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Applications of Aqueous Equilibria

log fun with Ka:Henderson-Hasselbalch Equation

+ A-H+(aq) (aq)HA (aq)

Ka =

[H+] [A-][HA]

log Ka = log [H+] + log[A-]

[HA]

= [A-][HA]- log [H+] - log Ka log+

= [A-][HA]pH pKa log+

Ka = [H+] x [A-]

[HA]

= [conj.base ][acid]

pH pKa log+

Buffer Solutions

• contain a weak acid (or base) and its conjugate in a salt form.

• resist changes in pH upon the addition of small amounts of either acid or base

Example

Which of the following are buffer systems?

(b) KBr/ HBr

(a) KF / HF

(c) Na2CO3/ NaHCO3

yes; HF is a weak acid and F- is its conjugate base

no; HBr is a strong acid

yes; HCO3- is a weak acid and

CO32- is its conjugate base

(d) 0.5 mol HCl added to 1 mol KF solution

yes H+ + F- → HFI 0.5 1 0E 0 0.5 0.5

How a sodium acetate/ acetic acid buffer buffers

OAcH AcO-

the chemistry

AcO- OAcH

OH- H+

Weak acidIts conjugate base

in salt formbuffers have ~equal amounts of these to

start with

add baseadd acid

the mathematics

Ka =

[H+] [AcO-][HOAc]

How a sodium acetate/ acetic acid buffer buffers

= [AcO- ]

[HOAc]pH pKa log+

What pH’s do you get for this buffer when[AcO-]/[HOAc] = 1:1, 2:1, 3:1, 4:1, 5:1, 1:2, 1:3, 1:4? Group # 1 2 3 4 5 6 7 8

Rewriting in the form of Henderson-Hasselbalch equationWhen buffered (initial amounts of acid and its

conjugate base are present) this term stays pretty

constant even when acids or bases are added.

= 1.8 x 10-5

Calculate the pH of two 1.00-L beakers of water after adding a) 0.100 mol HCl to one and b) 0.100 mol

NaOH to the other. (both are strong so use stoichy!)

a) pH = 1

UNBUFFERED SOLUTIONS

+ Cl-H+ (aq) (aq)0.100 M 00

HClI

pH = -log[H+] = -log(0.100)

pOH = 1

+ OH-Na+(aq) (aq)0.100 M 00

NaOHI

pOH = -log[OH-] = -log(0.100)

E 0.100 M 0.100 M0

E 0.100 M 0.100 M0

pH + pOH = 14 b) pH = 13

BUFFERED SOLUTIONS

1. HOAc/OAc- buffer alone

2. #1 with 0.10 mol HCl added

3. #1 with 0.10 mol NaOH added

3 Scenarios to contrast with our unbuffered examples:

Calculate the pH of a buffer containing 1.0 M acetic acid and 1.0 M sodium acetate.

+ AcO-H+ (aq) (aq)OAc (aq)H

pH = 4.74

1.0 M 1.0 M+x-x + x

BUFFERED SOLUTIONS

Ka = [H+] [AcO-]

[HOAc]1.8 x 10-5 =

= [conj.base ][acid]

pH pKa log+

=(1.0 M + x)

pH -log(1.8 x 10-5) log+ (1.0 M - x)

1.0 M 1.0 M0IE

pH = 4.74What happens when 0.10 mol of gaseous HCl is added

to 1.0 L of this solution?


0.1 mol 1.0 mol 1.0 mol

≅ 0.0 mol 0.9 mol 1.1 mol

Key 1st step: Strong acid + weak base (OAc-) goes to completion. Use stoichiometry with acid/base

neutralization reaction.+ AcO-H+ (aq) (aq) OAc (aq)H

I

E

+H+ (aq) AcO-(aq) OAc (aq)H≅ 0.0 mol 0.9 mol 1.1 mol

Ka = [H+] [AcO-]

[HOAc] 1. 8 x 10-5 = x (0.9 + x)(1.1 - x)

x = [H+] = 2.2 x 10-5 M pH = 4.66

OAc (aq)H AcO-(aq) H+ (aq)+

E

E

I 1.1 mol 0.9 mol1L 1L

0 M

1.1 M - x 0.9 M + x +x

Key 2nd step: Now let system reach equilibrium using these as initial conditions in the acid dissociation (Ka) reaction.

