Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton

Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

1

Section 16

Acid-Base Equilibria


2

Solutions of a Weak Acid or Base

• The simplest acid-base equilibria are those in which a single acid or base solute reacts with water.– we will first look at solutions of pure weak

acids or bases.

– Next, we will also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water. Salts can be neutral, acidic, or basic.


3

Acid-Ionization Equilibria

• Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion (hydrolysis).

– Because acetic acid is a weak electrolyte, it ionizes to a small extent in water, hence double arrow; not 100% like strong acid which wrote single arrow.

(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232


4


• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant, Ka).

– Consider the generic monoprotic acid, HA.

)aq(A)aq(OH )l(OH)aq(HA 32


5


)(A)(OH )()( 32 aqaqlOHaqHA

– The corresponding equilibrium expression is:

]OH][HA[]A][OH[

K2

3c

– Since the concentration of water remains relatively constant (liquid = 1), we rearrange the equation to get:

]HA[]A][OH[

K]OH[K 3c2a


6


• acid + water is Ka

– Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.

]HA[]A][OH[

K 3a


7

Experimental Determination of Ka

• The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions.

– Electrical conductivity or some other colligative property can be measured to determine the degree of ionization.

– With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution because it gives the concentration of hydronium ions at equil.


8

A Problem To Consider

• Nicotinic acid is a weak monoprotic acid with the formula HC6H4NO2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C.– It is important to realize that the solution was made

0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M, just a small amount less.


9


Starting 0.012 0 0Change

Equilibrium

– Let x be the moles per liter of product formed.

)()(OH )()( 24632246 aqNOHCaqlOHaqNOHHC


10


– We can obtain the value of x from the given pH.


11


– Substitute this value of x in our equilibrium expression.

– Note first, however, that

012.001159.0)00041.0012.0()x012.0(

the concentration of unionized acid remains virtually unchanged. Important point to remember for later.

– Substitute this value of x in our equilibrium expression.

522

a 104.1)012.0()00041.0(

)x012.0(x

K

HW 28


12


– To obtain the degree of ionization:

%4.3%1000.012

0.00041 ionization %

100% x ][

][ ionization %

x

acid

H

initial

ionizedacid

– weak acids typically less than 5%. The lower the %ionized, less H+ formed, the lower Ka, weaker the acid. Ka is measure of strength of acid or Kb if base. Strength based on ionization not pH directly. Lower pH more acidic, does not mean stronger acid


13

Ka is a measure of strength of acid. It indicates the amount ionized to form hydronium ions. The more hydronium ions formed, naturally the more acidic and lower pH. However, strength of acid is measure of amount ionized, not conc of acid. Conc can vary which ultimately will vary the pH of solution. To determine the strength of an acid you must compare Ka values not pH of solution. pH is dependent on Ka as well as conc. of solution. Must compare apples and apples.

10M HC2H3O2 Ka = 1.8 x 10-5 pH = 1.87

0.1M HF Ka = 6.9x10-4 pH = 2.08

Which stronger acid?


14

Calculations With Ka

• Once you know the value of Ka, you can calculate the equilibrium concentrations of species HA, A-, and H3O+ for solutions of different molarities.– The general method for doing this is same way we

did Kc or Kp except we can simplify the math because of percent ionize is so small compared to initial conc of acid; remember 0.012-x=0.012. This wasn't the case for Kc or Kp problems so can't neglect in those type problems


15


• Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid.

• It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression.

• The degree of ionization of a weak acid depends on both the Ka and the concentration of the acid solution


16


• How do you know when you can use this simplifying assumption?

– then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5% otherwise must do quadratic formula. Another way look at it factor of two - three difference in power of 10 is needed between Ka and conc of acid.

– It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is,

100[HA] aKif


17

ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO2, for which Ka = 6.0 x 10-4.

HNO2 + H2O <--> H3O+ + NO2-


18

A Problem To Consider• Really two equil: Kw as well but neglect most

of the time as long as [H+] is greater than

10-6 M from the acid.

