chapter 18 advanced aqueous equilibria - suffolk county … · · 2014-02-05chapter 18 —...

1 Chapter 18 — Advanced Aqueous Equilibria

Jeffrey Mack

California State University,

Sacramento

Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

More About Chemical Equilibria: Acid–Base & Precipitation Reactions

Stomach Acidity & Acid–Base Reactions

• In the previous chapter, you examined the

behavior of weak acids and bases in terms of

equilibrium involving conjugate pairs.

• The pH of a solution was found via Ka or Kb.

• What would happen if you started with a

solution of acid that was mixed with a solution

of its conjugate base?

• The change of pH when a significant ammout

of conjugate base is present is an example of

the “Common Ion Effect”.

The Common Ion Effect

What is the effect on the pH of a 0.25M NH3(aq)

solution when NH4Cl is added?

NH4+ is an ion that is COMMON to the equilibrium.

Le Chatelier predicts that the equilibrium will shift to

the left to reduce the disturbance.

This results in a reduciton of the hydroxide ion

concentration, which will lower the pH.

Hint: NH4+ is an acid!

3 2 4NH (aq) H O NH (aq) OH (aq)


First let’s find the pH of a 0.25M NH3(aq)

Solution:

[NH3] [NH4+] [OH-]

Initial 0.25 0 0

Change x +x +x

Equilibrium 0.025 x x x



First let’s find the pH of a 0.25M NH3(aq)

Solution:

[NH3] [NH4+] [OH-]

Initial 0.25 0 0

Change x +x +x

Equilibrium 0.025 x x x

- 25 4

b

3

[NH ][OH ] xK 1.8 x 10

[NH ] 0.25 - x


First let’s find the pH of a 0.25M NH3(aq) Solution:

Assuming x is << 0.25, x = [OH] =

pOH = 2.67

pH = 14.00 2.67 = 11.33 for 0.25 M NH3

- 25 4

b

3

[NH ][OH ] xK 1.8 x 10

[NH ] 0.25 - x

5OH 0.25 1.8 10 0.0021 M


What is the pH of a solution made by adding

equal volumes of 0.25M NH3(aq) and 0.10M

NH4Cl(aq)?

Since the solutions are mixed with one

another, the effect of dilution is cancelled out.

One can use the initial concentrations of each

species in the reaction without calculating new

molarities.

This also works with mole ratios.


What is the pH of a solution made by adding equal

volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?

Since there is more ammonia than ammonium

present, the RXN will proceed to the right.

3 2 4NH (aq) H O NH (aq) OH (aq)

[NH3] [NH4+] [OH-]

Initial 0.25 0.10 0

Change x + x +x

Equilibrium 0.025 x 0.10 + x x


What is the pH of a solution made by adding equal

volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?

[NH3] [NH4+] [OH-]

Initial 0.25 0.10 0

Change x + x +x


-5 4

b

3

[NH ][OH ] (0.10 x) xK 1.8 x 10

[NH ] 0.25 x


What is the pH of a solution made by adding equal volumes of

0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?

Assuming x is << 0.25 or 0.10

pOH = 4.35 pH = 9.65

The pH drops from11.33 due to the common ion!

-5 4

b

3

[NH ][OH ] (0.10 x) xK 1.8 x 10

[NH ] 0.25 x

5 50.25x OH 1.8 10 4.5 10 M

0.10



HCl is added to pure water.

HCl is added to a solution of a weak acid H2PO4

- and its conjugate base HPO4

2-.

Controlling pH: Buffer Solutions

• A “Buffer Solution” is an example of the

common ion effect.

• From an acid/base standpoint, buffers are

solutions that resist changes to pH.

• A buffer solution requires two components

that do not react with one another:

1. An acid capable of consuming OH

2. The acid’s conjugate base capable of

consuming H3O+


Consider the acetic acid / acetate buffer system.

• The ability for the acid to consume OH is seen

from the reverse of the base hydrolysis:

• Krev is >> 1, indicating that the reaction is product

favored.

• An hydroxide added will immediately react with

the acid so long as it is present.

