Equilibrium 21. The Equilibrium StateIn the previous (stoichiometry) section we assumed that reactions proceeded until one or more of the reactants was completely consumed; for many reactions this works well. However, in some reactions there is a significant back reaction. An example is the reaction of nitrogen and hydrogen to form ammonia. Note the double arrow, which indicates the occurence of both a forward and back reaction.

N2(g) + 3 H2(g) 2 NH3(g)

Here nitrogen and hydrogen are gradually consumed and ammonia is formed until, as shown at the dashed line on the right, the concentrations cease to change. At this point the rates of the forward reaction and the reverse reaction are equal. The system is then said to be at equilibrium.

2. The Equilibrium ConstantThe extent to which a reaction proceeds before it reaches equilibrium is characterized by an expression known as the mass action expression. This is set equal to a quantity known as K, the equilibrium constant.

The mass action expression is a fraction with the products on the top and the reactants on the bottom, each raised to an exponent equal to its coefficient in the balanced equation. Thus for the formation of ammonia (above):

In expressions like the one above, concentrations of gases are ususally given as pressures, with units of atmospheres. However the brackets shown above indicate that the concentrations are in moles per liter. Brackets should not be used for pressures, but sometimes they are.

It bears repeating that equilibrium is a dynamic and not a static process. In the rection A B, A is constantly being converted to B and B is constantly being converted to A. The equilibrium concentrations are those at which the rates of the forward and back reactions are equal. What makes equilibrium seem static is that since the rates of the forward and back reactions are equal, the reactant (and product) concentrations do not change.

3. Gas Phase Equilibrium CalculationsConsider the ammonia equilibrium, shown above. Start with N2 = 0.35 atm; H2 = 0.25 atm; and NH3 = 0.40 atm. To help set up the calculation, we will use something called an ICE diagram.

Figure 1. A reaction reaching equilibrium

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An ICE diagram gives the pressure or concentrations of each material present in the reaction. It gives the initial concentration, the change in concentration and the concentration at equilibrium.In most cases the Initial concentrations are given; the Changes are obtained from the stoichiometry of the reaction; the Equilibrium values are obtained by adding the Change to the Initial.

Thus the mass action expression would be written as:

K=¿¿

Of course this is mathematically rather messy. So let's try another equilibrium calculation which works out more easily.

Consider the reaction 2 A + B 2 C.

The system starts with 0.50 atm. of A, 0.50 atm. of B and 0.00 atm of C. If the system ends up with 0.20 atm of C, what is the value of K? First we set up an ICE table. Here the changes are determined from the fact that 0.20 atm. of C are formed.

Now we use the equilibrium pressures from the ICE chart to calculate the value of K.K=¿¿

Now consider the reaction

H2(g) + I2(g) 2 HI(g)

The system starts with [H2] = [I2] = 0.50 atm. and no HI. If K = 0.0100 we calculate:

Take the square root of both sides

2x = 0.050 - 0.10 x and x = 0.024

4. The Mathematics of Equilibrium ConstantsAlthough units are not usually given in equilibrim constants, most do have units associated with them. Consider, for example:

Here K will have the units of atm-2 or kPa-2.

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N2 H2 NH3

Initial 0.35 atm. 0.25 atm. 0.40 atm.

Change -x -3x +2x

Equilibrium 0.35 - x 0.25 - 3x 0.40 + 2x

A B C

Initial 0.50 0.50 0

Change -0.20 -0.10 +0.20

Equilibrium 0.30 0.40 0.20

H2(g) I2(g) HI(g)

Initial 0.50 0.50 0

Change -x -x +2x

Equilibrium 0.50 - x 0.50 - x 2x

However, suppose that the concentrations were given in moles/L instead of atm. Now the value of K must be in units of L2/mol2. Thus a gas phase equilibrium can have two different equilibrium constants. The term KP is used when the units are atmospheres (pressure) and the term KC is used when the units are moles/L (concentration). If your concentrations are in one set of units and your equilibrium constant uses another, K will have to be changed. This can be done with the equation:

Here the gas constant, R, is 0.08206 L atm/mol K (or 8.31 L kPa/mol), T is the Kelvin temperature, and Δn is the change in the number of moles of gas as the reaction is written. For the formation of ammonia, Δn is -2. This formula is given in the AP Chemistry formula sheet.

In addition to changing the units of equilibrium constants, you must also know when and how to combine them. Suppose we want to combine the reactions:

A + B C K = 50A + C 2D K = 300

If we combine (add) the two reactions we get a third reaction:2 A + B 2 D

What is K for the combined reaction?

The rule is that when you add reactions, you multiply the equilibrium constants. So K for the third reaction would be 15,000. If you subtract the second reaction from the first, then you would divide the first K by the second. If you double a reaction, you square its K. If you reverse a reaction, the new K is the reciprocal of the K for the forward reaction. Thus:

H2 + X2 2 HX K = 2500½ H2 + ½ X2 HX K = 50

Although we say that the products are on the top of the mass action expression, an equilibrium does not have a reactant or product since it goes in both directions. What we actually mean is that the compounds on the right side of the equation are on the top and the compounds on the left side of the equation are on the bottom.