Same calculation using the Henderson-Hasselbalch equation

= [c.base ][acid]pH pKa log+

= (0.9 + x)(1.1 - x)pH - log Ka log+

pH = 4.74 - 0.09 = 4.65

OAc (aq)H AcO- (aq) H+ (aq)+1.1 mol - x 0.9 mol + xE +x


pH = 4.74

What happens when 0.10 mol of gaseous HCl is added to 1.0 L of this solution?

pH = 4.66

What happens when 0.10 mol of solid NaOH is added to 1.0 L of this solution?

add 0.10 mol of NaOH(s) to 1.0 L of a solution that is 1.0 M in NaOAc and HOAc

+ AcO-OH- OAcH

0.1 mol 1.0 mol 1.0 mol

≅ 0.0 mol 0.9 mol 1.1 mol

+H2O

Key 1st Step: Strong base + weak acid goes to completion. Use stoichiometry with acid/base neutralization reaction.

Ka = [H+][AcO- ]

[HOAc] 1.8 x 10-5 = [H+](0.9 - x)

(1.1 + x)

[H+] = 1.47 x 10-5 M pH = 4.83

+ AcO-OH- OAcH≅ 0.0 mol 0.9 mol 1.1 mol

+H2O

AcO-OAcH +H+

E 0.9 mol 1.1 molx1L 1L

- x + x

Key 2nd step: Now let system reach equilibrium using these as initial conditions for the acid dissociation (Ka) reaction.

Same calculation using the Henderson-Hasselbalch equation

= [c.base ][acid]pH pKa log+

= (0.9 - x)(1.1 + x)pH - log (1.8 x 10-5) log+

pH = 4.74 + 0.09 = 4.83

0.9 mol - x 1.1 mol + xAcO-OAcH +H+

+xE


pH = 4.74

What happens when 0.10 mol of gaseous HCl is added to 1.0 L of this solution?

pH = 4.66

What happens when 0.10 mol of solid NaOH is added to 1.0 L of this solution?

pH = 4.83Compare to pH = 1 and pH = 13 for unbuffered solution!

Show the fraction of each species present as a function of pH

Distribution Curves

acid and conjugate base

base and conjugate acid

Buffer range = pKa ± 1.00

CH3COOH

100

75

25

50

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14distribution curves for acetic acid and

acetate ion as a function of pH

pH = pKa = 4.74

pH

%

CH3COO -

It is important because it contributes to controlling the pH of blood at 7.4.

The bicarbonate/carbonate acid buffer is more complicated because carbonic acid is

a diprotic acid.

pH

100

75

25

50

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

pKa =

pH =

6.83 10.32pKa =pH =

distribution curves for carbonic acid and bicarbonate ion as a function of

pH

CO32-HCO3

-H2CO3

%

Preparing a Buffer Solution with a Specific pH

= [c.base ][acid]

pH pKa log+

• center of buffer range: [acid] = [conj. base]What’s true about pH then?

• choose a weak acid/conjugate base combination where the pKa of the acid

corresponds to the pH at which you want the solution buffered

pH = pKa

How would you prepare a liter of “carbonate buffer” at a pH of 10.10 ?

you have available:

H2CO3

NaHCO3

Na2CO3

pKa 6.38

pKa 10.32

Pick NaHCO3 and its conjugate base.

How would you prepare a liter of “carbonate buffer” at a pH of 10.10 ?

=[c.base ]

[acid]pH pKa log+

=[Na2CO3

]

[NaHCO3]10.10 10.32 log+

[Na2CO3 ]

[NaHCO3]= 10-0.22 = 0.60

Acid-Base Titrations

Titration

a solution of known concentration, called a standard solution is added gradually to another solution of

unknown concentration until the chemical reaction between the two solutions is complete

equivalence point• the exact number of moles required for the

reaction

• detected by a color change in an acid-base indicator or electronically

Acid-Base Titrations

How does the pH change during the course of the following kinds of titration?

strong acid-strong base

weak acid-strong base

weak base-strong acid

Titrations Involving aStrong Acid and a

Strong Base

Titrations Involving a Strong Acid and a Strong Base

titrate 25.0 mL of 0.100 M HCl with 0.100 M NaOH

mmol molNote:mL and L

are equivalent

And for strong w/ strong, simply use stoichy! No Keq.


