Chapter 17

Additional Aspects of Equilibria

Salt Solutions in Water

Salt solutions completely ionize in H2O. If their ions can react with H2O to form H+ of OH-, the pH of the solution will change.

Anions that are the conjugate bases of weak acids will react with H2O to form OH- ions.

Anions that are conjugate bases of strong acids are very weak bases and aren’t strong enough to react with H2O. No pH change.

Anions that have ionizable protons (HSO4

-) are amphoteric. Can act as an acid or a base and must be determined based on the Ka and Kb for the ion.

Will K2HC6H5O7 form an acidic or basic solution in H2O?

--37562

-2756

-375632

-2756

OHOHCOHOHHC

OHCOHOHOHHC

or

Must look at Ka and Kb.

From table 16.3 in book get Ka3 = 4.0x10-7

Must calculate Kb from appropriate Kax.

-3756

-7562a2

-7562

-2756

OHHCHOHCHK

OHCHHOHHC

+2 +13-

Kw = Ka x Kb

b10-

5-

-14

a2

w K10 x 9.510 x 1.7

10 x 1

K

K

Now compare for our situation

Ka = 4.0x10-7

Kb = 5.9x10-10

Ka is 700 times larger than Kb so the acid reaction will predominate and solution will be acidic.

All cations (except alkali and heavy alkaline earth metal ions) act as weak acids.

Rules

• Salts derived from strong acid and strong base from neutral solutions in H2O.

ex. NaCl Na+ from NaOH

Cl- from HCl

• Salt derived from strong base/weak acid. Anion will be relatively strong conjugate base. pH > 7.

ex. Ba(C2H3O2)2

• Salt derived from weak base/strong acid. Strong conjugate acid. pH < 7.

ex. Al(NO3)3

Al3+ + 3H2O Al(OH3) + 3H+

• Salt derived from weak base/weak acid. ex. NH3C2H3O2. Both conjugate acid and base will be fairly strong in solution. pH depends on which ion hydrolyzes H2O better. Must compare Ka and Kb.

Additional aspects of Aqueous Equilibria

1. Common Ion Effect

Remember LeChatlier

-2(aq)32(aq)

equil

2(aq)32 OHCHOHHC

What happens if I add more C2H3O2- to

the solution?

The dissociation of a weak electrolyte is decreased by adding a strong electrolyte that has an ion in common with the weak electrolyte.

Ex. ? pH of solution of 0.30 mol acetic acid (HC2H3O2) and 0.30 mol NaC2H3O2 (sodium acetate) in 1.0L?

Steps

1. Identify major species and are they acid or base.

H+

H+

H+

C2H3O2-

C2H3O2-

C2H3O2-

C2H3O2-

C2H3O2-

C2H3O2-

Na+

Na+

Na+

HC2H3O2

HC2H3O2

HC2H3O2

Weak acidWon’t dissociatecompletelyso: C2H3O2

- + H+ + HC2H3O2

NaC2H3O2

Strong *electrolyteso: Na+ + C2H3O2

-

* How do I know that?

2) Identify important equilibrium reaction.

-2(aq)32(aq)322(aq)32 OHCOHOHOHHC

H+ is the same thing as H3O+ in aqueous solution (at this point in your chemistry life)

3) Calculate concentrations of equilibrium species.

Remember: C2H3O2- comes from HC2H3O2 and

NaC2H3O2

x)-(0.30

x)x)(0.30(

]OH[HC

]OH][C0[H1.8x10K

x)M(0.30xMx)M-(0.30

xMxMxM-

M30.000.30M

Eq

IH3OCOHOHOHHC

232

-23235-

a

AN

AN

AN

-2232232

Can we make the assumption that ‘x’ can be ignored here? How do you know?

2) Acid-Base Titrations

Quantitatively add an acid (or base) to a base (or acid).

Strong acid/strong base titrations:

The graph of pH vs. ml of titrant is called a titration curve.

pH before equivalence point is determined by concentration of acid NOT YET NEUTRALIZED.

pH at equivalence point is pH of the salt solution.

pH past equivalence point is determined by concentration of excess base.

? pH when 0.100M NaOH solution is added to 20.0 ml 0.100M HBr

a) 10.0 ml NaOH

H mol 002.0HBr mol 1

H mol 1HBr mol 0.002

HBr mol 002.0L

mol 0.100HBr L020.0

NaOH Na+ + OH- in waterSo ALL these will react with H+ until one or the other runs out!