To summarize what is used in the mass action expression:- the concentration of a solute is given in moles/L- the concentration of a gas is either moles/L or atmospheres- the concentration of an undissolved solid is exactly one.

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5. Le Chatelier's PrincipleLe Chatelier's Principle says that "if a stress is applied to a system at equilibrium, the system adjusts to reduce the stress." After you are accustomed to using Le Chatelier's Principle, it seems so obvious that one wonders why it had to be given a name. Yet it often shows up on the multiple choice portion of every AP Chemistry test.

Consider the reaction:N2 + 3 H2 2 NH3 + heat

Adding ammonia to a mixture of nitrogen, hydrogen and ammonia at equilibrium will place a stress on the system. The system will relieve this stress by converting some of the ammonia to nitrogen and hydrogen.

Adding nitrogen to the system will cause the equilibrium to shift to the right, consuming some of the added nitrogen as well as some of the hydrogen you started with.

If the temperature is raised, the system will remove some of the excess heat by shifting to the left. If the volume of the system is decreased, the equilibrium will shift to the right, toward fewer moles of gas, in order to remove some of the excess pressure.

A common "trick" is to ask about raising the pressure by adding a nonreactive gas (e.g. argon). Although the total pressure is increased, the partial pressures of the reactants are not affected and the equilibrium is not shifted. Except for changing the temperature, none of the changes we have discussed affect the value of K. Yet it would require a change of K in order to alter the concentrations of any of the reactants.

Consider the equilibrium in which an excess of solid naphthalene (moth balls) is in equilibrium with its vapor. How is the vapor concentration affected by adding more naphthalene? It isn't! Whether a system contains more or less solid, the concentration of the solid is still considered to be one. It hasn't changed.

6. Solubility EquilibriaWhen a soluble or slightly soluble salt is added to water it dissolves initially. Eventually, however, the solution reaches its capacity and further dissolution stops. What has happened is that the rate at which the salt dissolves has become equal to the rate at which the dissolved salt precipitates. We say that the solution is now "saturated." The system has reached equilibrium.

Since salts break into their component ions when they dissolve, this must be seen in the mass action expression. Consider, for example, the dissolution of lead(II) chloride in water:

PbCl2(s) Pb2+(aq) + 2 Cl–(aq)(Note that you don't get Cl2 from PbCl2). The appropriate mass action expression would be:

However, since lead chloride is a solid, it will have a concentration of unity and thus can be removed as a term in the equation. The equilibrium constant for a dissolution reaction is normally referred to as the "solubility product" and is abbreviated KSP. Thus:

Ksp = [Pb+2][Cl-]2

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In addition to the solubility product, a second quantity is often used in solubility calculations. This is the molar solubility, the solubility of the substance in moles per liter. The simplest solubility calculations are those which involve the conversion of the molar solubility to solubility product and the solubility product to the molar solubility. Be certain you know the difference!

Given that the KSP of lead chloride is 1.6  10–5, what is its molar solubility? If the molar solubility of lead chloride is x, then in a saturated solution the concentration of Pb2+ is x and the concentration of Cl– must be 2x. Plugging these terms into the mass action expression gives:

Ksp = (x)(2x)2

1.6  10–5 = 4x3

solubility = x = 0.016 M

Be careful that when you calculate the value of (2x)2 you remember to square both the "2" and the "x." An astonishing number of students make mistakes here.

Given that its molar solubility is 1.0  10–15, what is the KSP of Bi2S3? [Bi+3] = 2 1.0  10–15 = 2.0  10–15 M[S-2] = 3 1.0  10–15 = 3.0  10–15 M

Therefore:Ksp = [Bi+3]2[S-2]3 = [2.0  10–15]2[3.0  10–15]3 = 108  10–75 = 1.1  10–73

To make it easy for the grader to give you partial credit, always write out the mass action expression and then plug in the numbers.

7. The Common Ion EffectIf a 0.10 M sodium fluoride solution is used to dissolve calcium fluoride, substantially less calcium fluoride will dissolve than would have dissolved in pure water. This is caused by fluoride, an ion which is common to both compounds. This phenomenon is therefore known as the common ion effect. Let us calculate how much calcium fluoride actually dissolves.

If the molar solubility of calcium fluoride is x, we have:[Ca+2] = x and [F-] = 0.10 + 2x

This gives the following mass action expression:Ksp = 4.0  10–11 = (x)(0.10 + 2x)2

We can solve the cubic equation or we can make the simplifying assumption that 0.10 >> 2x. If we make this assumption, as we usually do, the 2x can be neglected and the equation becomes:

4.0  10–11 = (x)(0.10)2

x = [Ca2+] = 4.0  10–9 M

Was neglecting 2x a valid assumption? The criterion we will use is called "the 5% rule." If a variable is less than 5% of the constant to which it is added, the variable can be neglected. Is2x = 2 4.0  10–9 less than 5% of 0.10? It certainly is!