What is the pH after the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl ?

0.1mmol NaOH10ml x

mL= 1.0mmol OH-

0.1mmol HCl25ml x

mL= 2.5mmol H+

H2O (l )H+ (aq) + OH- (aq)

1.0mmol2.5mmol H+I 0

1.0mmol1.5mmol H+E 0

HCl (aq) NaCl (aq)H2O (l ) ++ NaOH(aq)



H2O (l )H+ (aq) + OH- (aq)

1.5mmol H+

35.0 mL (solution)[H+] = = 4.28 x 10-2 M

pH = 1.37



0.1mmol NaOH25ml x

mL= 2.5mmol OH-

0.1mmol HCl25ml x

mL= 2.5mmol H+

H2O (l )H+ (aq) + OH- (aq)

2.5 mmol2.5 mmol H+I 0

2.5 mmol0E 0



[H+] = 1.0 x 10-7 M

pH = 7.0

H2O (l ) + H3O+ (aq)+ OH- (aq)H2O (l )

pH versus ml 0.10 M NaOH added

(NaOH)volume

(total)volume

(HCl)mmol

M[H+] pH

0 25.0ml 2.5 0.10 1.0010.0 ml 35.0 ml 1.5 0.043 1.37

20.0 ml 45.0 ml 0.5 0.011 1.95

24.0 ml24.5 ml25.0 ml

49.0 ml49.5 ml50.0 ml

0.1

0.050

0.0020.00110-7

2.693.007.00

_

_

_

_

15

10

5

0

25 500

_ __ __

(NaOH)volume

pH7.0 at

equivalence point

Titration of 25.00 ml of

0.100 M HCl with 0.100 M

NaOH



0.1mmol NaOH35ml x

mL= 3.5mmol OH-

0.1mmol HCl25ml x

mL= 2.5mmol H+

H2O (l )H+ (aq) + OH- (aq)

3.5 mmol2.5 mmolstart 0

2.5 mmol0end 1.0 mmol


H2O (l )H+ (aq) + OH- (aq)

1.0 mmol OH-

60.0 mL (solution)[OH- ] = = 1.66 x 10-2 M

pH = 12.22

pOH = 1.78



(NaOH)volume

(total)volume

(HCl)mmol

M[H+] pH

25.0 ml 50.0ml 0 10-7 7.00

25.5 ml 50.5 ml 0.05 0.001 11.0026.0 ml 51.5 ml 0.10 0.002 11.2930.0 ml40.5 ml50.0 ml

55.0 ml65.0 ml75.0 ml

0.501.50

2.50

0.0090.0230.033

11.9612.3612.52

(NaOH)mmol

M[OH-]

_

_

_

_

15

10

5

0

25 500

_ __ __

(NaOH)volume

pH


0.100 M HCl with 0.100 M

NaOH

7.0 at equivalence

point

Sketch the titration curve if strong base is in the beaker and strong acid is in the buret. _

_

_

_

13

10

5

0

25 500

_ __ __

(HCl)volume

pH7.0 at

equivalence point


0.100 M HCl with 0.100 M

NaOH

Titrations Involving a weak Acid and a Strong Base

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH

after adding (a) 10.0 mL of 0.100 M NaOH(b) 25.0 mL of 0.100 M NaOH(c) 35.0 mL of 0.100 M NaOH


CH3COOH H2OCH3COONa ++ NaOH

CH3COOH H2OCH3COO - ++ OH -

Molecular Eqn

Net Ionic Eqn



1.0mmol2.5mmol 0 0

1.0mmol1.5mmol 0 1.0mmol

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding (a) 10.00 mL of 0.100 M NaOH. Key 1st step: Strong base (sodium hydroxide) + weak acid

(acetic) goes to completion. Use stoichiometry with neutralization reaction.


