Chemistry part 28, Julia Burdge,2e (2009) pps

Because acetate ion is a product in the ionization of acetic acid, the addition of acetate ion causes the equi-librium to shift to the left.. Sample Problem 17.1 shows how an equilibrium

682 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria

The percent ioniza t ion of acetic acid is

13~~~~ 3 M X 100% = 1.3%

By adding so dium acetate, we also add so d ium

ions to the solutio n, However, sodium ions do

not interact w ith wate r or w ith any of the other

species present [ ~ Section [6 10]

Remember that prior to the ionizati o n of a

wea k acid, the concentrat io n of hydrogen ion

in water at 25°( is 1 0 x 1 0-7 M H owever,

becau se this concentration is insignificant

comp ared to the conc e ntration re su lting

from the i oniza ti on, we can neglect it in our

equilibrium table

The Common Ion Effect

Up until now, we have discussed the properties of solutions containing a single solute In this tion, we will examine how the properties of a solution change when a second solute is introduced

sec-Recall that a system at equilibrium will shift in response to being stressed and that stress can be applied in a variety of ways, including the addition of a reactant or a product [I •• Section 15.5] Consider a liter of solution containing 0.10 mole of acetic acid Using the Kafor acetic acid

(1.8 X 10-5) and an equilibrium table [ ~ Section 16.5], the pH of this solution at 25°C can be deter mined:

CH3COOH(aq) +: ===z: H +(a q) + CH 3 COO - (aq)

[CH3COOH] = 0.09866 M , [H+] = [CH3COO- ] = 1.34 X 10- M and pH = 2.87

Now consider what happens when we add 0.050 mole of sodium acetate (CH3COONa) to the solution Sodium acetate dissociates completely in aqueous solution to give sodium ions and acetate ions:

CH 3 COONa(aq) H 20 , Na \aq) + CH 3 COO - (aq)

., -Thus, by adding sodium acetate, we have increased the concentration of acetate ion Because acetate ion is a product in the ionization of acetic acid, the addition of acetate ion causes the equi-librium to shift to the left The net result is a reduction in the percent ionization of acetic acid

Addition

/

H + (aq) + CH3COO - (aq)

Equilibrium is driven toward reactant

Shifting the equilibrium to the left consumes not only some of the added acetate ion, but also some ofthe hydrogen ion This causes the pH to change (in this case the pH increases) Sample Problem 17.1 shows how an equilibrium table can be used to calculate the pH of a solution of acetic acid after the addition of sodium acetate

Sample Problem 17.1 ", "

Determine the pH at 25 °C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of

0.10 M acetic acid (Assume that the addition of sodium acetate does not change the volume of the

so lution )

Strategy Construct a new eqUilibrium table to solve for the hydrogen ion concentration

Setup ' We' ~se ' the ' s ' tat~d concei;t~~iion ci ~~etic ' a c i'd : o.io M : ~ nd [H + ] = 0 M as the initial

concentrations in the table:

CH 3 COOH(aq) :.;:: =::t, H+(aq) + CH 3 COO - (aq)

SECTION 17.2 Buffer Solutions 683

Therefore, the equilibrium expression s implifies to

~5 ( x)(0.050)

1.8 X 10 = 0.10

and x = 3.6 X 1O ~5 M According to the equilibrium table, [H + ] = x, so pH = - log (3.6 X 1O ~5) = 4.44

Think About It The equilibrium concentrations of CH3COOH, C H 3 COO ~, and H + are the same

regardles s of whether we add so dium acetate to a so lution of acetic acid, a dd ace tic acid to a

solution of so dium acetate, or dissolve both s pe c ie s at the same tim e We could have constructed an

equilibrium table s tarting with th e e quilibrium concentrations in the 0.10 M acetic acid so lution:

Th i s is cons i derably sma lle r than the per cent

ionization prior to the addition of so diu m

acetate

Initial conce ntration (M): 0.09866 1.34 X 1O ~3 5.13 4 X 1O ~2 This is the sum of the equ ilib rium concentra

tio-Change in concentration (M) : - - - -- - - + -.::.y + -- - t - - - I - - - -- y" , -t - :o.y _ (1.3of acetate ion in a 4 x 1 0~3 M) and 0.10 the M soadded acelution of t ate aceion tic aoc

Equilibrium con centration (M): 0.09866 + Y 1.34 X 1O ~3 - Y 5.134 X 1O~ 2 - Y (0.0 50 M)