• Really [H+]tot = [H+]acid +[H+]H2O (water so small neglect)


19

ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid , HNO2, diluted to 100.0 mL with water for which Ka = 6.0 x 10-4.

HNO2 + H2O <--> H3O+ + NO2-


20


• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetyl salicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

– The molar mass of HC9H7O4 is 180.2 g.


21


• What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC9H7O4, in 0.500 L of water? The acid is monoprotic and Ka=3.3 x 10-4 at 25°C.

Starting[] 0.003607 0 0Change []

Equilibrium[]

)()(OH )()( 47932479 aqOHCaqlOHaqOHHC

– These data are summarized below.


22


– Note that

which is less than 100 or tell factor 10-3 vs 10-4

not factor 2-3; therefore, won't work, so we must solve the equilibrium equation exactly .

11103.3

0036.0K

C4

a

a


23

– If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get

Starting[] 0.003607 0 0

Change [] -x +x +xEquilibrium[] 0.003607 –x x x

)()(OH )()( 47932479 aqOHCaqlOHaqOHHC


24

– You can solve this equation exactly by using the quadratic formula.


25

– Taking the upper sign, we get and neglect water OK


– Now we can calculate the pH.

HW 29


26

Calculate the pH of a solution that has a concentration of 1.0 x 10-8 M NaOH.


27

Polyprotic Acids• Some acids have two or more protons

(hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.– Sulfuric acid, for example, can lose two protons in

aqueous solution.

)aq(HSO)aq(OH)l(OH)aq(SOH 43242

)aq(SO)aq(OH )l(OH)aq(HSO 24324


28

Polyprotic Acids

– For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered.

)aq(HCO)aq(OH )l(OH)aq(COH 33232

)aq(CO)aq(OH )l(OH)aq(HCO 23323


29

– Each equilibrium has an associated acid-ionization constant.

Polyprotic Acids

– For the loss of the first proton

7

32

331a 103.4

]COH[]HCO][OH[

K

– For the loss of the second proton

11

3

233

2a 108.4]HCO[

]CO][OH[K


30

– In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2.

Polyprotic Acids

– In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1. Harder to pull proton off charged species. Total H+ is sum of all equil H+ but usually neglect all but the principal rxn.


31

– However, reasonable assumptions can be made that simplify these calculations as we show in the next example.

Polyprotic Acids

– When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.


32

– For diprotic acids, Ka2 is so much smaller than Ka1 that the smaller amount of hydronium ion produced in the second reaction can be neglected.

A Problem To Consider• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6.

What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.


33

– The pH can be determined by simply solving the equilibrium problem posed by the first ionization.


(aq)(aq)OH )()( 666326662 OHHClOHaqOHCH

– the first ionization is:


Equilibrium


34


– Assuming that x is much smaller than 0.10 (1265), you get


35

– The ascorbate ion, C6H6O62-, which we will call y, is

produced only in the second ionization of H2C6H6O6.


• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C6H6O6

2- ?

The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

(aq)(aq)OH )()( 266632666

OHClOHaqOHHC


36

– Assume the starting concentrations for HC6H6O6- and

H3O+ to be those from the first equilibrium.

Starting [] 0.0028 0.0028 0Change []

Equilibrium []

(aq)(aq)OH )()( 266632666

OHClOHaqOHHC


37

– Substituting into the equilibrium expression

– Assuming y is much smaller than 0.0028 (1.75 x 109 > 100), the equation simplifies to

– Hence,

– typically, the concentration of the anion ion equals Ka2 if starting with reactants of first equil. HW 30


38

Base-Ionization Equilibria

• Equilibria involving weak bases are treated similarly to those for weak acids.– Ammonia, for example, ionizes in water as

follows.