1.0mmol + x1.5mmol - x +x

I

E

CH3COOH CH3COO - + H+

Ka = 1.8 x 10-5

E

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding (a) 10.00 mL of 0.100 M NaOH.


1.0mmol1.5mmol

35.0 mL 35.0 mL= 0.0043 M 0.0028 M

pH = 4.57

Plan: Assume x is small since Ka is small and turn mmol into M .

= (0.0043 - x)(0.0028 + x)

pH - log (1.8 x 10-5) log+

1.0mmol + x1.5mmol - x +xCH3COOH CH3COO - + H+

Ka = 1.8 x 10-5

E

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding (b) 25.0 mL of 0.100 M NaOH.



2.5mmol2.5mmol 0 02.5mmol0 0 2.5mmol

50.0 mL= 5.0 x 10-2 M

2.5mmol

H2OCH3COO - + CH3COOH + OH - 5.6 x 10-10Kb =

Key 1st step: Strong base (sodium hydroxide) + weak acid (acetic) goes to completion. Use stoichiometry with neutralization reaction.

Key 2nd step: Now let system reach equilibrium using these as initial conditions for the base dissociation (Kb) reaction.

IE

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding

(b) 25.0 mL of 0.100 M NaOH.


H2OCH3COO - + CH3COOH + OH -

5.0 x 10-2 M - x x x

(5.0 x 10-2 - x)5.6 x 10-10 =

x2

x = 5.29 x 10-6 [OH -] =

pOH = 5.27

pH = 8.72

E

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding

(c) 35.0mL of 0.100 M NaOH.



3.5mmol2.5mmol 0 02.5mmol0 1.0mmol 2.5mmol

1.0mmol OH -

60.0 mL (solution)

= 1.67 x 10-2 MpH = 12.22

pOH = 1.78

strong base!

Strong base dominates, so we can ignore weak base and directly determine pH.

IE


(NaOH)volume

(total)volume

(NaOAc)mmol [H+] pH

0 ml 25.0ml 0 1.3 x 10-3 2.87

(HOAc)mmol

2.5

10.0 ml 35.5 ml 1 2.7 x 10-5 4.571.5

(NaOH)mmol [OH-]

25.0 ml 50.0 ml 2.5 8.720 1.9 x 10-9

35.0 ml 60.0 ml 1 12.220 1.7 x 10-2

_

_

_

_

15

10

5

2.5

25 500

_ __ __

(0.10 M NaOH)volume

pH


0.10 M acetic acid with 0.10

M NaOH

_

_7.5

12.5

_

++ +

++

+++++++

+

+

++++

++++++

+4.75

8.72

Maximum bufferWhy?

Why?

Titrations Involving a Strong Acid and a Weak Base

Calculate the pH in the titration of 25.0 mL of 0.100 M HCl with 0.100 M NH3 after adding

(a) 10.0 mL of 0.100 M NH3(b) 25.0 mL of 0.100 M NH3(c) 35.0 mL of 0.100 M NH3


HCl Cl -NH4+ ++ NH3

H + NH4++ NH3

Calculate the pH in the titration of 25.0 mL of 0.100 M HCl with 0.100 M NH3 after adding (a) 10.00 mL of 0.100 M NH3.


H + NH4++ NH3

1.0mmol2.5mmol 0

1.0mmol1.5mmol 0

1.5 mmol H + pH = 1.3735.0 mL

= 4.29 x 10-2M[H +] =

E

I

Key 1st step: Strong acid (HCl) + weak base (ammonia) goes to completion. Use stoichiometry with neutralization reaction.

strong acid!

Strong acid dominates, so we can ignore weak acid NH4+ and directly determine pH.

2.5mmol

50.0 mL= 5.0 x 10-2M

Calculate the pH in the titration of 25.0 mL of 0.100 M HCl with 0.100 M NH3 after adding (b) 25.0 mL of 0.100 M NH3.