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Most of the difficulties which you will have with solubility products can be summarized in the following questions. Why did we double the x but not double 0.10? When do you double the fluoride concentration? The answer is..............never. You are not doubling the fluoride concentration. You are doubling x, the molar solubility, in order to get the fluoride concentration.

8. Selective PrecipitationIn these problems two different compounds precipitate from the same solution. There are all sorts of questions which can be asked. Consider, for example, a solution containing 0.010 M Ag+ solution and 0.010 M Ba2+. You are told that for silver chromate Ksp = 9.0  10–12 and for barium chromate Ksp = 8.5  10–11.

1) Na2CrO4 is added gradually and with stirring. Assuming that the chromate solution does not affect the volume, which compound precipitates first? We will imagine the chromate starting at zero and gradually rising until one compound — silver chromate or barium chromate — starts to precipitate. To do this we calculate the chromate concentration at which the precipitation starts.

For BaCrO4

Ksp = [Ba2+][CrO42–]

8.5  10–11 = (0.010)(x)x = [CrO4

2–] = 8.5  10–9 M

For Ag2CrO4

Ksp = [Ag+]2[CrO42–]

9.0  10–12 = (0.010)2(x)x = [CrO4

2–] = 9.0  10–8 M

The answer is barium chromate because it starts to precipitate at a lower chromate concentration.

2) What is the concentration of the first cation to precipitate when the second cation has just started to come out of solution?

Since the first cation to precipitate is barium,the second cation must be silver. We previously calculated that when silver chromate starts to precipitate the chromate concentration will be 9.0  10–

8 M. What barium concentration is consistent with that? We can calculate that as follows:Ksp = [Ba2+][CrO4

2–]8.5  10–11 = [Ba2+](9.0  10–8)[Ba2+] = 9.4  10–4 M

9. Mixtures of Two Solutions and Bounce BackA common type of solubility problem involves the mixing of two solutions which might or might not form a precipitate. The first question you will have to answer is whether a precipitate forms.

Suppose, for example, we mix 40 mL of 6.5  10–3 M Pb(NO3)2 with 60 mL of 0.30 M KI solution. Knowing that for PbI2 the solubility product, KSP = 1.4  10–8, does a precipitate form? To answer that question we must first calculate the concentrations of Pb2+ and I– after dilution.

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“Bounce back,” which you use when two solutions are mixed to form a precipitate, has three steps.1. Dilution2. Reaction3. Equilibration

If we assume that the volumes of the two solutions are additive, which we always do, then the final volume of the solution will be 100 mL. Thus, after dilution:¿

¿

Then take the concentrations of the diluted ions and use them to calculate Q, the equilibrium quotient. You may recall the Q has the same form as K but consists of concentrations as they really are and not the equilibrium values. So at equilibrium Q = K.

Q = [Pb2+][I–]2 = [2.6  10–3][0.18]2 = 8.42  10–7

Since Q > K, that is 8.42  10–7 > 1.4  10–8, a precipitate forms.

Next calculate the concentrations of ions after the precipitation has occured and the precipitate has been removed. It would not work to set up an equation like:

1.4  10–8 = (2.6  10–3 - x)(0.18 - 2x)2

This gives x = 2.6  10–3 and, therefore, [Pb2+]= 0. Certainly the lead concentration is close to zero, but it is not exactly zero and the exact value is what we need to calculate. The calculation is done by a method called "bounce back."

In bounce back we assume that the concentration of limiting reactant, in this case Pb2+, really does go all the way to zero. If necessary we correct the concentration of the other ion, iodide in this case. Then we calculate how much of the slightly soluble salt redissolves. The concentration of iodide is 0.18 - (2.6  10–3)/2 = 0.17 M. So we calculate the concentration of Pb2+ using the mass action expression.

1.4  10–8 = [Pb2+][I–]2 = [Pb2+](0.17)2

[Pb2+] = 4.8  10–5 M

You may also be asked the concentrations of the other ions, K+ and NO–

3 in this case. To calculate them, go back to the initial concentrations after dilution. If after dilution but before reaction [Pb2+] = 2.6  10–3, then [NO–

3] = 2 2.6  10–3 = 5.2  10–3. Similarly the K+ concentration in KI solution will be equal to the I– concentration before precipitation occurs. Thus [K+] = 0.18 M.

10. Strong Acids and BasesExamples of strong acids are H2SO4, HNO3, HCl, HBr, and HI. (There are a few others, e.g. HClO4, which we will not discuss.) Since strong acids dissociate completely in water, the H+ concentration (or the H3O+ concentration) is equal to the concentration of acid. The exception is sulfuric acid which, since it is diprotic, gives an H+ concentration which is as twice the acid concentration (or would be twice the concentration if the second proton came off completely).Similarly strong bases, the hydroxides of Group IA and Group IIA metals, dissociate completely. So in a solution of potassium hydroxide, the hydroxide concentration is equal to the concentration of potassium hydroxide. In a solution of calcium hydroxide, Ca(OH)2, the hydroxide concentration is

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twice the concentration of calcium hydroxide.