3 2 2 3 2

10

b

CH CO (aq) H O(l) CH CO H(aq) OH (aq)

K 5.6 10

9

rev

b

1K 1.8 10

K


Consider the acetic acid / acetate buffer

system.

• Similarly, the conjugate base (acetate) is

readily capable of consuming H3O+

• Krev is >> 1, indicating that the reaction is

product favored.

• An hydronium ion added will immediately

react wit the acid so long as it is present.

4

rev

a

1K 5.6 10

K


Problem:

What is the pH of a buffer that has [CH3CO2H] =

0.700 M and [CH3CO2] = 0.600 M?

Buffer Solutions

Problem:


0.700 M and [CH3CO2] = 0.600 M?

Since the concentration of acid is greater than the

base, equilibrium will move the reaction to the right.

3 2 2 3 2 3CH CO H(aq) H O CH CO (aq) H O (aq)

Buffer Solutions


Problem:


0.700 M and [CH3CO2] = 0.600 M?

Assuming that x << 0.700 and 0.600, we find:

[CH3CO2H] [CH3CO2] [H3O

+]

Initial 0.700 0.600 0

Change x + x +x


5 3a

[H O ] 0.600K 1.8 10

0.700

Buffer Solutions

Problem:


0.700 M and [CH3CO2] = 0.600 M?

5 3a

[H O ] 0.600K 1.8 10

0.700

Buffer Solutions

[H3O+] = 2.1 105

pH = 4.68

The expression for calculating the [H+] of the buffer reduces to:

The H3O+ concentration depends only Ka and the

ratio of acid to base.

3 23 a

3 2

Orig. conc. of CH CO H[H O ] K

Orig. conc. of CH CO

3 a

[Acid][H O ] K

[Conj. base]

Buffer Solutions

Similarly for a basic solution the [OH] of the buffer reduces to:

The OH concentration depends only Kb and the ratio of base to acid.

3 2b

3 2

Orig. conc. of CH CO[OH ] K

Orig. conc. of CH CO H

b

[Base][OH ] K

[Conj. acid]

Buffer Solutions

3 a

3 a

a

[Acid]log [H O ] K

[Conj. base]

[Acid]log[H O ] log K log

[conj. Base]

conj. BasepH pK log

Acid

The result is known as the “Henderson-Hasselbalch”

equation.

The pH of a buffer can be adjusted by manipulating

the ratio of acid to base.

Buffer Solutions: The Henderson-Hasselbalch Equation

24

What is the pH of a solution containing 0.30 M HCOOH and

0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect

0.30 – x 0.30

0.52 + x 0.52

pH = pKa + log [HCOO-]

[HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log [0.52]

[0.30] = 4.01

Mixture of weak acid and conjugate base!


• Suppose you wish to prepare a buffer a solution at

pH of 4.30. How would you proceed?

• A pH of 4.30 corresponds to an [H3O+] = 5.0105 M

Preparing a Buffer Preparing a Buffer




• First choose an acid with a pKa close to the desired

pH





pH

• Next adjust the ratio of acid to conjugate base to

achieve the desired pH.

a

conj. BasepH pK log

Acid

Preparing a Buffer





pH

• Acetic acid is the best choice.

Possible Acids Ka

1.2 102

1.8 105

4.0 1010

2

4 4HSO / SO

3 2 3 2CH CO H/CH CO

HCN/CN

Preparing a Buffer




3 a

5 5

[Acid][H O ] K

[Conj. base]

[Acid]5.00 10 1.8 10

[Conj. base]

[Acid] 2.78

[Conj. base] 1

Preparing a Buffer




• Therefore, if you start with 0.100 mol of acetate ion

then add 0.278 mol of acetic acid, will result in a

solution with a pH of 4.30.

[Acid] 2.78

[Conj. base] 1

Preparing a Buffer


Preparing a Buffer



• Therefore, if you start with 0.100 mol of acetate ion

then add 0.278 mol of acetic acid, will result in a

solution with a pH of 4.30.

• Since both species are in the same solution (the

same volume), the mole ratios are equal to the

concentration ratios!



• So, by adding 8.20 g of sodium acetate to 2780 ml

of a 0.100 M acetic acid solution, one would make a

buffer of pH 4.30.

3 2 3 2

3

3 2

82.03g0.100 mol NaCH CO 8.20g NaCH CO

1mol

1L 10 mL0.278 mols CH CO H 2780 mL

0.100mol 1L

Preparing a Buffer

Buffer prepared from

8.4 g NaHCO3

weak acid

16.0 g Na2CO3

conjugate base

What is the pH?

Preparing a Buffer

What is the pH when 1.00 mL of 1.00 M HCl is added to:

a) 1.00 L of pure water

b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and

[CH3CO2] = 0.600 M (pH = 4.68)

Adding an Acid to a Buffer


a) 1.00 L of pure water

mL L mols H3O+ M(H3O

+) pH

333

1 mol H O1 L 1.00 mols HCl 11.00 mL 0.00100 M H O

10 mL 1 L 1 mol HCl 1.00 L

pH log(0.00100) 3.00

Adding an Acid to a Buffer Adding an Acid to a Buffer



[CH3CO2] = 0.600 M (pH = 4.68)

1. The acid added will immediately be consumed by

the acetate ion.

2. This in turn increases the acetic acid concentration.

3. The acetic acid then will react with water to

reestablish equilibrium.




[CH3CO2] = 0.600 M (pH = 4.68)

1. The acid added will immediately be consumed by

the acetate ion.

3 3 2 3 2H O (aq) CH CO (aq) CH CO H(aq)


What is the pH when 1.00 mL of 1.00 M HCl is added to


[CH3CO2] = 0.600 M (pH = 4.68)

3 3 2 3 2H O (aq) CH CO (aq) CH CO H(aq)

[H3O+] [CH3CO2

] [CH3CO2H]

Before 0.00100 0.600 0.700

Change 0.00100 0.00100 + 0.00100

After 0 0.599 0.701




[CH3CO2] = 0.600 M (pH = 4.68)

2. This in turn increases the acetic acid

concentration.

3. The acetic acid then will react with water to

reestablish equilibrium.

3 2 2 3 3 2CH CO H(aq) H O(l) H O (aq) CH CO (aq)




[CH3CO2] = 0.600 M (pH = 4.68)

3 2 2 3 3 2CH CO H(aq) H O(l) H O (aq) CH CO (aq)

[CH3CO2H] [H3O+] [CH3CO2

]

Initial 0.701 0 0.599

Change x + x +x

Equilibrium 0.700 x x 0.599 + x

-5

3 a-

[HOAc] 0.701[H O ] K (1.8 x 10 )

[OAc ] 0.599




[CH3CO2] = 0.600 M (pH = 4.68)

The pH does not change! The solution absorbs the added

acid. It is a buffer!

5

3 a-

5

3

[HOAc] 0.701[H O ] K (1.8 10 )

[OAc ] 0.599

[H O ] 2.1 10

pH 4.68


• The solid acid and

conjugate base in the

packet are mixed with

water to give the specified

pH.

• Note that the quantity of

water does not affect the

pH of the buffer.

Commercial Buffers


43

Which of the following are buffer systems? (a) KF/HF

(b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) KF is a weak acid and F- is its conjugate base

buffer solution

(b) HBr is a strong acid

not a buffer solution

(c) CO32- is a weak base and HCO3

- is its conjugate acid

buffer solution

44

45 46

= 9.20

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer

system. What is the pH after the addition of 20.0 mL of

0.050 M NaOH to 80.0 mL of the buffer solution?

NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log [NH3]

[NH4+]

pKa = 9.25 pH = 9.25 + log [0.30]

[0.36] = 9.17

NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (moles)

end (moles)

0.029 0.001 0.024

0.028 0.0 0.025

pH = 9.25 + log [0.25]

[0.28] [NH4

+] = 0.028

0.10

final volume = 80.0 mL + 20.0 mL = 100 mL

[NH3] = 0.025

0.10

Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.

Acid–Base Titrations

Additional NaOH is added. pH rises as equivalence point is approached.

Acid–Base Titrations


Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.

Acid–Base Titrations Titration of a Strong Acid with a

Strong Base

The reaction of a strong acid and strong base

produces a salt and water.

The Net Ionic Equation is:

At the equivalence point, the moles of base added

equal the moles of acid titrated.

3 2H O (aq) OH (aq) 2H O(l)

3

3 w

3 3 w

14

3 w

H O OH

H O OH K

H O H O K

H O K 1.00 10

pH 7

Titration of a Strong Acid with a Strong Base

Problem:

100. mL of a 0.025 M solution of benzoic acid is

titrated with 0.100 M NaOH to the equivalence point.

What is the pH of the final solution?

Titration of a Weak Acid with a Strong Base

Equivalence Point

(moles base = moles acid)

At the equivalence

point, all of the acid is

converted to its

conjugate base.

The conjugate base

will then react with

water to reestablish

equilibrium.

The pH can be

determined from Kb.


HBz (aq) + OH(aq) Bz(aq) + H2O(l)

C6H5CO2H = HBz Benzoate ion = Bz-




Problem:




Solution:

At the equivalence point, all of the HBz is converted

to Bz- by the strong base.

The conjugate base of a weak acid (Bz-) will

hydrolyze to reform the weak acid (Kb). The pH will be

> 7

This will yield the [H3O+] and pH.

Problem:




Volume of OH- added to the eq. point:


The new total volume of the solution is 125 mL

3

3

1L 0.025mol HBz 1mol OH 1L 10 mL100.0mL 25mL

10 mL 1L 1molHbz 0.100molOH 1L


Problem:




Moles of OH- & Bz- at the eq. point:


The concentration of Bz- at the eq. point is:

3

1 L 0.025 mol HBz 1 mol OH 1 mol Bz100.0 mL 0.0025 mols Bz

10 mL 1 L 1 mol Hbz 1 mol OH

30.0025 mols Bz 10 mL0.020 M Bz

125 mL 1 L


Problem:




[OH-] at the eq. point:

10

2 bBz (aq) H O(l) HBz (aq) OH (aq) K 1.6 10

[Bz-] [HBz] [OH-]

Initial 0.020 0 0

Change - x + x + x

Equilibrium 0.020 - x x x


Problem:





[Bz-] [HBz] [OH-]

Initial 0.020 0 0

Change - x + x + x

Equilibrium 0.020 - x x x

210

b

xK 1.6 x 10

0.020 x


Problem:





Assuming that x << 0.020,

210

b

6

xK 1.6 10

0.020 x

X = [OH ] =1.8 10 pOH = 5.75 pH = 8.25



Conclusion: At the equivalence

point of the titration, unlike the

titration of a strong acid and

strong base, the pH is > 7.

This is due to the production of

the conjugate base of a week

acid.

Equivalence point

pH = 8.25


Conclusion: What would the pH

equal at the half-way point of the

titration?

Hint: Only ½ of the moles of

weak acid have been converted

to its conjugate base!

Half-way point

pH = ??



Problem:




At the half-way point, moles of Hbz and Bz- are

equal.

This is a BUFFER SOLUTION!

Problem:





equal.


a

conj. BasepH pK log

Acid


Problem:





equal.


a a a a

5

conj. BasepH pK log pK log(1) pK 0 pK

Acid

pH log 6.3 10 4.20


Acetic Acid Titrated with NaOH


In the case of a

titration of a weak

polyprotic acid (HnA)

there are “n”

equivalence points.

In the case of the

diprotic oxalic acid,

(H2C2O4) there are

two equivalence

points.

Titration of a Weak Polyprotic Acid with a Strong Base

The titration of a polyprotic weak acid follows the same

process a monoprotic weak acid.

As the acid is titrated, buffering occurs until the last eq. point is

reached.

The pH is relative to the amounts of conjugate acids and

bases.

At the second eq. point all of the acid has been converted to

A2-, pH is determined by Kb.

a(1)

a(2)

K

2 2

K2

2

H A(aq) OH (aq) HA (aq) H O(l)

HA (aq) OH (aq) A (aq) H O(l)

wb

a(2)

KK

K

Titration of a Weak Polyprotic Acid with a Strong Base

In the case of a

titration of a weak

base, the process

follows that of a weak

acid in reverse.