H + NH4++ NH3

2.5 mmol2.5mmol 02.5 mmol00E

I

Key 1st step: Strong acid (HCl) + weak base (ammonia) goes to completion. Use stoichiometry with neutralization reaction.


NH3 H+Ka = 5.6 x 10-10

NH4+ +

E 2.5mmol - x +x +x

Calculate the pH in the titration of 25.0 mL of 0.100 M HCl with 0.100 M NH3 after adding (b) 25.0 mL of 0.100 M NH3.


(5.0 x 10-2 - x)5.6 x 10-10 =

x2

x = 5.29 x 10-6 [H+] =pH = 5.28

NH3 H+Ka = 5.6 x 10-10

NH4+ +

E 2.5mmol - x +x +x

Calculate the pH in the titration of 25.0 mL of 0.100 M HCl with 0.100 M NH3 after adding (c) 35.0mL of 0.100 M NH3.


3.5 mmol2.5mmol 0

H + + NH3 NH4+

2.5 mmol0 1.0 mmol

NH3 H+

Ka = 5.6 x 10-10

NH4+ +

2.5mmol 1.0mmol

60.0 mL= 4.1 x 10-2M

60.0 mL= 1.66 x 10-2M

Calculate the pH in the titration of 25.0 mL of 0.100 M HCl with 0.100 M NH3 after adding (c) 35.0mL of 0.100 M NH3.


NH3 H+

Ka = 5.6 x 10-10

NH4+ +

4.16 x 10-2 1.66 x 10-2

(4.16 x 10-2)=

1.66 x 10-2

(5.6 x 10-10 )= 1.4 x 10-9

pH = 8.85

[NH4+]

[H+] =[NH3]

Ka

pH versus ml 0.10 M NH3 added

(NH3)

volume (total)volume [H+] pH

(HCl)mmol

0 ml 25.0ml 0.1 1.02.5

10.0 ml 35.0 ml 0.0429 1.371.5

25.0 ml 50.0 ml 5.282.5 0

35.0 ml 60.0 ml 8.8512.5

(NH4+ )

mmol (NH3

)mmol

+

_

_

_

_

10

5

2.5

20 400

_ __ _

(0.10 M NH3)volume

pH


0.10 M HCl acid with 0.10

M NH3

_

_

7.5

12.5

_

+ + ++

+

++++

+++

60 80

+++++

++++

5.36 at equivalence point

+

_

_

_

_

10

5

2.5

20 400

_ __ _

(0.10 M NH3)volume

pH

_

_

7.5

12.5

_

+ + ++

+

++++

+++

9.25 when [NH3] = [NH4+]

60 80

+++++

++++


0.10 M HCl acid with 0.10

M NH3

Solubility Equilibria

Applies the principles of equilibrium to ionic solids of low solubility in water

• Milk of Magnesia demo

Solubility product (Ksp)

in a saturated solution, the species in solution is in equilibrium with undissolved

material

is characterized by an equilibrium constant Ksp called the solubility product

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

Examples:

= [Cl-]Ksp [Ag+] = 1.6 x 10-10

2Ag+Ag2CO3 (s) (aq) + CO32- (aq)

= [CO32-]Ksp [Ag+]2 = 8.1 x 10-12

3Ca2+Ca3(PO4)2 (s) (aq) + 2PO43- (aq)

= [PO43-]2 Ksp [Ca2+] 3 = 1.2 x 10-26

When ions in solution reform solid they do so on the surface of the solid.

NaCl ( s )

Na+Cl- Cl-Na+

The amount of excess solid present has no effect on the position of solubility equilibrium.

NaCl ( s )

Na+Cl- Cl-Na+

Doubling the surface area of the solid doubles both the rate of reforming and dissolving. position of equilibrium unchanged

Molar Solubility and Solubility

Molar solubility: the number of moles solute in one liter of a saturated solution (mol/L)

Solubility: the number of grams of solute in one liter of a solution (g/L)

Depending on the text being used

Solubility v.s. Solubility Product

• solubility product is an equilibrium constant and thus has only one value at a certain temperature

• solubility is an equilibrium position and has an infinite number of possible values at a given

temperature depending on the conditions

ie: the common ion effect

given Ksp you should be able to calculate

molar solubility

concentration of anion in a saturated solution

concentration of cation in a saturated solution

molar solubilities when common-ion effect comes into play

Practice Exercise

The solubility of PbCrO4 is 4.5 x 10-5 g/L. What is its solubility product ?