Because of their complete dissociation, calculating the pH of a strong acid or a strong base solution is simple. Let us calculate, for example, the pH of a 2.0  10–4 M solution of HBr. The H+ concentration is 2.0  10–4 and the log of 2.0  10–4 is -3.70. (Recall that in a logarithm only the digits after the decimal are significant.) Since pH is -log[H+], the pH of the solution is 3.70.

As an example of a strong base, suppose we have a solution which is 2.0  10–4 M in Ca(OH)2. This leads to a hydroxide concentration of 4.0  10–4. The log of 4.0  10–4 is -3.40. However since this is based on hydroxide ion and not hydrogen ion, the log corresponds to something we call the pOH and not the pH. Here the pOH will be 3.40. Because [H+][OH–] = 1.00  10–14 we know that the log of H+ plus the log of OH– add up to 14 (i.e. pH + pOH = 14.0). Thus, pH = 14.0 - pOH = 14.00 - 3.40 = 10.60

11. Weak Acids and Weak BasesAcids which are not listed among the strong acids (i.e. acids which are not H2SO4, HNO3, HCl, HBr, or HI) are classed as weak acids. Bases which are not strong (i.e. bases other than the hydroxides of Group IA and Group IIA metals) are weak. In contrast to a strong acid, which is completely dissociated in aqueous solution, a weak acid dissociates only partly. That is, the dissociation of a weak acid or base is an equilibrium. In order to calculate the results of that equilibrium, we need to use the equilibrium constant.

Let us consider the dissociation of a 0.50 M HF, a weak acid which reacts as below.HF H+ + F–

The dissociation has an equilibrium constant, referred to as "Ka", equal to 7.2  10–4.K a=7.2×10−4=¿¿

The reaction starts with [HF] = 0.50 M and [H+] = [F–] = 0. If the amount of HF which dissociates is x, then [HF] = (0.50 - x) and [H+] = [F–] = x. Therefore:

7.2×10−4=(x )(x)

(0.50−x)

Clearing the fraction and combining like terms leads to a quadratic equation. Generally, however, it is possible to make an approximation which allows us to avoid using the quadratic formula. If K is small, x should be small. In fact it will probably be so small that (0.50 - x) will be equal to 0.50, and the x on the bottom can be ignored. Neglecting x gives the following equation.

7.2×10−4= x2

0.50

This is easily solved to give x = 0.019, which means that:[H+] = [F–] = 0.019 M and[HF] = 0.50 - 0.019 0.50 M

As before we use the "5% rule" to determine whether neglecting x was valid. Since 0.019 is less than 5% of 0.50, the assumption is valid. Of course in some problems you know the value of x and are using it to calculate K. In such a case you don't neglect x — even if it is less than 5%.

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Weak base calculations typically involve ammonia or a similar compound (e.g. trimethylamine, N(CH3)3). Let us calculate the pH of a 0.10 M ammonia solution, which reacts with water:

NH3 + H2O NH4+ + OH–

The equilibrium constant, Kb, is equal to 1.8  10–5. The equilibrium expression is, therefore:1.8×10−5=¿¿

Notice that water is in the reaction but not in the mass action expression. Substituting the concentrations gives:

1.8×10−5= x x0.10−x

Neglecting x and multiplying gives:

x2 = 1.8  10–6 M

x = [OH–] = 1.34  10–3 M

pOH = 2.87

pH = 11.13

12. Polyprotic AcidsA polyprotic acid has more than one removable proton. An example would be hydrogen sulfide:

H2S H+ + HS– Ka1 = 1.1  10–7

HS– H+ + S2– Ka2 = 1.2  10–13

Why are the equilibrium constants for the removal of the two protons different? This is because the first proton is removed from a neutral molecule while the second is removed from an anion.

As an example, we will calculate the ion concentration in 0.100 M H2S solution.

1.1×10−7=[ x ] [x ]

[0.100−x ]

After neglecting the x in the denominator, we solve to get:x = [H +] = [HS -] = 1.0  10–4 M

Additional hydrogen ions will be formed in the second ionization. However, the H+ formed in later steps is ALWAYS neglected. Let us now determine the sulfide concentration:

1.2×10−13=[1.0×10−4+ y ] [ y ][1.0×10− 4− y ]

Notice that we used "y" instead of "x" in order to lessen confusion. After neglecting the additive y's, the problem becomes enormously easy, and we get:

y = [S -2] = 1.2  10–13 MNeglecting the y was clearly justified. Also note that "y" is equal to the H+ which comes from the second dissociation. As stated above, y is always small and can be ignored in calculating H+.

13. Hydrolysis

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Hydrolysis occurs when the conjugate base of a weak acid or the conjugate acid of a weak base is dissolved in water. To calculate the pH of a sodium acetate solution, one calculates the hydroxide concentration caused by the hydrolysis of the acetate.

C2H3O–2 + H2O HC2H3O2 + OH–

You will notice that in this reaction acetate functions as a base. Therefore to solve the problem you must use Kb for the acetate ion — which is Kw divided by Ka. Thus for 0.15 M acetate you calculate:

Kb=K w

K a=1.0×10−14

1.8×10−5 =5.55×10−10

Setting x = [OH–] = [HC2H3O2]K b=5.55×10−10=¿¿

Neglecting the x in the denominator gives:

x2 = 8.32  10–11

x = [OH–] = 9.12  10–6 M

pOH = 5.04 and pH = 8.96

In hydrolysis only the conjugate base (or conjugate acid) is present. If the conjugate base is accompanied by the acid — if acetate is accompanied by acetic acid — the solution is a buffer. (See below.) The following solutions should be treated as hydrolysis problems:

Sodium carbonateAmmonium chlorideEqual moles of acetic acid and sodium hydroxideA titration at its equivalence point

The following solutions should not be treated as hydrolysis problems:Acetic acidAmmoniaUnequal numbers of moles of acetic acid and sodium hydroxideA titration anyplace except at its equivalence point

For qualitative questions — that is whether a given compound forms an acidic, basic or neutral solution — you must know the following: The salt of a strong acid and a strong base is neutral; the salt of a strong acid and a weak base is acidic; the salt of a weak acid and a strong base is basic; the salt of a weak acid and a weak base is a nasty question which no one will ask you. Using this logic, you should predict that ammonium nitrate is acidic, calcium chloride is neutral, potassium phosphate is basic and aluminum sulfate is acidic.

14. BuffersA buffer is an aqueous solution which contains roughly equal concentrations of a weak acid and its

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conjugate base or a weak base and its conjugate acid. A buffer resists pH change because no matter whether H+ or OH– is added, there is something in the solution which will react with the added ion. Consider a buffer made from the weak acid HF and its conjugate base F–. If an acid is added, it reacts with the fluoride:

H+ + F– HF

If a base is added, it will react with the hydrofluoric acid:OH– + HF H2O + F–

The pH of a buffer is easy to calculate. Consider a buffer based on HF, which has a KA = 7.2  10–4. In a solution where concentrations of both the weak acid and its conjugate base are equal to 0.10 M, the pH can be calculated from the mass action expression.K a=7.2×10−4=¿¿

[H+] = 7.2  10–4 M

pH = 3.14

If H+ is added to the buffer, it all reacts with F– to form HF. (This is important!) So if we add (without dilution) 0.01 M H+, we end up with 0.11 M HF and 0.90 M F–. Therefore:K a=7.2×10−4=¿¿

[H+] = 8.8  10–4 M

pH = 3.05

Now let us try another buffer calculation, based on hypochlorous acid, HOCl, for whichKa = 3.5  10–8.

Suppose we start with 1 liter of 0.50 M HOCl and add, without diluting the solution, 0.20 moles of KOH. This is typical of how buffer problems start; so it is important you understand it. The strong base (KOH) reacts completely with the weak acid (HOCl) to form H2O and OCl–.

HOCl + OH– OCl– + H2OThis can be summarized in the ICE chart on the right:The H+ is formed by "bounce back" and its concentration is determined by the equilibrium expression:

3.5×10−8=¿¿

[H+] = 5.25  10–8 M

pH = 7.28

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Remember that adding a strong base to a mixture of HA and A- both decreases the HA and increases the A-.

HOCl OH– OCl–

Initial .50 .20 0

Change -.20 -.20 +.20

Equilibrium .30 0 .20

Now consider a buffer made by mixing 70.0 mL of 0.100 M ammonia with 30.0 mL of 0.100 M HCl. We will assume the volumes are additive. After dilution, the concentrations are:

[NH3] = 0.070 M

[HCl] = 0.030 M

The HCl (actually the H+ from the HCl) reacts with the NH3 as shown below:H+ + NH3 NH4

+

This gives:

[NH4+] = 0.030 M

[NH3] = 0.040 M

Plugging this in to the mass action expression gives us:K b=¿¿

1.8×10−5=(0.30 ) ¿¿

[OH–] = 2.4  10–5 M

pOH = 4.62

pH = 9.38

Many textbooks explain buffers in terms of the Henderson-Hasselbalch Equation:

This is equivalent to what we have been doing, and many students find Henderson-Hasselbalch makes simple problems even simpler. However, it makes hard problems even more difficult. So we are not using it here.

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15. IndicatorsHow do people determine the pH of a solution? An electrochemical device called a pH meter can be used. However, these are delicate and expensive and they need frequent calibration. A simpler method involves the use of an organic acid or base which has different colors in its acidic and basic forms. Such a material, an example of which is phenolphthalein, is called an indicator.

Phenolphthalein is colorless in its acid form (Figure 1) but becomes pink when it loses a proton. The equilibrium constantfor the loss of a proton by this indicator (HIn H+ + In–) isKa = 1  10–9 — which gives a pKA of 9. The pKA of an acid is significant since it equals the pH at which half the material is in the acid form and half is in the basic form. Thus the pK roughly equals the pH at which color change occurs. Of course one can see the pink color of the basic form long before the indicator is half converted. Consider what happens at pH 8.K a=¿¿10−9=10−8 ¿¿

0.1=¿¿

Dropping the pH by one unit causesthe fraction of indicator in the basic form to decrease from 50% to about 9%. However, a tinge of pink in an otherwise colorless solution can probably be seen at even less than9%. For most indicators the color changes over a range of about onepH unit on either side of the pK. Fortunately, this is narrow enoughfor most purposes.

