely dissociate in water.
Strong Base - A base with a pK b less than zero. Strong bases completely dissociate in water.
Titration - An experiment that neutralizes an unknown amount of acid or base with a known volume and concentration of acid or base to determine the amount of unknown acid or base.
Weak Acid - An acid with a pK a greater than zero. Weak acids do not completely dissociate in water.
Weak Base - A base with a pK b greater than zero. Weak bases do not completely dissociate in water.

Buffered Solutions

How Buffers Work

As you have seen in calculating the pH of solutions, only a small amount of a strong acid is necessary to drastically alter the pH. For certain experiments, however, it is desirable to keep a fairly constant pH while acids or bases are added to the solution either by reaction or by the experimenter. Buffers are designed to fill that role. Chemists use buffers routinely to moderate the pH of a reaction. Biology finds manifold uses for buffers which range from controlling blood pH to ensuring that urine does not reach painfully acidic levels.

A buffer is simply a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffers work by reacting with any added acid or base to control the pH. For example, let's consider the action of a buffer composed of the weak base ammonia, NH3, and its conjugate acid, NH4 +. When HCl is added to that buffer, the NH3 "soaks up" the acid's proton to become NH4 +. Because that proton is locked up in the ammonium ion, it proton does not serve to significantly increase the pH of the solution. When NaOH is added to the same buffer, the ammonium ion donates a proton to the base to become ammonia and water. Here the buffer also serves to neutralize the base.

As the above example shows, a buffer works by replacing a strong acid or base with a weak one. The strong acid's proton is replaced by ammonium ion, a weak acid. The strong base OH- was replaced by the weak base ammonia. These replacements of strong acids and bases for weaker ones give buffers their extraordinary ability to moderate pH.

Calculating the pH of Buffered Solutions

Buffers must be chosen for the appropriate pH range that they are called on to control. The pH range of a buffered solution is given by the Henderson- Hasselbalch equation. For the purpose of the derivation, we will imagine a buffer composed of an acid, HA, and its conjugate base, A-. We know that the acid dissociation constant pK a of the acid is given by this expression:

The equation can be rearranged as follows:

Taking the -log of this expression and rearranging the terms to make each one positive gives the Henderson-Hasselbalch equation:

Figure %: The Henderson-Hasselbalch Equation

Note that the sample species HA and A- in the above Expression are generalized to the terms acid and base, respectively. To use the equation, place the concentration of the acidic buffer species where the equation says "acid" and place the concentration of the basic buffer species where the equation calls for "base". It is essential that you use the pK a of the acidic species and not the pK b of the basic species when working with basic buffers--many students forget this point when doing buffer problems.

A buffer problem can be fairly simple to solve, provided you don't get confused by all the other chemistry you know. For example, let's calculate the pH of a solution that is 0.5 M acetic acid and 0.5 sodium acetate both before and after enough SO3 gas is dissolved to make the solution 0.1 M in sulfuric acid. Before the acid is added, we can use the Henderson-Hasselbalch equation to calculate the pH.

Figure %: Note that the pK a of acetic acid is 4.75.

This part of the problem does not require us to do the sort of equilibrium calculations that we must use for Non-Buffered Solutions, but many students still try to do it the hard way. The hard way is a correct way of doing the problem, but it may cost you valuable time on a test.

To calculate the pH after the acid is added, we assume that the acid reacts with the base in solution and that the reaction has a 100% yield. Therefore, we say that 0.1 moles per liter of acetate ion reacts with 0.1 moles per liter of sulfuric acid to give 0.1 moles per liter of acetic acid and hydrogen sulfate. Here, we ignore the second dissociation of sulfuric acid because it is minor in comparison to the first. So the final concentration of acetic acid is 0.6 M and acetate is 0.4M. Plugging those values into the Henderson-Hasselbalch equation gives a pH of 4.57. Note that a 0.1 M solution of strong acid would give to a pH of 1 but the buffer gives a pH of 4.57 instead.

To probe the useful range of the buffer, let's calculate the pH of the solution resulting from the same situation above but with different concentrations of the buffer. If the buffer is 1.0 M in both acetate and acetic acid, then the pH of the resulting solution after the introduction of acid is 4.66. However, if we make the solution only 0.11 M in acetic acid and acetate, then we calculate a pH of 3.45! Therefore, if you want a more effective buffer, make sure that the concentration of the buffering agents is large in comparison to the added acid or base.