Pb2+PbCrO4 (s) + CrO42- (aq)(aq)

4.5 x 10-5g

Lx

323.2 g

1 mol= 1.4 x 10-7 mol/L

First convert to mol/L

Practice Exercise

The solubility of PbCrO4 is 4.5 x 10-5 g/L. What is its solubility product ?

Pb2+PbCrO4 (s) + CrO42- (aq)(aq)

1.4 x 10-7 M 1.4 x 10-7 M

[CrO42 -] = Ksp [Pb2+]

(1.4 x 10-7) (1.4 x 10-7)= Ksp

= 1.9 x 10-14Ksp

Practice Exercise

Calculate the molar solubility of PbCO3. (Ksp = 3.3 x 10-14 )

Pb2+PbCO3 (s) + CO32- (aq)(aq)

E +x +x

I 0.00 M 0.00 M

[CO32 -] = Ksp [Pb2+]

(x) (x)= Ksp = (x)2

x = 1.8 x 10-7mo l /L

3.3 x 10-14=

Practice Exercise

The Ksp value for Cu(OH)2 = 2.2 x 10-20 at 25ºC. Calculate its solubility in g/L.

Cu2+Cu(OH)2 (s) + 2OH- (aq)(aq)

E +x +2xI 0.00 M 0.00 M

[OH-]2 = Ksp [Cu2+]

(x)(2x)2= Ksp = 4x3 2.2 x 10-20== (x)(4x2)

4x3 2.2 x 10-20=

2.2 x 10-20

4( (1/3

x =

x = 1.77 x 10-7 molL

x97.55 g

1 mol

x = 1.72 x 10-5 g/L

Relative Solubilities

CaSO4

SaltAgCl

CuI

Ksp

8.3 x 10-17

5.1 x 10-12

6.1 x 10-5

all Ksp expressions are of the form[Y ] = Ksp [X ]

Therefore, solubility decreases in the order

CaSO4 AgClCuI> >


Bi2S3

SaltCuSAg2S

Ksp

8.3 x 10-45

1.6 x 10-49

1.1 x 10-73

Therefore, need to do a calculation

Ksp expressions are of the form[Cu2+][S2-], [Ag+]2[S2-], [Bi3+]2[S2-]3


CuS

8.3 x 10-45 [Cu2+][S2-]

Cu2+ + S2 - (aq)(aq)

= = x2

Ag2S

1.6 x 10-49 [Ag+]2 [S2-]

2Ag+ + S2 - (aq)(aq)

= = 4x3

Bi2S3

1.1 x 10-73 [Bi3+]2 [S2-]3

2Bi3+ + 3S2 - (aq)(aq)

= = 108x5


Bi2S3

Salt

CuS

Ag2S

Ksp

8.3 x 10-45

1.6 x 10-49

1.1 x 10-73

Molar solubility

9.2 x 10-23

3.4 x 10-17

1.0 x 10-15

Predicting Precipitation Reactions

Ion Product

Qc > Ksp: precipitate will form

Qc = Ksp: saturated solution

Qc < Ksp: no precipitate will form; substance dissolves

Q = [products] x [reactants] y

• analogous to reaction quotient

• tells us whether a precipitate will form under a given set of reaction conditions

Practice Exercise

If 2.00 ml of 0.200 M NaOH are added to 1.0 L of 0.100 M CaCl2, will a

precipitate form?