So 10 ml

NaOH mol 001.0NaOH L

mol 0.100 NaOH L010.0

0.001 mol NaOH yields 0.001 mol OH- in H2O.

OH- + H+ H2O

H mol 001.0

OH mol 1

H mol 1 OH mol 001.0

--

Consumed by OH-

Started with 0.002 mol H+

That leaves 0.001 mol H+

pH = 31.5 pH

H M033.0L 0.030

H mol 0.001

B) 20.0 ml NaOH total

NaOH mol 002.0NaOH L

mol 0.100L 020.0

H mol 0.000

by consumed H mol 002.0

H mol 002.0

So, is our pH 14?

NOAutoprotolysis H2O H+ + OH-

pH would be that of the salt + autoprotolysis 7

C) 30.0 ml NaOH

OH mol 0.001

OH mol 002.0

OH mol 003.0

Consumed by

0.002 mol H+ in

initial solution

pOH = 3pH = 11

12.3 pH

1.70 pOH

OH M 02.0L 0.050

OH mol 0.001 --

Buffered Solutions

Solutions that resist change in pH when small amounts of acid or base are added.

Many neutral systems are buffered systems.

Buffers contain both an acidic species and a basic species.

A weak acid/base conjugate pair is common.

NaC2H3O2 and HC2H3O2

NH4Cl and NH3

Choose appropriate components and adjust concentrations to get buffer at any pH.

][X

[HX]K]H[

[HX]

]][X[HK

XHHX

-a

-

a

-

Ratio of conjugate acid/base pair

OH- + HX H2O + X-

[HX] [X-]

But if [HX] and [X-] are large enough, a small amount of [OH-] won’t matter.

Na HCl NaCl

H+ + X- HX

Same goes for a small amount of acid.

When [HX] and [X-] are approximately equal, buffer is most effective for pH change in either direction. Choose buffer whose acid has a pKa close to desired pH.

Buffer Capacity and pH

In General:

[acid]

[base]logpK pH a

Henderson-Hasselbach equation

Use starting concentration of acid and base components of the buffer.

? pH of Buffer0.15M NaHCO3

0.10M Na2CO3

[base]

[acid]K][H a

pH1.10

)176.0()25.10(07.10

(0.15)

(0.10)log)0-log(5.6x1

[acid]

[base]logpKpH

COHHCO

COHCOCOH

11-

a

-23

-3

-23

-332

Ka1 Ka2

Strong Acid-Weak Base

and

Weak Acid-Strong Base

Titrations

* pH > 7 at stoichiometric endpoint for weak acid/strong base titration.

* pH < 7 at stoichiometric endpoint for strong acid/weak base titration.

Example

30.0ml sample of 0.20M C6H5COOH is titrated with 0.30M KOH.

Ka C6H5COOH = 6.5x10-5

Calculate pH at stoichiometric endpoint.

C6H5COOH weak acid

C6H5COOH C6H5COO- + H+

Ka = 6.5x10-5 in water

But what about with a strong base?

OH- will deprotonate all of the C6H5COOH

So, at the endpoint, enough OH- has been added to react with all the C6H5COOH.

Up to this point, the problem looks just like a strong acid/strong base titration!

What’s different?

Weak acid yields a strong conjugate base.

So,

C6H5COO- + H2O C6H5COOH + OH-

That’s the logic.

COOHHC mol 0.006

L

COOHHC mol 0.20L030.0

56

56

Must have same moles of OH-

needed KOH vol L 0.02

KOH mol 0.30

L 1OH mol 006.0 -

Total volume = 0.03L + 0.02L

= 0.05L = 50ml

M12.0L 0.050

mol 0.006]COOH[C -

56

What is happening in solution?

Relatively strong base in water:

C6H5COO- + H2O C6H5COOH + OH-

ba

wb

-56

-56

b

KKK?K

]COOH[C

]COOH][OHH[CK

10-5-

-14

b x1054.16.5x10

1x10K

10-b

AN

AN

AN

-56256

x1054.1x)-(0.12

x)(x)(K

xxx)M-(0.12Eq

xxx-

000.12MI

OHCOOHHCOH-COOHC

10-2

x1054.10.12

x

x2 = 1.85x10-11

x = 4.3x10-6

x = [C6H5COOH] = [OH-]pOH = 5.37pH = 8.6

? pH

10.0g KCH3CO2 dissolved in 250ml solution.