In thi s case, the reaction proceed s to ,the left (The acetic acid concentration increase s, and the

concentrations of hydrogen and acetate ion s decrea s e.) Solving for y gives 1.304 X 1O~ 3 M

[H + ] = 1.34 X 1O~ 3 - Y = 3.6 X 1O ~ 5 M and pH = 4.44 We get the sa me pH either way • • • • • •• •••

so dium acetate in 1.0 L of 0.25 M acetic acid (Ass ume that th e addition of so dium acetate doe s not

change the volume of the s olution )

so dium acetate in 1.0 L of 0.25 M acet ic acid (Ass ume that the addit i o n of sodi um acetate does not

change the volume of the so lution )

An aqueous solution of a weak: electrolyte contains both the weak: electrolyte and its tion products, which are ions If a soluble salt that contains one of those ions is added, the equi-

ioniza-librium shifts to the left, thereby suppressing the ionization of the weak: electrolyte In general,

when a compound containing an ion in common with a dissolved substance is added to a solution

at equilibrium, the equilibrium shifts to the left This phenomenon is known as the common ion

effect

17 1 1 Which of the following would ca u se

a decrease in the percent ioni z ation of

nitrous acid ( HN02) when adde d to a

so lution of nitrou s acid at e quilibrium ? (Se lect all that appl y.)

17 1 2 Wh at i s the pH of a so lution prepar ed

b y addi ng 0.05 mole of NaF to 1.0 L

of 0.1 M HF a t 2YC? (Ass ume that th e

addition of NaF does not change the volume of the so lution ) (Ka for HF =

A solution that contains a weak acid and its conjugate base (or a weak: base and its conjugate

acid) is a buffer solution ' or 'slmpiy a ' buffer: B'uffer'soiutioiis; by ' vli-tiie ' o(ihe1r 'comPOSItIon; ' r(dst '

changes in pH upon addition of small amounts of either an acid or a base The ability to resist pH

change is very important to chemical and biological systems, including the human body The pH

of blood is about 7.4, whereas that of gastric juices is about 1.5 Each of these pH values is crucial

for proper enzyme function and the balance of osmotic pressure, and each is maintained within a

very narrow pH range by a buffer

Trea ti ng the problem as though both CH3

COm-a nd C H3COO- are added at the same tim e a nd

•

the reaction proceeds to th e ri ght simp li fies the

solut ion

The common ion can also be H + or O H ~ F or

example, addition of a strong acid to a so l uti on

of a weak acid supp r esses i o niza tio n of the ~

acid Similarly, addition of a st rong base to a solu t ion of weak base suppre sses ionizati on of

the weak base

Any solution of a weak acid co nt a in s some

co njugate base I n a buffer so luti on, t houg h, ;;-::

amounts of weak acid and conj ug ate base m=

be comparable, meaning that th e con ju ga te hax

must be sup plied by a dissolved salt

684 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria

Remember that sod ium acetate is a strong

electrolyte [ ~ ~ Sec t ion 4 , 1] so it dissociates

completely in water to give sodium ions and

acetate ions:

,, '

Multimed i a

Acids and Bases effect of addition of a

strong acid and a strong base on a buffer ,

The forward reaction is suppressed by the

presence of the common ion, CH3COO - ,

and the reverse process is suppressed by the

presence of CH3COOH

As long as th e amount of strong acid added

to the buffer does not exceed the amount of

conjugate ba se originally present, all the added

acid will be consumed and converted to weak

acid,

Calculating the pH of a Buffer

., C'onsider' a ;;olution th'at i's 'i :6 M Ii; aceti'c ' aClC!" and 'i :6 M sodium acetate If a small amount of acid

is added to this solution, it is consumed completely by the acetate ion,

thus converting a strong acid (H+) to a weak acid (CH3COOH) Addition of a strong acid lowers the pH of a solution However, a buffer's ability to convert a strong acid to a weak acid minimizes the effect of the addition on the pH

Similarly, if a small amount of a base is added, it is consumed completely by the acetic

acid,

thus converting a strong base (OH-) to a weak base (CH3COO-) Addition of a strong base increases the pH of a solution Again, however, a buffer's ability to convert a strong base to a weak

base minimizes the effect of the addition on pH

To illustrate the function of a buffer, suppose that we have 1 L of the acetic acid-sodium acetate solution described previously We can calculate the pH of the buffer using the procedure

At equilibrium, therefore, [H+] = 1.8 X 10- 5 M and pH = 4.74

Now consider what happens when we add 0.10 mole of HCI to the buffer (We assume that the addition of HCI causes no change in the volume of the solution.) The reaction that takes place

, ' ~ "

when we add a strong acid is the conversion of H to CH3COOH The added acid is all consumed"

along with an equal amount of acetate ion We keep track of the amounts of acetic acid and acetate ion when a strong acid (or base) is added by writing the starting amounts above the equation and the final amounts (after the added substance has been consumed) below the equation:

Upon addition of H+: 1.0 mol 0.1 mol 1.0 mol

CH 3C OO - (aq) + H +( aq) , CH 3 COOH(aq)

After H+ has been consumed: 0.9 mol o mol 1.1 mol

We can use the resulting amounts of acetic acid and acetate ion to construct a new equilibrium table:

CH 3 COOH(aq) +: =::z:' H + (aq) + CH3COO-(aq)

SECTION 17.2 Buffer Solutions 685

the small amount of weak acid that ionizes (x) because ionization is suppressed by the presence of

a common ion Similarly, we ignore the hydrolysis of the acetate ion because of the presence of

acetic acid This enables us to derive an expression for determining the pH of a buffer We begin

and acetate ion were 1.1 M and 0.9 M, respectively Using these concentrations in the

Henderson-Hasselbalch equation gives

The small difference between this pH and the 4.66 calculated using an equilibrium table is due to

•

Had we added 0.1 0 mole of H e I to 1 L of p U lE water, the pH would have gone from 7 00 t o

1 00!

Media Player /MP EG Animation :

Figure 17.1, Buffer Solutions, pp 686-68 7

buffer contains (0.100 mollL)(O.lO L) = 0.010 mol

each acetic acid and acetate ion

Water

pH = 7.00

aO

When we add 0.001 mol of strong acid, it is completely

consumed by the acetate ion in the buffer

Before reaction: 0.001 mol 0.010 mol 0.010 mol

•

H + (aq) + CH 3 COO-(aq) • CH 3 COOH(aq)

After reaction: 0 mol 0.009 mol

2S.0 · c

When we add 0.001 mol of strong base, it is completely consumed by the acetic acid in the buffer

0.011 mol

N

7

•

~

There is nothing in pure water to consume strong

acid Therefore, its pH drops drastically

H = -1 0.001 mol = 200

Ther€ is nothing in pure water to Gonsume strong

base Therefore, its pH rises drastically

0.001 mol pOH = -log 0.10 L = 2.00, pH = 12.00

What's the point?

A buffer contains both a weak acid and its conjugate base *

Small amounts of strong acid or strong base are consumed by the buffer components, thereby preventing drastic pH changes

Pure water does not contain species that can consume acid

or base Even a very small addition of either causes a large change in pH

* A buffer could also be prepared using a weak base and its conjugate acid

687

688 CHAPTER 17 Acid-Base Equilibria and Solubility Equilibria

The volume of the buffer is 1 L i n thi s e x ampl e,

so the number of moles o f a substance i s equal

t o the molar con c entrati o n I n c ases where the

buffer volume is something other than 1 L,

however, we can still use molar amounts in the

Henderson-Hasselbalch equat i on becau s e the

volume would cancel in the top and bottom of

the log term

Think About It Always do a

"reality check" on a calculated pH

Although a buffer does minimize

the effect of added base, the pH

does increase If you find that

you've calculated a lower pH after

the addition of a base, check for

errors like mixing up the weak acid

and conjugate base concentrations

or losing track cif a minus sign

volume of the solution )

Strategy Added base will react with the acetic acid component of the buffer, converting OH- to

CH 3 COOH(aq) + OW(aq) - H20(/) + CH 3 COO - (aq)

Thus , the pH of the buffer after addition of 0.10 mole of NaOH is 4.83

Practice Problem A Calculate the pH of 1 L of a buffer that is 1.0 M in acetic acid and 1.0 Min

s odium acetate after the addition of 0.25 mole of NaOH

Practice Problem B Calculate the pH of 1 L of a buffer that is 1.0 M in acetic acid and 1.5 Min sodium acetate after the addition of 0.20 mole of HCI

Preparing a Buffer Solution with a Specific pH

A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base

is added If the concentrations of a weak acid and conjugate base differ by more than a factor of

10, the solution does not have this capacity Therefore, we consider a solution a buffer, and can use Equation 17.1 to calculate its pH, only if the following condition is met:

First, we choose a weak acid whose pK a is close to the desired pH Next, we substitute the

pH and pKa values into Equation 17.1 to obtain the necessary ratio of [conjugate base ]/[ weak acid] This ratio can then be converted to molar quantities for the preparation of the buffer Sample Problem 17.3 demonstrates this procedure

Select an appropriate weak acid from the table in the margin, and describe how you would prepare a buffer with a pH of 9.50

SECTION 17.2 Buffer Solutions 689

Strategy Select an acid with a pKa within one pH unit of 9.50 Use the pKa of the acid and Equation

17.1 to calculate the neces sary ratio of [conjugate base]/[ weak acid] Select concentrations of the

buffer components that yield the calculated ratio

Setup Two of the acid s listed in Table 16.6 have pKa values in the desired range: hydrocyanic acid

(HCN, pK a = 9.31) and phenol (C6HsOH, pK a = 9.89)

Solution Plugging the values for phenol into Equation 1 7.1 gives

Therefore, the ratio of [C6HsO - ] to [C6HsOH] mu s t be 0.41 to 1 One way to achieve this would be

to dissolve 0.41 mole of C6HsONa and 1.00 mole of C6HsOH in 1 L of water

Practice Problem A Select an appropriate acid from Table 16.6, and de scri be how yo u would

prepare a buffer with pH = 4.5

Practice Problem B What range of pH values could be achieved with a buffer consisting of nitrou s

acid (HN02) and sodium nitrite (N02 )7

Bringing Chemistry to Life

Maintaining the pH of Blood

There are about 5 L of blood in the average adult Circulating blood keeps cells alive by providing

them with oxygen and nutrients and by removing carbon dioxide and other waste materials The

efficiency of this enolmously complex system relies on buffers

Blood plasma contains many compounds, including proteins, metal ions, and inorganic phosphates

The erythrocytes contain hemoglobin molecules, as well as the enzyme carbonic anhydrase, which

catalyzes both the formation and the decomposition of carbonic acid (H2C03) :

The substances inside the erythrocytes are protected from extracellular fluid (blood plasma) by a

semipermeable cell membrane that allows only certain molecules to diffuse through it

prin-cipal buffer systems are HCO;-IH 2 C 0 3 and hemoglobin The hemoglobin molecule is a complex

where HHb represents the hemoglobin molecule and Hb - is the conjugate base of HHb

than HHb:

Carbon dioxide produced by metabolic processes diffuses into the erythrocyte, where it is rapidly

converted to H2C03 by carbonic anhydrase:

Think About It There is an

infinite number of combinations of

[conjugate base] and [weak acid]

that will give the nece ssary ratio

Note that this pH could also be

achieved u s ing HCN and a cyanide salt For most purpose s, it is best

to u se the least toxic compounds available

690 CHAPTER 17 Acid-Base Equi libria and Solubility Equilibria

has two important consequences First, the bicarbonate ion diffuses out of the erythrocyte and

is carried by the blood plasma to the lungs This is the major mechanism for removing carbon dioxide Second, the H+ ions shift the equilibrium in favor of the un-ionized oxyhemoglobin

molecules:

Because HHb02 releases oxygen more readily than does its conjugate base (Hb0 2 ), the formation

of the acid promotes the following reaction from left to right:

HHbO zeaq) +: ==:!:' HHb(aq ) + 0 2(a q)

The O2 molecules diffuse out of the erythrocyte and are taken up by other cells to carry out metabolism

When the venous blood returns to the lungs, the preceding processes are reversed The bonate ions now diffuse into the erythrocyte, where they react with hemoglobin to form carbonic acid:

bicar-HHb (aq) + HC0 3 (a q) +: ==:!: ' Hb - (aq) + H 2 C0 3 (aq)

Most of the acid is then converted to CO2 by carbonic anhydrase:

The carbon dioxide diffuses to the lungs and is eventually exhaled The formation of the Hb - ions

(due to the reaction between HHb and HC03") also favors the uptake of oxygen at the lungs,

Hb -(aq) + O ? (aq) :;:::: ===z:' Hb0 2 (aq)

because Hb- has a greater affinity for oxygen than does HHb '

When the arterial blood flows back to the body tissues, the entire cycle is repeated

17.2.1 Which of the following combinations

ca n be used to prepare a buffer ? a) HCIICI -

b) HFIF

-c) CH3COOHlOH

-d) HN0 2 IN0 2

e) HN0 3 IN0 3

17.2.2 What i s the pH of a buffer that is

0.76 M in HF and 0 98 Min NaF?

a) 3.26

b) 3.04 c) 3.15

d ) 10.85 e) 10.74

Acid-Base Titrations

17.2.3

17.2.4

Consider 1 L of a buffer that i s 0.85 M

in formic acid (HCOOH) and 1.4 M in

sodium formate (HCOONa) Calculate the pH after the addition of 0.15 mol HCI (Assume the a d dition causes no