The strips of pH paper which most of you have used for measuring pH are based on indicators. Typically a mixture of three indicators, which change color at different pH values, is used to give a solution which changes color continuously over a wide pH range. Litmus is also an indicator, one which turns red in acidic solution and blue in basic solution. (Remember bbblue for bbbasic.)

Indicators are widely used to determine the end point in a titration. What you need to know, how to choose which indicator to use in a given titration, will be discussed in a later section.

16. TitrationsTitrations are used to determine the amount of acid or base in a sample. If the sample to be analyzed is an acid, a base solution of known concentration (the titrant) is added. The point at which the moles of acid and the moles of base are equal is known as the "end point" or "equivalence point." Calculations are based on the volume of titrant needed to reach the end point. Thus one can calculate

Figure 1. The acidic form of phenolphthalein

Figure 2. Properties of Acid-Base Indicators at 25°C

Indicator pH Range pKa Acid Form Base FormMethyl violet 0.0 - 1.6 0.8 yellow blueMethyl yellow 2.9 - 4.0 3.3 red yellowMethyl orange 3.1 - 4.4 4.2 red yellowMethyl red 4.2 - 6.2 5.0 red yellowChlorophenol red 4.8 - 6.4 6.0 yellow redBromothymol blue 6.0 - 7.6 7.1 yellow bluePhenol red 6.4 - 8.0 7.4 yellow redCresol purple 7.4 - 9.0 8.3 yellow purpleThymol blue 8.0 - 9.6 8.9 yellow bluePhenolphthalein 8.0 - 9.8 9.0 colorless red

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Figure 3. The titration of a strong acid with a strong base.

the moles of acid in a sample:MolesA = MB VB

If the acid is in a solution of known volume, its molarity can be calculated from:MA VA = MB VB

Suppose a sample of organic acid weighing 0.255 g was found to require 35.6 mL of 0.0110 M KOH to reach the equivalence point. What is the molecular weight of the unknown acid? We know the molecular weight = grams/moles and that at the end point molesA = molesB. So:

M .W .=massAmoles A

=mass AmolesB

=mass AMBV B

= 0.2550.0110×0.0356

=651g /mol

What is the molarity of an acid solution if 5.0 mL are titrated by 28.8 mL of 0.0110 M KOH?

MA VA = MB VB

MA 5.0 = 0.0110 28.8

MA = 0.0634 M

Since volume is stated in mL on both sides of the equation, the units cancel.

All this assumes a monoprotic acid. If the acid is diprotic, and assuming that both protons are titrated, it is necessary to multiply the acid molarity by two. Thus:

2 MA VA = MB VB

17. Titration CurvesTo understand what happens during a titration it is helpful to graph the way pH changes as the titration proceeds. This graph is known as a titration curve.

First look at what happens when a strong acid is titrated with a strong base (Figure 3). The pH starts low. As the titration proceeds (as base is added) the pH increases slowly. Suddenly, as the titration nears the equivalence point, the slope increases and the pH changes more rapidly. The equivalence point is, in fact, the point at which the titration curve has its greatest slope. This is why titrations work so well. At the equivalence point, the very point you want to determine, a small amount of titrant will cause a large

change in pH. A single drop could cause the pH to change from 4 to 10, for example. This is fortunate, since indicators change color gradually over a range of several pH units. This titration is

oftendone using phenolphthalein (pKA = 9) as an indicator.

The titration of a weak acid with a strong base gives a similar curve (Figure 4), but there are differences. The initial pH is higher, since

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the acid is weaker, and the pH at the equivalence point will be higher as well. What this means is that the vertical portion of the titration curve, the segment near the inflection point, is smaller than with a strong acid. This means that determining the end point is more difficult with a weak acid than with a strong acid, and choice of indicator is more critical.

This is a good time to address a point which some students find confusing. The simple comparison between strong and weakacids given in the previous paragraph works only if the concentrationsare the same. A concentrated solution of a weak acid can be more acidic than a dilute solution of strong acid.

You must know what species are present at various points on the titration curve. Consider the titration of an HF solution with a solution of KOH. At first the solution contains only HF. As the reaction proceeds, the HF reacts with OH– to form H2O and F–. Thus, the principal species are K+, HF and F–. At the equivalence point, all the HF has been converted to F– and the principal species present are K+ and F–. What happens when the volume of titrant is half of that needed to reach the equivalence point? (This is called the half equivalence point). Here the concentrations of HF and F– are equal. If [HF] = [F-], we get the following:

and Therefore the pH is equal to the pK. This is not a calculation which a practicing chemist does often, but it occurs frequently on the AP test. Watch for the half-equivalence point!