Titration

Throughout the series of SparkNotes on acidsbases, we have focused on calculating the pH of solutions. Now, we'll take a look at a more practical side of acid-base chemistry--how to measure the pH of a solution. The experiment that measures the pH of a solution is called a titration. Titrations involve the addition of a known amount of an acid or base to neutralize an unknown amount of acid or base. From the amount of acid needed to neutralize the base, or base to neutralize the acid, one can calculate the pH of the original solution. To detect the endpoint of a titration, an indicator is used. An indicator is an acid or base whose conjugate acid or conjugate base has a different color from that of the initial form. As the pH of the solution comes to equal the pK a of the indicator, a sharp color change is observed. We will examine how the shift of the indicator to its conjugate form is related to the Henderson-Hasselbalch equation.

An interesting property of titration is the titration curve--a plot of solution pH versus titrant volume. We can extract useful information from the titration curve of a solution. In this SparkNote, we will study the different types of titration curves and discuss the chemistry behind them.

Acid-Base Titrations

The Titration Experiment

Titration is a general class of experiment where a known property of one solution is used to infer an unknown property of another solution. In acid-base chemistry, we often use titration to determine the pH of a certain solution.

A setup for the titration of an acid with a base is shown in :

Figure %: A titration setup

We use this instrumentation to calculate the amount of unknown acid in the receiving flask by measuring the amount of base, or titrant, it takes to neutralize the acid. There are two major ways to know when the solution has been neutralized. The first uses a pH meter in the receiving flask adding base slowly until the pH reads exactly 7. The second method uses an indicator. An indicator is an acid or base whose conjugate acid or conjugate base has a color different from that of the original compound. The color changes when the solution contains a 1:1 mixture of the differently colored forms of the indicator. As you know from the Henderson-Hasselbalch equation, the pH equals the pK a of the indicator at the endpoint of the indicator. Since we know the pH of the solution and the volume of titrant added, we can then deduce how much base was needed to neutralize the unknown sample.

Titration Curves

A titration curve is drawn by plotting data attained during a titration, titrant volume on the x-axis and pH on the y-axis. The titration curve serves to profile the unknown solution. In the shape of the curve lies much chemistry and an interesting summary of what we have learned so far about acids and bases.

The titration of a strong acid with a strong base produces the following titration curve:

Figure %: Titration curve of a strong base titrating a strong acid

Note the sharp transition region near the equivalence point on the . Also remember that the equivalence point for a strong acid-strong base titration curve is exactly 7 because the salt produced does not undergo anyhydrolysis reactions.

However, if a strong base is used to titrate a weak acid, the pH at the equivalence point will not be 7. There is a lag in reaching the equivalence point, as some of the weak acid is converted to its conjugate base. You should recognize the pair of a weak acid and its conjugate base as a buffer. In , we see the resultant lag that precedes the equivalence point, called the buffering region. In the buffering region, it takes a large amount of NaOH to produce a small change in the pH of the receiving solution.

Figure %: Titration curve of a strong base titrating a weak acid

Because the conjugate base is basic, the pH will be greater than 7 at the equivalence point. You will need to calculate the pH using the Henderson-Hasselbalch equation, and inputting the pK b and concentration of the conjugate base of the weak acid.

The titration of a base with an acid produces a flipped-over version of the titration curve of an acid with a base. pH is decreased upon addition of the acid.

Note that the pH of a solution at the equivalence point has nothing to do with the volume of titrant necessary to reach the equivalence point; it is a property inherent to the composition of the solution. The pH at the equivalence point is calculated in the same manner used to calculate the pH of weak base solutions in Calculating pH's.

When polyprotic acids are titrated with strong bases, there are multiple equivalence points. The titration curve of a polyprotic acid shows an equivalence point for the each protonation:

Figure %: Titration curve of a strong base titrating a polyprotic acid

The titration curve shown above is for a diprotic acid such as H2SO4 and is not unlike two stacked . For a diprotic acid, there are two buffering regions and two equivalence points. This proves the earlier assertion that polyprotic acids lose their protons in a stepwise manner.


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