chapter 18 principles of chemical reactivity: other aspects of aqueous equilibria

Jeffrey MackCalifornia 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 0Change x +x +xEquilibrium 0.025 x x x



[NH3] [NH4+] [OH-]

Initial 0.25 0 0Change x +x +xEquilibrium 0.025 x x x

- 25 4

b3

[NH ][OH ] xK 1.8 x 10 [NH ] 0.25 - x



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

pOH = 2.67 pH = 14.00 2.67 = 11.33 for 0.25 M NH3

- 25 4

b3

[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 0Change x + x +xEquilibrium 0.025 x 0.10 + x x



[NH3] [NH4+] [OH-]


-5 4

b3

[NH ][OH ] (0.10 x) xK 1.8 x 10 [NH ] 0.25 x



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

b3

[NH ][OH ] (0.10 x) xK 1.8 x 10 [NH ] 0.25 x

5 50.25x OH 1.8 10 4.5 10 M0.10


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

10b

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

K 5.6 10

9rev

b

1K 1.8 10K


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.

4rev

a

1K 5.6 10K


Problem:What is the pH of a buffer that has [CH3CO2H] = 0.700 M and [CH3CO2

] = 0.600 M?

Buffer Solutions


] = 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


] = 0.600 M?

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

[CH3CO2H] [CH3CO2] [H3O+]


5 3a

[H O ] 0.600K 1.8 10 0.700

Buffer Solutions


] = 0.600 M?

5 3

a[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 ] KOrig. 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 ] KOrig. 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 effect0.30 – x 0.300.52 + 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• 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• First choose an acid with a pKa close to the desired

pH


• A pH of 4.30 corresponds to an [H3O+] = 5.0105 M• First choose an acid with a pKa close to the desired

pH• Next adjust the ratio of acid to conjugate base to

achieve the desired pH.

a

conj. BasepH pK log

Acid

Preparing a Buffer


• A pH of 4.30 corresponds to an [H3O+] = 5.0105 M• First choose an acid with a pKa close to the desired

pH

• Acetic acid is the best choice.

Possible Acids Ka

1.2 102

1.8 105

4.0 1010

24 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• Suppose you wish to prepare a buffer a solution at

pH of 4.30. How would you proceed?• 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 CO1mol

1L 10 mL0.278 mols CH CO H 2780 mL0.100mol 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 waterb) 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

What is the pH when 1.00 mL of 1.00 M HCl is added to: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 O10 mL 1 L 1 mol HCl 1.00 L

pH log(0.00100) 3.00


Adding an Acid to a BufferWhat is the pH when 1.00 mL of 1.00 M HCl is added to:b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and

[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.

What is the pH when 1.00 mL of 1.00 M HCl is added to:b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and

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

[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 0.700Change 0.00100 0.00100 + 0.00100After 0 0.599 0.701


What is the pH when 1.00 mL of 1.00 M HCl is added to:b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and

[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 0.599Change x + x +xEquilibrium 0.700 x x 0.599 + x

-53 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!

53 a-

53

[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 basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is its conjugate acidbuffer solution

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+] =

0.0280.10

final volume = 80.0 mL + 20.0 mL = 100 mL

[NH3] = 0.0250.10

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.


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


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

143 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-




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 > 7This will yield the [H3O+] and pH.


Volume of OH- added to the eq. point: HBz (aq) + OH(aq) Bz(aq) + H2O(l)

The new total volume of the solution is 125 mL

3

31L 0.025mol HBz 1mol OH 1L 10 mL100.0mL 25mL

10 mL 1L 1molHbz 0.100molOH 1L


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

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

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

31 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 mL 0.020 M Bz125 mL 1 L


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? [OH-] at the eq. point:

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

[Bz-] [HBz] [OH-]Initial 0.020 0 0Change - x + x + xEquilibrium 0.020 - x x x



[Bz-] [HBz] [OH-]Initial 0.020 0 0Change - x + x + xEquilibrium 0.020 - x x x

210

bxK 1.6 x 10

0.020 x



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: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? At the half-way point, moles of Hbz and Bz- are equal.

This is a BUFFER SOLUTION!



a

conj. BasepH pK log

Acid




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

K 22

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

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

wb

a(2)

KKK

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.


NeutralpH 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 solubleIf Ksp << 1 then MX is considered to be insolubleIf 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.

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.