Practice Exercise

If 2.00 ml of 0.200 M NaOH are added to 1.0 L of 0.100 M CaCl2, will a

precipitate form?= 8.0 x 10-6 Ksp

Ca2+(aq)Ca(OH)2 (s) + 2OH- (aq)

Ca2+(aq)Ca(OH)2 (s) + 2OH- (aq)

Ca2+ = 1.0 L x0.10 mol

L= .1 mol

OH- = 2.0 x 10-3 L x0.20 mol

L= 4.0 x 10-4 mol

.1 mol

1.002 L

4.0 x 10-4 mol

1.002 L

[Ca2+] = 0.100 M [OH ] = 0.0004 M

Ca2+(aq)Ca(OH)2 (s) + 2OH- (aq).100 M .0004 M

[OH-]2 = Q [Ca2+]

< KspQ

(.0004 )2 (.100 )= Q

= 8.0 x 10-6 Ksp

= 1.6 x 10-8 Q

; precipitate will not form

Separation of Ions by Fractional Precipitation

When a solution contains several ions can one be removed by precipitation, leaving

all the others in solution?

Practice Exercise

(a) AgCl

Calculate the concentration of Ag+

necessary to cause precipitation of

(Ksp = 1.6 x 10-10 )

(b) Ag3PO4 (Ksp = 1.8 x 10-18 )

from a solution in which Cl- and PO43-

are each present at a concentration of 0.10 M.

Practice Exercise

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

Ksp 1.6 x 10-10=[Cl-] = 0.1 M ;

3Ag+Ag3PO4 (s) + PO43- (aq) (aq)

Ksp 1.8 x 10-18=[PO43- ] = 0.1 M ;

Practice Exercise

calculate the concentration of Ag+at which AgCl begins to precipitate

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

Ksp 1.6 x 10-10=[Cl-] = 0.1 M ;

1.6 x 10-10 = [Cl-] [Ag+]

1.6 x 10-10 = (0.1) [Ag+]

[Ag+] = 1.6 x 10-9

Practice Exercise calculate the concentration of Ag+ at which

Ag3PO4 begins to precipitate

1.8 x 10-18 = [Ag+] 3 [PO43- ]

1.8 x 10-18 = (0.10) [Ag+]3

[Ag+] = 2.6 x 10-6

3Ag+Ag3PO4 (s) + PO43- (aq) (aq)

Ksp 1.8 x 10-18=[PO43- ] = 0.1 M ;

[Ag+]3 = 1.8 x 10-17

Practice Exercise

[Ag+] to begin precipitation of:

[PO43-] = 2.6 x 10-6 M

[Cl-] = 1.6 x 10-9 M

Who precipitates first?AgCl precipitates before Ag3PO4

Practice Exercise

What is [Cl-] when Ag3PO4 begin precipitate ?

PO43- begins to precipitate when [Ag+]

reaches 2.6 x 10-6 M.

Ksp = 1.6 x 10-10 = [Ag+] [Cl-]

1.6 x 10-10 = (2.6 x 10-6) [Cl-]

[Cl-] = 6.2 x 10-5

The Common Ion Effect and Solubility

A common ion suppresses the solubility of an ionic substance

Practice Exercise Calculate the solubility of AgBr in g/L in:

Ksp = 7.7 x 10-13

Ag+AgBr (s) + Br- (aq)(aq)

(a) pure water

(b) 0.0010 M NaBr

Practice Exercise


(a) pure water

(b) 0.0010 M NaBr

Le Chatelier’s principle: by adding NaBr, the equilibrium should shift to the _____and the

solubility of AgBr is ______.

Calculate the solubility of AgBr in g/L in: Ksp = 7.7 x 10-13

leftless

Practice Exercise


(a) pure water

x2 7.7 x 10-13=x 8.8 x 10-7 M=

[Br-] = Ksp [Ag+] 7.7 x 10-13=

= 1.7 x 10-4 g/Lsolubility of

AgBr


[Br-] = Ksp [Ag+] 7.7 x 10-13=

Practice Exercise


(b) 0.0010 M NaBr

[Ag+] 7.7 x 10-10 M=

7.7 x 10-13=[Ag+] ( 0.0010)

= 1.4 x 10-7 g/L solubility of AgBr


pH and Solubility

pH and Solubility

Mg2+Mg(OH)2 + 2OH-

Ksp = 1.2 x 10-11 = [Mg2+] [OH-]2

consider Mg(OH)2

solubility decreases in basic solution because increase in [HO-] requires decrease in

[Mg2+]

pH and Solubility

Mg2+Mg(OH)2 + 2 HO-

Ksp = 1.2 x 10-11 = [Mg2+] [HO-]2

consider Mg(OH)2

solubility increases in acidic solution because [HO-] is

decreased

H+

H2O

Practice Exercise

Calculate its molar solubility of Cr(OH)3 (Ksp = 3.0 x 10-29 ) in a pH 10 buffer.