To calculate the pH at various points on the curve you should remember that the initial solution is a weak acid, a calculation which you know well. The solution at the half-equivalence point (or at any other point part way through the titration) is a buffer. Finally, at the equivalence point all the weak acid and strong base have reacted. All the HF has been converted to F– and therefore the [H+] and [OH–] concentrations are based on the reaction:

F– + H2O HF + OH–

The titration of a weak base with a strong acid is simply the reverse of this. It starts at a high pH (but not as high a pH as it would if it were a strong base) and has its equivalence point below 7. Similarly the titration of a strong base with a strong acid gives a curve which is the same as the titration of a strong acid with a strong base. They both have their equivalence points at 7, but in one the pH increases and in the other it decreases.

Students worry about what happens when you titrate a weak acid with a weak base. Don't! The experimenter gets to choose the titrant and no one would choose to titrate with a weak base.

The titration of a polyprotic acid is more complex. Consider a diprotic acid with: KA1 = 10–3 and Ka2 = 10–10. Since the acid has two K's, the titration curve has two inflection points. When the K's of a polyprotic acid are close, one deprotonation starts before the other has finished and determining the end point is difficult.

18. Mixing Different Types of Acid/Base Problems

Figure 5. Titration Curve for a diprotic acid.

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Acid/base problems are not difficult. Unfortunately students often don't know what category a problem falls into and try to solve the wrong type. There are four types of acid/base problems. We talk about acids, but everything here is equally applicable to bases.

Strong Acid -- nothing present except for a strong acid

Weak Acid -- nothing present except for the weak acid and its dissociation products. Finding the pH of an acetic acid solution is a weak acid problem.

Buffer -- a weak acid and its conjugate base. Acetic acid and sodium acetate would make a buffer. Acetic acid and sodium hydroxide would also make a buffer, because the hydroxide would react with the acetic acid to form acetate. However hydrochloric acid and sodium chloride would not form a buffer, because hydrochloric acid is strong.

Hydrolysis -- only the conjugate base of a weak acid. Thus, sodium acetate solution is a hydrolysis problem. However, if you add an acid to this solution, it becomes a buffer.

Some examples:1.0 M SODIUM FLUORIDE. Forms sodium ions and fluoride ions in solution. The fluoride ions

react: F– + H2O HF + OH–, which is hydrolysis.1.0 M SODIUM FLUORIDE + 0.5 M HYDROFLUORIC ACID. These dissolve to form F– and HF. They

don't react; they form a buffered equilibrium.1.0 M HYDROFLUORIC ACID + 1.0 M SODIUM HYDROXIDE. They form HF and OH-, which react

according to HF + OH– F– + H2O. Because these are present in equal concentrations, you end up with 1.0 M fluoride which then hydrolyzes.

1.0 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. The solution contains 1.0 M F– and 1.0 M H+. They react to form 1.0 M HF which then dissociates in a typical weak acid reaction.

0.5 M SODIUM FLUORIDE + 1.0 M HYDROCHLORIC ACID. These form H+ and F– ions. All the fluoride is protonated by 0.5 M of the H+ and the remaining 0.5 M H+ is left to act as a strong acid.

1.0 M HYDROFLUORIC ACID + 0.5 M SODIUM HYDROXIDE. These form HF and OH-. All of the hydroxide reacts with 0.5 M HF to form 0.5 M F-. This mixture, which now contains 0.5 M fluoride and 0.5 M HF, acts as a buffer.

1.0 M SODIUM FLUORIDE AND 0.5 M HYDROCHLORIC ACID. These dissolve to give a solution containing H+ and F– ions. All the H+ from the hydrochloric acid is used to protonate 0.5 M of the fluoride forming 0.5 M HF. This leaves a mixture containing 0.5 M HF and the remaining 0.5 M F-, which is a buffer.

19. Bronsted and Lewis ModelsAfter all we have learned about acids and bases and their properties, it is strange to realize that the definition of acids and bases which we have been using is only one of several. The definition which we have used, that an acid supplies H+ ions and a base gives OH– ions, is the original definition as put forth by the Swedish chemist Svante Arrhenius. It is, therefore, called the Arrhenius model or the Arrhenius definition. It is very useful, but there are others.

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The Bronsted Model, or more properly the Bronsted-Lowry Model, defines an acid as a proton donor and a base as a proton acceptor. Consider the reaction of hydrofluoric acid with ammonia

HF + NH3 F– + NH4+

The hydrofluoric acid donates a proton to the ammonia, making the hydrofluoric acid an acid and the ammonia a base. Fluoride, which results from the loss of a proton by hydrofluoric acid, is considered a base since in the reverse reaction it gains a proton. It is said to be the "conjugate base" of hydrofluoric acid. Similarly the ammonium ion,

Altlhough this is less obvious, the dissolving of hydrogen chloride in water can also be considered

HCl + H2O H3O+ + Cl–

Here HCl is the acid, water is the base, hydronium ion (H3O+, essentially another way of writing H+) is the conjugate acid and chloride is the conjugate base.