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


AgCl(s) Ag+(aq) + Cl-(aq)Saturated solution has

[Ag+] = [Cl-] = 1.67 10-5 MUse this to calculate Kc

Kc = [Ag+] [Cl-] = (1.67 10-5)(1.67 10-5) = 2.79 10-10


AgCl(s) Ag+(aq) + Cl-(aq)Kc = [Ag+] [Cl-] = 2.79 10-10

Because this is the product of “solubilities”, we call it:

Ksp = solubility product constant


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


Recall that lead (II) iodide dissociates via:

spK 22

2 2sp

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

K [Pb ][I ]



From the reaction stoichiometry,

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

spK 22PbI (s) Pb (aq) 2I (aq)


Problem:The solubility of lead (II) iodide is found to be 0.00130M. What is the Ksp for PbI2?Entering these values into the Ksp expression,

For PbI2, Ksp = 4 (solubility)3

2 2sp

2sp

3 9sp

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 2spK [Mg ][F ]


Ksp for MgF2 = 5.2 1011. Calculate the solubility in:(a) moles/L & (b) in g/L (a) The solubility of the salt is governed by equilibrium

so let’s set up an ICE table:

2 2spK [Mg ][F ]


[MgF2(s)] [Mg2+] [F]Initial - 0 0Change - + x + 2xEquilibrium - x 2x

Ksp for MgF2 = 5.2 1011. Calculate the solubility in:(a) moles/L & (b) in g/L (a) Entering the values into the Ksp expression:

The solubility of MgF2 = 2.4 104 mols/L

2 2spK [Mg ][F ]


22sp

2 3sp

11sp 43 3

K Mg I

K x 2x 4x

K 5.2 10x 2.4 104 4

Ksp for MgF2 = 5.2 1011. Calculate the solubility in:(a) moles/L & (b) in g/L (b) the solubility of the salt in g/L is found using the

formula weight:

42 2 2

2.4 10 mols MgF 62.3g MgF MgF0.015g LL mol

2 2spK [Mg ][F ]


What is the maximum [Cl] in solution with 0.010 M Hg2

2+ without forming Hg2Cl2 (s)? 2

2 2 2

18sp

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

K 1.1 10



2+ without forming Hg2Cl2 (s)?

Precipitation will initiate when the product of the concentrations exceeds the Ksp.

22 2 2

18sp


K 1.1 10



2+ without forming Hg2Cl2 (s)?

The maximum chloride concentration can be found from the Ksp expression.

22 2 2

218 2sp 2

18sp 822


K 1.4 10 = Hg Cl

K 1.4 10Cl 1.2 100.010Hg


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

22

5sp

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 24 4

2 2 2sp 4

5sp

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

K Ba SO x

x K 1.0 10 M



(b)


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

Initial - 0.010 0Change - + x + xEquilibrium - 0.010 + x x


(b)


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

Initial - 0.010 0Change - + x + xEquilibrium - 0.010 + x x

2 2sp 2

sp

K Ba SO

K 0.010 x x


(b)

Since x << 0.010,

2 2sp 2

sp

K Ba SO

K 0.010 x x

sp

10sp

8

K 0.010 x

K 1.1 10x0.010 0.010

x 1.1 10 M



(b)

Since x << 0.010,

2 2sp 2

sp

K Ba SO

K 0.010 x x

sp

10sp

8

K 0.010 x

K 1.1 10x0.010 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



The concentration of

X drops




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 QuotientSuppose 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.Solution: Calculate “Q” and compare to Ksp.

22

22

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

Q Mg OH



22

26 4

14


Q 1.5 10 1.0 10

Q 1.5 10



Since Q < Ksp, no precipitation will occur.

14 12spQ 1.5 10 K 5.6 10


What concentration of hydroxide ion will precipitate a 1.5 106 M Mg2+ solution.

22

26sp


K 1.5 10 x


What concentration of hydroxide ion will precipitate a 1.5 106 M Mg2+ solution.

12sp

6 6

3

K 5.6 10x OH1.5 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:

22 6

23

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

13f

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

14D 13

f

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

1 1K 4.8 10K 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 10net f sp

+3 2 3

net 23


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 23


Ag(NH ) ClK 2.9 10

NH



+3 3 2

+3 2 3

net 23


Ag(NH ) ClK 2.9 10

NH

[NH3] [[Ag(NH3)2]+] [Cl]Initial 0.010 0 0Change - 2x + x + xEquilibrium 0.010 – 2x x x



23

netxK 2.9 10

0.010 2x

[NH3] [[Ag(NH3)2]+] [Cl]Initial 0.010 0 0Change - 2x + x + xEquilibrium 0.010 – 2x x x



Solving using the quadratic equation:X = 0.0032M = [Cl]

23

net

2 3 5

xK 2.9 100.010 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 g1 L 1 mol Cl 1 mol AgCl


Ksp ValuesAgCl 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 principles of chemical reactivity: other aspects of aqueous equilibria

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