Ka(CH3CO2) = 1.8x10-5

10.0g KCH3CO2 = 0.093mol KCH3CO2

0.093mol KCH3CO2 yields 0.093mol CH3CO2

-

10-5-

14-

a

wb

-232

-23

-23

-23

x106.51.8x10

x101

K

KK

OHHCOCHOHCOCH

COCH M37.00.250L

COCH 0.093mol

xxx)M-(0.37Eq

xxx-

000.37MI

OHHCOCHOHCOCH

AN

AN

AN

-232

-23

9.11pH

4.89pOH

][OH1.3x10x

1.68x10x

x106.50.37

x

x106.5)x - (0.37

(x)(x)K

-5-

10-2

10-2

10-b

Solubility Equilibria

Ksp = the degree to which a solid is soluble in water.

]][SO[BaK

SOBaBaSO-2

42

sp

-24(aq)

2(aq)4(s)

Table for Ksp at 25ºC

Appendix D

Ksp BaSO4 = 1.1x10-10

The smaller Ksp is means less will dissolve in H2O

Ksp Ca3(PO4)2 = [Ca2+]3 [PO43-]2 =

2.0x10-29

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

Converting between Solubility and Ksp

Solubility Molar solubility Molarof compound of compound [ions] Ksp

(g/L) (mol/L)

17.34) PbBr2 molar solubility = 1.0x10-2 mol/L

? Ksp

Lmol2--

(aq)

Lmol2-2

(aq)

-(aq)

2(aq)2(s)

2.0x10][Br

1.0x10][Pb

2BrPbPbBr

Ksp = [Pb2+][Br-]2

Ksp = (1.0x10-2)(2.0x10-2)2

Ksp = 4.0x10-6

Common Ion Effect

Remember Le Châtlier!-(aq)

2(aq)2(s) 2FCaCaF

17.36) ? solution of CaF2 in g/L in 0.15M KF solution-(aq)

2(aq)2(s) 2FCaCaF

Ksp = [Ca2+][F-]2

molar solution CaF2 = [Ca2+] = X

[F-] = 2X

another source of F- = 0.15M F-

So [Ca2+] = X

[F-] = 2X + 0.15

Ksp = (X)(2X + 0.15)2

* assume X is small compared to 0.15

Ksp = 3.9x10-11 = (X)(0.15)2 = 0.0225X

X = 1.7x10-9 M Ca

22

7-

22-9

CaF ppb 13.0L

g0.13

L

CaF g1.3x10

mol 1

CaF 78.1g

1L

CaF mol1.7x10

x

Precipitation will happen when

Q > Ksp

Q = ion product

if Q = Ksp, equilibrium exists, saturated solution

if Q < Ksp, solid dissolves until Q = Ksp

Hg2+

Solubility and pH

Very important in environmental systems

Consider Hg(OH)2 Ksp = 3.0x10-26

Hg(OH)2 Hg2+ + 2OH-

Ksp = [Hg2+][OH-]2 = 3.0x10-26

at pH = 3, pOH = 11 [OH] = 1.0x10-11

[Hg2+][1.0x10-11]2 = 3.0x10-26

[Hg2+] = 3x10-4M = 6.0x10-2g/L = 60mg/L = 60ppm

Selective Precipitation

Say you have Cu2+ and Zn2+ in solution

You can ppt one out of solution without the other use S2- by adding H2S(g)

CuS Ksp = 6.3x10-36

ZnS Ksp = 1.1x10-21

Must adjust pH to accomplish this.

][Zn

1.1x10][S

1.1x10K]][S[Zn

2

21--2

-21sp

-22

Say the solution in 0.10M in Zn2+ and 0.10 in Cu2+

M1.1x100.10

1.1x10

][Zn

1.1x10 20--21

2

-21

Now look at how H2S affects pH[H+]2[S2-] = 7x10-22

0.6)0-log(2.4x1pH

0.24M6x10][H

6x101.1x10

7x10

][S

7x10][H

1-

2-

2-20-

-22

-2

-222

ZnS won’t precipitate at pH below 0.6

Complex formation

Ions in solution come together and form a soluble complex so they can no longer be directly measured

422

3(aq)2(s)(aq) H2C3Fe6HClHCl2C3Fe12H

Should be able to use Cl- analysis to monitor reaction

NO WAY MAN

Can (and does) form a soluble FeCl2/FeCl3 complex