)aq(OH)aq(NH )l(OH)aq(NH 423

– The corresponding equilibrium constant is:

]OH][NH[]OH][NH[

K23

4c


39


• Equilibria involving weak bases are treated similarly to those for weak acids.– Ammonia, for example, ionizes in water as

follows.


– The concentration of water is nearly constant.

]NH[

]OH][NH[K]OH[K

3

4c2b


40


• Equilibria involving weak bases are treated similarly to those for weak acids. Higher Kb stronger base, ionize more OH-, more basic.

– In general, a weak base B with the base ionization

)aq(OH)aq(HB )l(OH)aq(B 2

has a base ionization constant equal to

]B[

]OH][HB[Kb

HW 31


41


• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.

1. Write the equation and make a table of concentrations.

2. Set up the equilibrium constant expression.

3. Solve for x = [OH-].

– As before, we will follow the three steps in solving an equilibrium.


42


• What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9.– Pyridine ionizes by picking up a proton from water

(as ammonia does).

)aq(OH)aq(NHHC )l(OH)aq(NHC 55255


Equilibrium


43


– Note that

which is much greater than 100, so we may use the simplifying assumption that (0.20-x) (0.20).

89 104.1

104.120.0

b

bK

C


44


Starting 0.20 0 0Change -x +x +x

Equilibrium 0.20-x x x

– The equilibrium expression is

)aq(OH)aq(NHHC )l(OH)aq(NHC 55255


45


– Using our simplifying assumption that the x in the denominator is negligible, we get


46


– Solving for pOH

– Since pH + pOH = 14.00

HW 32


47

Acid-Base Properties of a Salt Solution

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.

– A 0.1 M solution has a pH of 11.1 and is therefore fairly basic.

– Consider a solution of sodium cyanide, NaCN.

)aq(CN)aq(Na)s(NaCN OH 2


48

– Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH- making it a basic solution

)aq(OH)aq(HCN )l(OH)aq(CN 2


– From the Brønsted-Lowry point of view, the CN- ion acts as a base, because it accepts a proton from H2O.


49

– You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic.


– The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.



50

• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

– The CN- ion hydrolyzes to give the conjugate acid and hydroxide.




51

– The hydrolysis reaction for CN- has the form of a base ionization so you write the Kb expression for it.




52

– The NH4+ ion hydrolyzes to the conjugate

base (NH3) and hydronium ion.



• An example of an cation that undergoes hydrolysis is the ammonium ion

)()( 44 aqClaqNHClNH


53

Predicting Whether a Salt is Acidic, Basic, or Neutral

• How can you predict whether a particular salt will be acidic, basic, or neutral and undergoes hydrolysis?

– The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs.

– Consequently, the anions of weak acids (poor proton donors) are good proton acceptors.

– Anions of weak acids are basic.


54


– On the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze.

– For example, the Cl- ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water. If did, strong acid breaks up 100% and hydroxide produced cancelled with hydronium of strong acid and make neutral solution

reaction no)()(

,

)()(

OH)()(

2

2

32

lOHaqCl

essencein

OHHCllOHaqCl

CllOHaqHCl


55


– Small highly charged (attract O more weakens OH bond) cations have an attraction for the O on water and cause the production of H+ while large low charged cations do not. Low charged large ions are group I and II metals (cations of strong bases) and do not undergo hydrolysis.

– For example,

reaction no)l(OH)aq(Na 2

In summary,

If cation is that of the strong base, it will have neutral hydrolysis; otherwise, cation will have acidic hydrolysis

If anion of monoprotic strong acid, it will have neutral hydrolysis; otherwise, anion will have basic hydrolysis


56

• Salts may be acidic/basic/neutral and are composed of cations (positive ions) and anions (negative ions)

cation anion

Acidic or neutral basic or neutralCations of strong bases Anions of monoprotic strongare neutral; otherwise, acids are neutral; otherwise,cation contributes acidity anion contributes basicity

Na+, K+, Li+, Ca2+, Sr2+, Ba2+ Cl-, Br-, I-, NO3-, ClO4

- add add neutral; rest of cations add neutral; rest of anions add acidity to salt basicity to salt

Depending on the two components (cation/anion) the overall salt will be acidic/neutral/basic – nn n; an a; nb b.