There exists a region

of buffering followed

by a rapid drop in pH

at the eq. point.

Titration of a Weak Base with a Strong Acid

pH Indicators for Acid–Base Titrations

• An acid/base indicator is a substance that changes

color at a specific pH.

• HInd (acid) has another color than Ind (base)

• These are usually organic compounds that have

conjugated pi-bonds, often they are dyes or

compounds that occur in nature such as red

cabbage pigment or tannins in tea.

• Care must be taken when choosing an appropriate

indicator so that the color change (end point of the

titration) is close to the steep portion of the titration

curve where the equivalence point is found.




Neutral

pH pH<<7 pH >7

Buffer

<7 pH >>7

Natural Indicators: Red Rose Extract in Methanol

• Prior to this chapter, exchange

reactions which formed ionic salts

were governed by the solubility rules.

• A compound was either soluble,

insoluble or slightly soluble.

• So how do we differentiate between

these?

• The answer lies in equilibrium.

• It turns out that equilibrium governs

the solubility of inorganic salts.

Solubility of Salts

Lead(II) iodide

The extent of solubility can be measured by the

equilibrium process of the salt’s ion

concentrations in solution, Ksp.

Ksp is called the solubility constant for an ionic

compound. It is the product of the ion’s

solubilities.

For the salt:

AxBy(s) xAy+(aq) + yBx-(aq)

Ksp = [Ay+]x[Bx-]y

Solubility of Salts Solubility of Salts

Consider the solubility of a salt MX:

If MX is added to water then:

Generally speaking,

If Ksp >> 1 then MX is considered to be soluble

If Ksp << 1 then MX is considered to be insoluble

If Ksp 1 then MX is slightly soluble

2H O(l)

sp

MX(s) M (aq) X (aq)

K [M ][X ]

Solubility of Salts

• All salts formed in this

experiment are said to

be INSOLUBLE.

• They form when

mixing moderately

concentrated solutions

of the metal ion with

chloride ions.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Analysis of Silver Group


• Although all salts formed in this experiment

are said to be insoluble, they do dissolve to

some SLIGHT extent.

AgCl(s) e Ag+(aq) Cl-(aq)

• When equilibrium has been established, no

more AgCl dissolves and the solution is

SATURATED.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2


AgCl(s) Ag+(aq) + Cl-(aq)

When solution is SATURATED, expt. shows

that [Ag+] = 1.67 x 10-5 M.

This is equivalent to the SOLUBILITY of AgCl.

What is [Cl-]?

[Cl-] = [Ag+] = 1.67 x 10-5 M

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2



Saturated solution has

[Ag+] = [Cl-] = 1.67 10-5 M

Use this to calculate Kc

Kc = [Ag+] [Cl-]

= (1.67 10-5)(1.67 10-5)

= 2.79 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2



Kc = [Ag+] [Cl-] = 2.79 10-10

Because this is the product of “solubilities”,

we call it:

Ksp = solubility product constant

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2


PbCl2(s) Pb2+(aq) + 2 Cl-(aq)

Ksp = 1.9 10-5 = [Pb2+][Cl–]2

Lead(II) Chloride

Problem:

The solubility of lead (II) iodide is found to be

0.00130M. What is the Ksp for PbI2?

Relating Solubility & Ksp


Problem:



Recall that lead (II) iodide dissociates via:

spK2

2

2 2

sp

PbI (s) Pb (aq) 2I (aq)

K [Pb ][I ]


Problem:



From the reaction stoichiometry,

[Pb2+] = 0.00130M & [I] = 2 0.00130M = 0.00260M

spK2

2PbI (s) Pb (aq) 2I (aq)


Problem:



Entering these values into the Ksp expression,

For PbI2, Ksp = 4 (solubility)3

2 2

sp

2

sp

3 9

sp

K [Pb ][I ]

K 0.00130 2 0.00130

K 4 0.00130 8.79 10


Ksp for MgF2 = 5.2 1011. Calculate the solubility in:

(a) moles/L & (b) in g/L

2 2

spK [Mg ][F ]




(a) The solubility of the salt is governed by equilibrium

so let’s set up an ICE table:

2 2

spK [Mg ][F ]


[MgF2(s)] [Mg2+] [F]

Initial - 0 0

Change - + x + 2x

Equilibrium - x 2x



(a) Entering the values into the Ksp expression:

The solubility of MgF2 = 2.4 104 mols/L

2 2

spK [Mg ][F ]


22

sp

2 3

sp

11sp 43 3

K Mg I

K x 2x 4x

K 5.2 10x 2.4 10

4 4




(b) the solubility of the salt in g/L is found using the

formula weight:

4

2 2 22.4 10 mols MgF 62.3g MgF MgF

0.015g LL mol

2 2

spK [Mg ][F ]


What is the maximum [Cl] in solution with 0.010 M

Hg22+ without forming Hg2Cl2 (s)?

2

2 2 2

18

sp

Hg Cl (s) Hg (aq) 2Cl (aq)

K 1.1 10




Precipitation will initiate when the product of the

concentrations exceeds the Ksp.

2

2 2 2

18

sp


K 1.1 10




The maximum chloride concentration can be found

from the Ksp expression.

2

2 2 2

218 2

sp 2

18sp 8

2

2


K 1.4 10 = Hg Cl

K 1.4 10Cl 1.2 10

0.010Hg


Adding an ion “common” to an equilibrium causes the equilibrium to shift towards reactants according to Le Chatelier’s principle.

Solubility & the Common Ion Effect

2

2

5

sp

PbCl (s) Pb (aq) 2Cl (aq)

K 1.7 10

Common Ion Effect


What is the the solubility of BaSO4(s) in

(a) pure water and (b) in 0.010 M Ba(NO3)2?

Ksp for BaSO4 = 1.11010





(a)

2 2

4 4

2 2 2

sp 4

5

sp

BaSO (s) Ba (aq) SO (aq)

K Ba SO x

x K 1.0 10 M





(b)


[BaSO4(s)] [Ba2+] [SO42]

Initial - 0.010 0

Change - + x + x

Equilibrium - 0.010 + x x




(b)


[BaSO4(s)] [Ba2+] [SO42]

Initial - 0.010 0

Change - + x + x

Equilibrium - 0.010 + x x

2 2

sp 2

sp

K Ba SO

K 0.010 x x




(b)

Since x << 0.010,

2 2

sp 2

sp

K Ba SO

K 0.010 x x

sp

10sp

8

K 0.010 x

K 1.1 10x

0.010 0.010

x 1.1 10 M





(b)

Since x << 0.010,

2 2

sp 2

sp

K Ba SO

K 0.010 x x

sp

10sp

8

K 0.010 x

K 1.1 10x

0.010 0.010

x 1.1 10 M


Solubility is

significantly

decreased by

the presence

of a common

ion.


Some anions of precipitates are the conjugate

bases of weak acids.

In solution these anions can hydrolyze to form the

weak acid.

Addition of additional base can shift equilibrium to

the right thus reducing the concentration of X.

This in turn would increase the solubility of the

ppt.

2X (aq) H O(l) HX(aq) OH (aq)

Effect of Basic Salts on Solubility


Decrease the

concentration of

HX

Add a base:



Eq. Shifts to the

right


Eq. Shifts to the

right

The

concentration of

X drops



Eq. Shifts to the

right

MX(s) M (aq) X (aq)

The solubility of MX increases!


Relating Q to Ksp

Q = Ksp The solution is saturated

Q < Ksp The solution is not saturated

Q = Ksp The solution is over saturated,

precipitation will occur

Ksp & the Reaction Quotient


Suppose you have a solution that is

1.5 106 M Mg2+.

Enough NaOH(s) is added to give a

[OH] = 1.0 106 M.

Will a precipitate form?

Ksp & the Reaction Quotient Ksp & the Reaction Quotient

Suppose you have a solution that is

1.5 106 M Mg2+.

Enough NaOH(s) is added to give a

[OH] = 1.0 106 M.

Will a precipitate form?

Solution: Calculate “Q” and compare to Ksp.

1.5 106 M Mg2+ & [OH] = 1.0 104 M.


2

2

22

Mg(OH) (s) Mg (aq) 2OH (aq)

Q Mg OH


1.5 106 M Mg2+ & [OH] = 1.0 104 M.


2

2

26 4

14


Q 1.5 10 1.0 10

Q 1.5 10


1.5 106 M Mg2+ & [OH] = 1.0 104 M.


Since Q < Ksp, no precipitation will occur.

14 12

spQ 1.5 10 K 5.6 10


What concentration of hydroxide ion will precipitate a

1.5 106 M Mg2+ solution.

2

2

26

sp


K 1.5 10 x



What concentration of hydroxide ion will precipitate a

1.5 106 M Mg2+ solution.

12

sp

6 6

3

K 5.6 10x OH

1.5 10 1.5 10

OH 1.9 10


Metal ions exist in solution as complex ions.

A complex ion involves a metal ion bound to

molecules or ions called “ligands”.

Ligands are Lewis bases that form “coordinate

covalent bonds” with the metal.

Examples are:

2

2 6

2

3

Ni(H O)

or

Cu(NH )

Equilibria & Complex Ions


The equilibrium constants for complex ion are

very product favored:

Kf is known as the formation equilibrium

constant.

2+ 2+

2 3 4

13

f

Cu (aq) + 4 H O(l) [Cu(NH ) ]

K 2.1 10


The extent of dissociation of a complex ions is

given by the “dissociation constant”, KD

2+ 2+

3 4 2

14

D 13

f

[Cu(NH ) ] Cu (aq) + 4 H O(l)

1 1K 4.8 10

K 2.1 10


The presence of a ligand dramatically affect

the solubility of a precipitate:

3 3 2AgCl(s) 2 NH (aq) Ag(NH ) Cl (aq)

Dissolving Precipitates by forming Complex Ions




+

3 3 2AgCl(s) 2NH (aq) Ag(NH ) Cl (aq)

Solubility of Complex Ions



sp

f

+

3 3 2

K+

K +

3 3 2

AgCl(s) 2NH (aq) Ag(NH ) Cl (aq)

AgCl(s) Ag (aq) Cl (aq)

Ag (aq) 2NH (aq) Ag(NH )




sp

f

+

3 3 2

K+

K +

3 3 2

7 10

net f sp

+

3 2 3

net 2

3


AgCl(s) Ag (aq) Cl (aq)

Ag (aq) 2NH (aq) Ag(NH )

K K K 1.6 10 1.8 10

Ag(NH ) ClK 2.9 10

NH


What is the solubility of AgCl(s) in grams per liter in a

0.010M NH3(aq) solution:

+

3 3 2

+

3 2 3

net 2

3


Ag(NH ) ClK 2.9 10

NH




+

3 3 2

+

3 2 3

net 2

3


Ag(NH ) ClK 2.9 10

NH

[NH3] [[Ag(NH3)2]+] [Cl]

Initial 0.010 0 0

Change - 2x + x + x

Equilibrium 0.010 – 2x x x




23

net

xK 2.9 10

0.010 2x

[NH3] [[Ag(NH3)2]+] [Cl]

Initial 0.010 0 0

Change - 2x + x + x

Equilibrium 0.010 – 2x x x





Solving using the quadratic equation:

X = 0.0032M = [Cl]

23

net

2 3 5

xK 2.9 10

0.010 2x

x 5.8 10 2.9 10 0




In pure water, the solubility of AgCl is only

0.0019 g/L

+

3 3 2AgCl(s) 2NH (aq) Ag(NH ) Cl (aq)

0.0032 mol Cl 1 mol AgCl 143.32g AgCl1 L 0.046 g

1 L 1 mol Cl 1 mol AgCl


Ksp Values

AgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrO4 1.8 x 10-14

Separating Metal Ions Cu2+, Ag+, Pb2+

chapter 18 advanced aqueous equilibria - suffolk county … · · 2014-02-05chapter 18 —...

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