Water is a base. How strange! Now look at what happens when ammonia is added to water:NH3 + H2O NH4

+ + OH–

Ammonia is a base, water is an acid, ammonium ion is the conjugate acid and hydroxide ion is the conjugate base. So water is either an acid or base, depending on the circumstances. In the Bronsted model acid/base reactions can be considered to be a competition between the baseand the conjugate as to which one wants the proton more.

Imagine a creature on some distant planet where lakes are made of ammonia instead of water. Acid/base chemistry in ammonia would be based on the reaction:

NH3 + NH3 NH4+ + NH2

–

Chemistry students on this planet might think it strange that acid/base neutralization reactions could occur in liquid water. Yet they could easily understand aqueous chemistry in terms of the Bronsted model.

Another way of defining acids and bases is seen in the Lewis model. Lewis acids are electron pair acceptors and Lewis bases are electron pair donors. This is consistent with the Arrhenius model, since a proton is itself an excellent acceptor of electrons while hydroxide is a good electron donor. Thus a neutralization reaction is water is still an acid base reaction.

H+ + OH– H2O

However, other reactions which we do not think of as acid/base reactions are acid/base reactions according to the Lewis model. Consider the reaction of trimethylamine, which we know to have an

Figure 6. A Lewis acid – Lewis base reaction

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unshared pair of electrons and to thus be a Lewis base. Suppose it reacts with BF3, a moleculewhich lacks an octet and is, therefore, capable ofacting as an electron acceptor. The molecule which results, BF3-N(CH3)3, is referred to as a "Lewis acid/Lewis base adduct."

20. Relation of Structure to AcidityA compound which contains a hydrogen atom will often give it up when dissolved, forming an acidic solution. However many compounds, for example organic compounds other than carboxylic acids, do not lose protons in water. The question of why some proton-containing compounds ionize easily (strong acids), others ionize with difficulty (weak acids) and still others don't ionize at all (non-acids) has to do with bond strength. Why bonds strengths vary as they do is complex.

The simplest example of a structure/acidity relationshipoccurs in the hydrogen halides — HF, HCl, HBr and HI.Since HF contains the most electronegative, and thereforethe most non-metallic, of the halogens, it seems that it should form the most acidic hydride. However that's not the way it works. Being the most electronegative element means it will form the strongest bond with hydrogen, which means it will form the weakest acid.

The next trend is among the oxy-acids, for example HClO4, HClO3, HClO2 and HClO. The trend is that the more oxygens there are, the stronger the acid. Thus the trend in acid strength is: HClO4 > HClO3 > HClO2 > HClO. To understand this you must first realize that the formulas written above give a misleading picture of the structures. In almost all oxy-acids the hydrogen is bound to an oxygen and not to the central atom. Thus, HClO2 is actually H-O-Cl-O. (That's why HCl, a strong acid, doesn't fit the trend. The hydrogen in HCl is bound directly to chlorine.)

Why do additional oxygen atoms make an oxy-acid stronger? The easiest way to view this (and there are several legitimate ways) is to say that oxygen, being very electronegative, withdraws electron density from the chlorine. Since losing a proton (a hydrogen cation) will leave behind a negative charge, something which withdraws negative charge will make the ion more stable. The acidity of other oxy-acids follows the same trend. Thus: H2SO4 > H2SO3 and HNO3 > HNO2.

Another trend you should know is that metal hydroxides are basic while non-metal hydroxides are acidic. Thus ClOH is an acid (usually written as HOCl) and KOH is a base. Similarly metal oxides generally react with water to form basic solutions, for example:

CaO + H2O Ca2+ + 2 OH–

Non-metal oxides, on the other hand, react with water to form acidic solutions, for example:SO2 + H2O H2SO3

Why is this? The answer is that an electronegative element, such as chlorine, forms a covalent bond with oxygen. This bond, like other covalent bonds, does not break easily in water. Thus the only bond which can break is the H-O bond. In metal hydroxides, however, the central atom is electropositive and thus forms an ionic bond. The ionic bond between the metal ion and the hydroxide ion is strong, but unlike the covalent bond it comes apart easily in a polar solvent such as water. (As you may

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Bond AcidStrength Strength

Acid (kJ/mol) in WaterHF 565 WeakHCl 427 StrongHBr 363 StrongHI 295 Strong

recall, polar water molecules surround the ions and stabilize the charges.)

The intermediate case, in which the central atom is neither strongly electronegative nor strongly electropositive, leads to a situation where both the X-O bond and the O-H bond are capable of breaking. This leads to a phenomenon known as "amphoterism," in which the oxide or hydroxide is capable of acting as either an acid or a base. It is difficult to predict whether a transition metal oxide will be acidic, basic or amphoteric. In general you will not have to do that. However, when we get to the reaction section of this course, you will learn that aluminum and zinc form amphoteric oxides. But that's for another day.

21. "How many acids" problemsOne good way of testing your knowledge of acid-base chemistry is to ask how many compounds in a list form acidic (or basic) solutions in water. Consider the following list of compounds.

CaBr2, NH4Cl, SO3, AlCl3, KCN

If you answered that "two form acidic solutions," you are wrong. The correct answer is "three," specifically NH4Cl, SO3, and AlCl3. Make sure you understand this.

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