57


• To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.


58



59


• AlCl 3

• Zn(NO3)2

• KClO4

• Na3PO4

• LiF

• NH4F Ka = 5.6x10-10 > Kb = 1.4x10-11

• NH4ClO Ka = 5.6x10-10 < Kb = 3.6x10-7

• NH4C2H3O2 Ka = 5.6x10-10 = Kb = 5.6x10-10

• one with larger K will dictate acidity/basicity - stronger species

HW 33


60

The pH of a Salt Solution

• To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion.

– The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.

– Thus the Kb for CN- is related to the Ka for HCN.


61


• To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

– When two reactions are added, their equilibrium constants are multiplied.

)()( )()( 322 aqOHaqOHlOHlOH Kw


62

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

– Therefore,

)()( )O( )( 322 aqOHaqOHlHlOH Kw

CxKKK owba [email protected] 14

HW 34


63


• For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases.– The only difference is first must obtain the

Ka or Kb for the ion that hydrolyzes.


64

Attack of hydrolysis problem

When attacking pH (hydrolysis problems) always ask yourself if acid, base, or salt -- if acid or base, is it strong (straight forward) or weak (Ka or Kb problem) -- If salt, write dissociation eq and examine cation/anions involved to determine if acid or base hydrolysis is available then Ka or Kb

problem.


65

A Problem To Consider• What is the pH of a 0.10 M NaCN solution

at 25 °C? The Ka for HCN is 4.9 x 10-10.


66



67

example: What is the pH of 0.10 M NH4I solution?

Kb NH3 = 1.8 x 10-5

HW 35


68

The Common Ion Effect

• The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.



69

The Common Ion Effect

– The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased, meaning less hydronium formed and higher pH of solution as compared to pure acid.


– This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. Another source of ion coming from an additional solute added to the solution.

)()( 232232 aqOHCaqNaOHNaC


70

Buffers• When common ion effect involves weak

acid/conjugate base it is referred to as a buffer. Buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

– Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Typically conj is in form of salt.

– Thus, a buffer contains both an acid species and a base species in equilibrium which allows it to absorb any added acid or base and keep pH of solution relatively constant


71

Buffers– Consider a buffer with equal molar amounts of HA

and its conjugate base A-.


72

A Problem To Consider• An aqueous solution is 0.025 M in formic acid, HCHO2 and 0.018 M

in sodium formate, NaCHO2. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4.


73



74

• A solution is prepared to be 0.15 M HC2H3O2 and 0.20 M NaC2H3O2. What is the pH of this solution at 25oC?

Ka HC2H3O2 @ 25oC = 1.8 x 10-5


75

• Calculate the pH of 75 mL of a buffer solution (0.15 M HC2H3O2 / 0.20 M NaC2H3O2 pH = 4.87) to which 9.5 mL of 0.10 M HCl is added. Ka HC2H3O2 @ 25oC = 1.8 x 10-5


76HW 36


77

Buffers

– Two important characteristics of a buffer are its buffer capacity and its pH.

– Buffer capacity depends on the amount of acid and conjugate base present in the solution.

– The next topic illustrates how to calculate the pH of a buffer by special eq but can do like we just did as well to keep with thought process and not memorize a equation.


78

The Henderson-Hasselbalch Equation• How do you prepare a buffer of given pH?– A buffer must be prepared from selecting a conjugate

acid-base pair in which the Ka of the acid is approximately equal to the desired H3O+ concentration.

– Note: you can add extra conjugate base, higher pH, or less base, lower pH, to get to exact pH wanted.

– To illustrate, consider a buffer of a weak acid HA and its conjugate base A-.

The acid ionization equilibrium is:



79

The Henderson-Hasselbalch Equation

– The acid ionization constant is:

– By rearranging, you get an equation for the H3O+ concentration.