Cr3+Cr(OH)3 (s) + 3OH - (aq)(aq)

= Ksp 3.0 x 10-29 = [OH -]3 [Cr3+]

3.0 x 10-29 = [10-4]3 [Cr3+]

= 3 x 10-17 mol/L[Cr3+]

pOH = 4

In General

salts of weak acids will be more soluble in acid solution than in pure water

because the anion of a weak acid (conjugate base) is basic

pH and Solubility

Ba2+BaF2 + 2F -Ksp = 1.7 x 10-6 = [Ba2+] [F -]2

BaF2 is a salt of a weak acid; its anion (F-) is basic

solubility increases in acid because F - concentration is

decreased

H+

HF

Complex Ion Equilibria and Solubility

“A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.”

demos - write the equilibria for each stepCuCl2 + NH3 + HCl

AgNO3 + NaCl + NH3 + HCl

kind of like Ksp except....

Formation constant

�

K f =[metal cation][neutral or anion ligand]

[complex ion]

the product is a dissolved ion complex, not a solid precipitate

Cu2+(aq) Cu(NH3)42+(aq)

[Cu(NH3)42+]

[Cu2+] [NH3]4Kf = = 5.0 x 1013

NH3 (aq)

• a complex ion is formed by Lewis acid-Lewis base reaction;

• the bond between the Lewis acid and Lewis base is covalent

• a complex ion is characterized by the formation constant ( Kf )

metal ion is the Lewis acid

the neutral molecule or ion that acts as the Lewis base is called a Ligand

• the number of ligands attached to the metal ion is called the coordination number

Some typical complex ions

complex ion Kf

Ag(NH3)2+ 1.5 x 107Ag+ 2NH3+

Cu2+ 4CN -+

Co3+ 6NH3+

Cu(CN)42- 1.0 x 1025

Co(NH3)63+ 5.0 x 1031

The effect of complex ion formation generally is to increase the solubility of a

substance.

AgCl (s) Ag(NH3)2+ (aq)

[Ag(NH3)2+]

[Ag+] [NH3]2Kf = = 1.5 x 107

NH3 (aq)

Practice exercise

Ag+(aq)AgBr (s) + Br - (aq)

Ksp = 7.7 x 10-13

What is the molar solubility of AgBr in a 1.0 M solution of NH3?

(aq)Ag+ 2NH3 Ag(NH3)2++(aq) (aq)

1.5 x 107Kf =

AgBr (s) 2NH3+ (aq) + Br - (aq)Ag(NH3)2

+ (aq)Kf Ksp = K

AgBr (s) 2NH3+ (aq)

+ Br - (aq)Ag(NH3)2+ (aq)

Practice exercise

What is the molar solubility of AgBr in a 1.0 M solution of NH3?

Kf Ksp = K [Ag(NH3)2

+] [Br-]

[NH3]2 = 12.3 x 10-6 =

12.3 x 10-6 =x2

1.0x = 0.0035 M

the solubility of AgBr in pure water

is 9 x 10-7 M

Some typical coordination numbersCoordination

numbers

Ag+

Cu+

Co3+

Au+

Mn2+

Fe2+

Co2+

Ni2+

Cu2+

Zn2+

Cr3+

Au3+

Sc3+

Coordination numbers

Coordination numbers

2,4

2

2,4

4,6

4,6

4,6

4,6

4,6

6

6

6

6

4

Some common ligands

CN-

NH3

H2O

CH3NH2

CO

NO

F-

Cl-

Br-

I-

SCN-

Indicators

applications of aqueous equilibriateachermarten.com/apchem/aplecturenotes_files/ch15zum7... ·...

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