Con s ider 1 L of a b u ffer that is 1.5 M

in hydrocyanic acid (HCN) and 1.2 M

in sodi u m cyanide (NaCN) Calcu l ate the pH after the addition of 0.25 mol NaOH (Assume the addition causes no

In Section 4.6 we introduced acid-base titrations as a form of chemical analysis Having discussed

buffer solutions, we can now look in more detail at the quantitative aspects of acid-base titrations

We will consider three types of reactions: (1) titrations involving a strong acid and a strong base,

SECTION 17.3 Acid-Base Titrations 6 9'

u

" ~

I

!

(2) titrations involving a weak acid 'and a strong base, and (3) titrations involving a strong acid and

titration

NaOH(aq) + HCI(aq) - - - NaCI(aq) + H 2 0(l)

•

R ecall that only the d i gits to the right of the

de cimal point are significant in a pH va l u e

hydrogen ion and hydroxide ion concentrations are very small at the equivalence point (roughly ···

Recall t h at for an acid and base that c om bine

increases slowly with the continued addition of NaOH

as a function of volume titrant adde d )

of a strong acid-strong base titration

A 0.100 M NaOH solution, the titran t ,

is added from a buret to 25.0 mL of a

0.100 M Hel solution in an Erlenm e y er

flask

692 CHAPTER 17 Acid-Base Equilibria and Solubility Equ i libria

These calculations could also be done using

moles, but using m ill imoles sim plifies the

ca l culations Remember that millimoles = M x

mL [ ~ Section 4.5]

OH-It is possible to calculate the pH of the solution at every stage of titration Here are three sample calculations

1 Consider the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCI: The total vciiume' of"ihe · scii utl ori · ls · 3S".() · fiC · The· number· of"millirnoles of NaOH in 10.0 mL is

1 mL

25.0 mL X 0.100 ~l HCI = 2.50 mIllol

Thus, the amount of HClleft after partial neutralization is 2.50 - 1.00, or 1.50 mIllol Next,

35.0 mL

2 Consider the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCI: This is a

volume of the solution is now 60.0 mL The number of rnillimoles of NaOH added is

added (mL) added (mmol) Excess OH- (mmol) volume (mL) (moI / L) pOH pH

SECTION 17.3 Acid-Base Tit r ati o ns 69 3

Weak Acid-Strong Base Titrations

This equation can be simplified to

follows:

calculations

1 Prior to the addition of any base, the pH is determined by the ionization of acetic acid We

a weak acid-stro n g base titration

A 0.100 M NaOH so l ution is added from a buret t o 25.0 mL of a 0 100 M

CH3COOH so l ution in an E r lenme y er fla s k Beca u se of the hydrolysis of th e

s alt formed, t h e pH at t h e equivalen ce

point is greater than 7

694 CHAPTER 17 Acid-Base Equil i bria and Solubi l ity Equi li bria

Remember that we can use mol or mmol

amounts in place of concentrations in the

Henderson-Hasse l balch equation

K a XKb=Kw [ ~ Section 16.7]

Prior t o the addit i on of any base, and at

th e eq ui val e nce point, this is an equilibrium

problem that is solved using a concentration, an

ionization constant, and an equilibrium table

Prior t o t he equivalence point, pH is determined

using the Hender son-Hasselbalch equation:

I [conjugate base]

pH = pK a + og ~ -= o -,-;: '

[weak acid]

After the equivalen ce point, the titration curve

of a weak acid is identical to th at of a strong

acid

2 After the first addition of base, some of the acetic acid has been converted to acetate ion via

CH 3 COOH(aq) + OH-(aq ) - -+ CH3COO - (aq) + H 2 0 ( I )

With significant amounts of both acetic acid and acetate ion in solution, we now treat the

calculated in this way

3 At the equivalence point, all the acetic acid has been neutralized and we are left with

acetate ion in solution (There is also sodium ion, which does not undergo hydrolysis and

concentration and the K b of acetate ion, The equivalence point occurs when 25.0 mL of

,

[CH COO-] = 2,5 mmol = 0050 M

As we did at the beginning of the titration, we construct an equilibrium table:

.,

1.5 1.0

0.5

0,0

1.0 1.5 2.0 2.5