]A[

]HA[K]O[H a3

]HA[]A][OH[

K 3a



80

The Henderson-Hasselbalch Equation

– Taking the negative logarithm of both sides of the equation we obtain:

)][

][log()Klog(]Olog[H-

)][

][Klog(]Olog[H-

a3

a3

A

HA

A

HA

– The previous equation can be rewritten

]HA[]A[

logKppH a


81

The Henderson-Hasselbalch Equation– More generally, you can write

– This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation (for buffers only). Also way to work previous buffer problem. Note pH = pKa when conc base=acid

],[

],[logKpH a HAacid

Abasep


82

• Let’s prepare a buffer with pH of 4.90

– So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pKa close to the desired pH or 10-4.90 to find Ka close to 1.25 x 10-5 (desired hydronium conc. equivalent to pH desired)

HC2H3O2 Ka = 1.8 x 10-5

H2CO3 Ka = 4.3 x 10-7

HNO2 Ka = 4.5 x 10-4

– The Ka for acetic acid is 1.8 x 10-5, and its pKa is 4.74 or pH of same value if equal conc of acid/salt are used.

– You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid], meaning more basic salt.


83

What if I wanted to use 0.100 M HC2H3O2 (Ka = 1.8 x 10-5), what concentration of NaC2H3O2 would I need to obtain a buffer with pH =4.90.

HW 37


84

Acid-Ionization Titration Curves (or mixtures)• An acid-base titration curve is a plot of the pH of a

solution of acid (or base) against the volume of added base (or acid).

– Such curves are used to gain insight into the titration process.

– You can use titration curves to choose an appropriate indicator that will show when the titration is complete.

Figure: Curve for the titration of a strong acid by a strong base.


86

• Figure shows a curve for the titration of HCl with NaOH.

– Note that the pH changes slowly until the titration approaches the equivalence point.

– The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added; contrary to end point (observable point).

– At this equivalence point of SA and SB, the pH of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze.

– However, the pH changes rapidly from a pH of about 3 to a pH of about 11.

– To detect the equivalence point, you need an acid-base indicator that changes color beyond equivalence point.

– Phenolphthalein can be used because it changes color in the pH range 8.2-10.


87

Previous example equivalent pt was at 7 but not all neutralization reactions eq pts are at 7. During a titration or just mixing an acid and base together to make a solution, the resulting solution will contain a salt and water (Acid + Base --> salt + water). One of the chemical properties of acids and bases is that they neutralize one another; however, that doesn’t mean that the product will be neutral.

The misconception is that if the acid and base are in stoich proportions that the resulting solution is neutral. This is not true. The salt formed may be a acidic, basic, or neutral salt and will dictate the pH of the solution. Common sense can help you predict the pH of a mixture.

SA + SB --> neutral salt only true neutralization pH =7

WA + SB --> basic salt original acid and base neutralize but product has basic properties and basic pH

SA + WB --> acidic salt original acid and base neutralize but product has acidic properties and acidic pH


88

When working acid-base titration problems or mixtures treat them all the same but must be able to examine the components remaining after neutralization to determine what affects pH or not (strong, weak, salts). They are all done the same. The problem will dictate where you go. We can derive titration curve by doing a calculation at different points and draw curve or measure with pH meter.

1.) Calculate mmols initially

2.) write the neutralization reaction and find mmols of components left after neutralization

3.) find new conc. of species that affect pH

4.) do hydrolysis of main species that affect pH (like earlier in chapter)


89

A Problem To Consider• Calculate the pH of a solution in which 50.0 mL of 1.000 M

NaOH is added to 50.0 mL of 1.000 M HCl.


90




91



HW 38


92

A Problem To Consider• Calculate the pH of a solution in which 20.00 mL of 1.000 M NaOH is

added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5


93

A Problem To Consider• continue


94




95



96




97


HW 39

Documents

Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton