Tài liệu Chapter 10: Polyprotic Acid-Base Equilibria pptx

Treating Complex Acid-Base Systems Complex systems are defined as solutions made up of: 1 ‘Two acids or bases of different strengths HCI + CH;COOH NaOH + CH;COƠ' 2 An acid or base tha

Chapter 10:

Polyprotic Acid-Base Equilibria Chapter 11:

Acid-Base Titrations

1 Treating Complex Acid-Base Systems

Complex systems are defined as solutions made up of:

1) ‘Two acids or bases of different strengths

HCI + CH;COOH NaOH + CH;COƠ'

2) An acid or base that has two or more acidic

protons or basic functional groups

H3PO, Ca(OH),

3) Anamphiprotic substance that is capable of acting

as both acid and base

HCO; + H,O <=> CO,” + H;0'°

HCO; + H›O <=> H.CO, + OH

NH; CH,COO' + H,0 <=> NH,CH,COO’ + H;0°

NH;°CH;COO + H,O <=> NH; CH,COOH + OH

2 Equations for more than one equilibrium are required to describe the characteristics of any of these systems

3 Applications

1) We encounter these systems in most biological and environmental matrices

2) We need to predict which species will be

present (and in what amounts) 1s important in

defining biological acid/base buffering

3) We can use the titration curve to examine how

the species will change as we slowly add titrant (in other words, change the pH).

4 Examples

A Mixtures of Strong and Weak Acids or Strong and Weak Bases

B Polyfunctional Acids and Bases:

C Buffer solutions Involving Polyprotic Acids:

D Calculation of the pH of Solutions of Amphiprotic Salts (NaHA):

E Titration Curves of Polyfunctional Acids:

A Mixtures of Strong and Weak Acids or Strong and Weak Bases

28

10.0

= 6.0

4.0

0 2U 30 0U 00 OO

Volume of 0.1000 M NaOH, mL

4 Examples

A Mixtures of Strong and Weak Acids or Strong and Weak Bases

Example 11-l1A: Derive a titration curve for a titration of a 50.00 mL solution containing 0.1000 M strong acid, HCl, and 0.1000 M weak acid

HA (K, = 1 X 10”) with 0.1000 NaOH (similar to Example 11-1)

The equilibria:

HCl + H,O => H;0°+ CI [1]

HA + HạO <=> HạO'"+ Aˆ [2]

2 HO <=> H:O” + OH |3 |

[H;0°] = Cyc: + [A] + [OH]

Since the HCI, which 1s completely dissociated, will repress the dissociation

Of HA and HO

Assume that [OH | and [A ]<<Cyc so that [H3;0+] = Cyc

1 Vxaon =0: The pH before the addition of titrant is determined by the

concentration of HCl alone: [H30+] = Cuc

2.0 <Vwaou <50.00 mL: After titrant has been added, the titration curve will be identical to that for the titration of the HCI alone and

pH will be determined by the remaining HCl in the solution

Moles of [H,O° |]— Moles of [OH | „.„

V total

3 VNaon =20.00 mL: When the HCI has been neutralized (the

first equivalence-point), the presence of the weak acid must be considered At this point, the equilibrium described in eq [2]:

4.50.00 mL<Vyaon <100.00mL HA is now reacting with the titrant

and the titration curve will be identical to that of the titration of a weak acid and

pH = pk, —log 7 = pK, +log LÔ = pK, +log Chúa

5 Vyaoy =100.00 mL At the point where both HCl and HA are

neutralized (the second equivalence-point), the titration solution now

contains A’ which reacts with water

6 Vyaoy 7100.00 mL After all of the acid has been neutralized,

further addition of titrant results in a mixture of a weak and strong base

pH of the solution is determined by the concentration of the strong

base

Moles of [OH] — Moles of [H,07 ]

V total

[OH ]=

In summary, the titration curve includes a titration curve of

HCI (a strong acid) with NaOH and a titration curve of HA

(weak acid) with NaOH

You still can apply the theories and equations of neutralization titration in Chapter 9 into this complex system.

EXAMPLE 11-1

Calculate the pH of a 25.00-mL mixture that is 0.1200 M in hydrochloric acid and 0.0800 M in the weak acid HA (K, = 1.00 * LO~*) during its ttra-

tion with 0.1000 M KOH Derive data for additions of the following mL of

base: (a) 0.00 and (b) 5.00 and (c) 29.00 mL

Cha VuaA=CkowVkon Yond bo Vist re-kort25-00x0.0800/0.1000= 50.00mL

| (a) 0.00 mL KOH Added

[H,07] = eye + [AT] = 0.1200 + [AT ]= Cy, = 0.1200 M

[Aq ] K, — 1.00 10”

0.0800 = cy, = [HAI + {|A- |

0.0800 = [A 148.33 x I0-#) +/A-| = (1.20 x I0?)(A- ]

[A-] = 6.7 x 1075

[H,07] = cue + [A>] = 0.1200 + [AT] = C, = 0.1200 M

(b) Upon Addition of 5.00 mL of Base

Before the 1%‘ equivalence point: 5.00 mL <30.00 mL

To determine whether our assumption is still valid, we compute [A™ ] as

we did in part (a), knowing that the concentration of HA is now 0.0800 % 25.00/30.00 = 0.0667, and find

[AT]= 8.0 x 103

(c) Upon addition of 29.00 mL of 0.1000 M KOH

Before the 1%‘ equivalence point: 29.00 mL <30.00 mL

In addition, from mass-balance considerations, we know that

[HA] + [A~] = eq, = 3.70 X 107? (11-2)

We rearrange the acid dissociation-constant expression for HA and obtain

A Mixtures of Strong and Weak Acids or Strong and Weak Bases

28

10.0

= 6.0

4.0

0 2U 30 0U 00 OO

Volume of 0.1000 M NaOH, mL

Any Questions?

Equilibria Constants for Overall Reactions H;PO, + H,O <=> H,0° + H,PO, [Eq 1]

Definition of K, and K,, for Polyfunctional Acids

Ka

H:POx + H;O <=> HO” + H;POx

H›POx' + HO <=> H:O” + HPO,Z

HPO,” + H›O <=> HO” + PO,”

Ky

PO, +H O <=> HPO,” + OH”

HPO,” + H.O <=> H»PO, + OH”

H»PO, + H›O <=> H:PO„ + OH

K,, Ky3 = Ky, Ky = K,3 y= [H30°][OH] = K,,

Any Questions?

C Buffer solutions Involving Polyprotic Acids:

(A) Three buffer systems can be described when using a

weak diprotic acid, H;A, and its salts, H,A™ HA” and A”

(B) Two buffer systems can be described when using a weak

diprotic acid, H,A, and its salts, HA’ and A’

Wecan treat these systems as separated one for calculation of

pH and species present

Example 11-2:

Calculate the hydronium ton concentration for a buffer solution that is”

2.00 M in phosphoric acid and 1.50 M in potassium dihydrogen phosphate

The principal equilibrium in this solution is the dissociation of H,PQ,

We now use the equilibrium-constant expression for K, to show that

[HPOZ” ] can be neglected:

[H,O0* )[HPOZ-] 9.48 * 10-°[HPO |

(H;PO; ] 1.50

[HPOZ2” ] = 1.00 x 1075

K = 6.34 X 10-4.=

D Calculation of the pH of Solutions of Amphiprotic Salts (NaHA) The pH is determined by the following equilibria:

Using the systematic method to solve for the [HO]:

b2 —

2 The mass balance equation:

3 The charge balance equation:

[H;O”] + [Na ]= [OH] + [HA] + 2[A“]

[Na ] — CNaHA

Hence [HạO”]| + Cxana = [OH] + [HA] + 2[A7]

Subtracting the mass balance from the charge balance equation:

[H;O”]= [A“] + [OH] - [H;A]

We can express all terms using [HA'], [H:O ], Ka, Ky and K,>

HO") = eta

Example 11-3:

Calculate the hydronium ion concentration of a 0.100 M NaHCO, solution

We first examine the assumptions leading to Equation 11-11 The dissoci- ation constants for H,CO, are K,, = 4.45 X 10°? and K,, = 4.69 * 107"

Clearly, ¢x4,/K,) in the denominator is much larger than unity; in addition,

K oCwana has a value of 4.69 X 10~'*, which is substantially greater than K,,- Thus, Equation 11-11 applies and

Titration curve of Polyfunctional Acid (H,A)

of HA- and A”~

E Titration Curves of Polyfunctional Acids:

For the titration of a diprotic acid (H2A) with a strong base,

the [H;O°] in the following regions of the titration curve are calculated as follows:

If CHoA/Kại >100, [H,O' ] — \JCu¿AKạ 2) 0< VTitrant < Vist equivalence point

4) Vist equivalence point < V Titrant < Vond equivalence point

pH = pK,, + log— Cụy, [H;O”]= sa [A>]

5) V Titrant — Vond equivalence point

(OH ]=,/C.K,, =.|C Aw

c — moles of acid (H,A) before titration _ Cia Vacid before titration

° total volume Vicia + Veitrant

6) V Titrant > Vond equivalence point

Moles of [OH |.„.„ —Moles of[H.,O” ]

V

total

[OH |=

Because K,, for maleic acid is large, we must solve the quadratic equation

exactly or by successive approximations, When we do so, we obtain

[H,0*] = 3.01 x 10-?

pH = 2 — log 3.01 = 1.52

First Buffer Region: 0 <V y,o4 < 25 00 mL VNaon =5 mLU

Cum = [HM] = ` = a0 ee ĐI 6.67 X I0?M

Substitution of these values into the equilibrium-constant expression for K al

yields a tentative yalue of 5.2 X 10-* M for [H,O*] It is clear, however,

that the approximation [H,O* | <= ey, 4 Or Cyyy- is not valid; therefore Equa-

tions 10-6 and 10-7 must be used, and

[HM~| = 1,67 * 10-* + [H,0*+] — JOH-]

[H;M] = 6.67 * 10” = [HOT] + [OHTT

Because the solution is quite acidic, the approximation that [OH~ ] is very

small is surely justified Substitution of these expressions into the dissociation-

constant relationship gives

[H,O° ](1.67 X 10-2 + [H,0*]) 6.67 X 10°? — [H,0* |

First Equivalence Point i

At the first equivalence point, Vv NaOH = 25 00 mL

25.00 < 0.1000 -

HM ~] = crx = [ |] = Craum 50.00 5.00 x 102?

Simplification of the numerator in Equation 11-10 is clearly justified, On the

other hand, the second term in the denominator is not <= I Hence,

Further additions of base to the solution create a new buffer system consist-

| ing of HM~ and M*~ When enough base has been added so that the reaction

of HM~ with water to give OH~ can be neglected (a few tenths of a milliliter

beyond the first equivalence point), the pH of the mixture is readily obtained

from K,, With the introduction of 25.50 mL of NaOH, for example,

Second Equivalence Point VNaon = 29 90 mL

After the addition of 50.00 mL of 0.1000 M sodium hydroxide, the solution is

0.0333 M in Na,M Reaction of the base M*~ with water is the predominant

equilibrium in the system and the only one that we need to take into account

_ moles of acid (H,M) before titration _ Cum Vacid before titration 0.1000x25.00

total volume V acid +VYV titrant —25.00+50.00

pH Beyond the Second Equivalence Point VNaOH> 50 00 mL |

Further additions of sodium hydroxide repress the basic dissociation of M*

The pH is calculated from the concentration of NaOH added in excess of that

required for the complete neutralization of H,M Thus, when 51.00 mL of NaOH have been added, we have a 1.00 mL excess of 0.1000 M NaOH and

H.M + H,0 =H,O* +HM-—sK,, = 1.3 X 10°?

HM~ + H,O == H,0° + M” Ki = 5.9 X 107?

Any Questions?

Example 11-4: Identify by letter the curve you would expect in the titra

tion of a solution containing

(a) disodium maleate, Na,M, with standard acid C curve

(b) pyruvic acid, HP, with standard base A curve

(c) sodium carbonate, Na,CO,, with standard acid C eyrve

F General Expression for a Values of a Polyprotic Acid (H,A)

(Fraction of species when pH is fixed and known)

For the denominator:

D=Cr =[H;O'”]" + Kai [H30"]""? + Kai Kao[H30 J + 4 Kar Kao-Kn

where n = number of acidic protons in formula

For a diprotic acid (H,M):

Summary

Y Derive titration curves of complex acid-base system

v Mixtures of strong and weak acids or bases

Y Polyfunctional acids and bases

Y Buffer solution involving polyprotic acids

v Amphiprotfic salts (e.ø., NaHCO2z)

Y General Expression for ơ values

Homework

> 10-A, B, D, 4, 11, 17, 18, 23

> 11-A,B, F, 3, 6, 14, 19, 23

Before working on Homework,

Practice with all examples that we discussed 1n the class and examples in the textbook!!!