10/15/2014

1

More Acid and Base Chemistry

Common-ion effect In the last chapter, we calculated the [H3O

+] of a 0.123M HClO as 6.02x10-5M. The percent dissociation for this solution would be:

0.0489% x1000.123

6.02x10 5

But what would happen if we place 0.123moles of HClO in a 1.000L container that already has 0.100moles of the conjugate base NaClO. Will this change anything?

I

E C

nitial

hange

quilibirum

0.123M 0M

-x +x

0.123M - x x

HClO(aq)+ H2O(l)⇌ H3O

+(aq) + ClO-(aq)

0.100M

+x

0.100+x

[HClO]

]][ClOO[H3a

K

x-0.123

x)(x)(0.1002.95x10 8

Hypoclorite ion already in solution before acid added

Assume these “x’s” will be very small

0.123

(x)(0.100)2.95x10 8

0.100x3.63x10 9

83.63x10x

0.0000295% x1000.123

3.63x10 8

Note the change in percent dissociation. The percent dissociation decreased because there

were already ClO- ions in solution. The reaction did not proceed as far toward the reactants to

reach equilibrium. This is the common-ion effect. Common ions in solution decrease

dissociation of weak acids and bases.

10/15/2014

2

Common-ion effect What would happen if the solution had a pH of 1.000 before adding 0.123M HClO?

0.0489% x1000.123

6.02x10 5

I

E C

nitial

hange

quilibirum

0.123M 0.100M

-x +x

0.123M - x x

HClO(aq)+ H2O(l)⇌ H3O

+(aq) + ClO-(aq)

0M

+x

0.100+x [HClO]

]][ClOO[H3a

K

x-0.123

x)(x)(0.1002.95x10 8

Hydromiun ion already in solution before acid added

Assume these “x’s” will be very small

0.123

(0.100)(x)2.95x10 8

0.100x3.63x10 9

83.63x10x

0.0000295% x1000.123

3.63x10 8

Both products of dissociation of HClO, ClO- and H3O

+, caused a decrease in the dissociation of the acid. This is the common

ion effect.

How would ionization be affected if both H3O+

and ClO- were 0.100M initially? Hint: think about Qa

Buffers Buffers have the ability to stabilize pH, even with the addition of acids or bases. Buffers are

composed of weak conjugate bases and acids in equilibrium.

Add HCl

Plain ole water

pH = 7 pH = 0

Cl- H+

ΔpH = -7

NH4+

NH3

Add HCl

NH3 /NH4+ buffer solution

NH4+

NH4+

NH4+

NH3

NH3

NH3

NH4+

NH3 NH4+

NH4+

NH4+

NH3

NH3

NH4+

pH = 9.25

Cl-

pH = 9.03

ΔpH = -0.22

HCl + H2O → H3O+ + Cl-

HCl + NH3 → NH4+ + Cl-

HCl adds H3O+ to the solution with its

reaction with water.

HCl reacts with NH3 to give more NH4+

ions. H3O+ can only be added to the

solution through the NH3/NH4+

equilibrium

Through the reaction, 1 mol of H3O+ is

added to the water.

Through the reaction, only 3.7x10-10 mol of H3O

+ is added to the water.

10/15/2014

3

Calculating pH of buffer solutions

Initial Change Equilibrium

5.0M 0M

-x +x

5.0M - x x

NH4+(aq)+ H2O(l)⇌

H3O+(aq) + NH3(aq)

3.0M

+x

3.0M+x

Calculate the pH of a buffer solution in a 1.0L container with 3.0 moles NH3 of and 5.0 moles of NH4+. Ka for NH4

+ is 5.8x10-10.

][NH

]][NHO[H

4

33a

K x-5.0

x)(x)(3.05.8x10 10

5.0

(x)(3.0)5.8x10 10

109.7x10x

pH = -log[H3O+] = -log(9.7x10-10) = 9.01

Because the [NH3] +

[NH4+] =

8.0M, this is an 8.0 M

buffer

acid] [conjugate

base] [conjugatelogppH a K

Henderson – Hasslebalch Formula •Change in acid/base concentrations by dissociation negligible. [acid/base]initial ≈ [acid/base]equilibrium

But instead of another I.C.E. table, we can use…

5.0M

3.0Mlog9.24pH pH=9.02

Let’s do the same calculation above, but using H-H

Adding Strong Acid/Base to a Buffer Solution

What happens when we add 0.50L of 2.00M HCl to the solution

First, we need to determine how adding HCl changes the NH3/NH4+ buffer. To do this, we will use a change table. It is best

to use moles not M in this table (you will see why later).

Let’s take our first buffer problem from the last page: “Calculate the pH of a buffer solution in a 1.0L container with 3.0 moles NH3 of and 5.0 moles of NH4

+. Ka for NH4+ is 5.8x10-10.”

H+ (aq) + NH3(aq) → NH4

+(aq) Note that the HCl (acid) will react with the NH3 (base) in the solution.

1.0mole 3.0moles 5.0moles The strong acid or base will always react completely.

Remember, 100% dissociation.

0mole 2.0moles 6.0moles

Before:

-1.0mole -1.0mole +1.0mole

After:

Change:

acid] [conjugate

base] [conjugatelogppH a K

6.0mol

2.0mollog9.24pH

pKa = -log Ka = -log(5.8x10-10) = 9.24

Since both NH3 and NH4+ are in the same

volume, we can use moles instead of concentration. If you divide both HA and A- by the same volume, the ratio stays the

same, pH=8.76

10/15/2014

4

A More Realistic Buffer Solution Problem Calculate the pH of a buffer solution in a 100.mL container with 0.100M of each component of a benzoic acid /

sodium benzoate buffer. Ka of benzoic acid is 6.4x10-5.

acid] [conjugate

base] [conjugatelogppH a K

0.100M

0.100Mlog19.4pH 19.4pH

What is the pH after adding 10.00mL of 0.500M NaOH?

OH- (aq) + HA(aq) → A- (aq)

0.00500mol 0.0100mol 0.0100mol

0mol 0.0150mol

Before:

After:

Change: -0.00500mol -0.00500mol +0.00500mol

0.00500mol

0.00500mol

0.0150mollog19.4pH 67.4pH

A-

A-

HA

A-

Since there are still appreciable amounts of both HA and A-, it still a buffer solution.

Thus, H-H can be used

Buffer Solution

10/15/2014

5

A-

A-

HA

A-

A-

A- A-

A-

A- A- OH-

HA H+ HA Na

+ Cl-

Cl- Na+

A More Realistic Buffer Solution Problem What is the pH after adding a total of 20.00mL of 0.500M NaOH to the original (100.0mL 0.100M each) buffer

solution? Ka 6.4x10-5.

OH- (aq) + HA(aq) → A- (aq)

0.0100mol 0.0100mol 0.0100mol

0mol 0.0200mol

Before:

After:

Change: -0.0100mol -0.0100mol +0.0100mol

0mol

Is this even a buffer solution any more? NO! Only A-! Then I cannot use H-H!

A-

A- A-

A-

Weak Base Solution

0.167M 0M

-x +x

F- (aq)+ H2O(l)⇌ OH-(aq) + HF(aq)

0M

+x

Initial Change

00.167M - x x x Equilibrium

--

A M167.00.1000L L02000.0

A mol0200.0

102

10x6.10.167

x 610x2.5x

6- 10x2.5x ][OH

)012.5log(14pOH41 pH -6x

72.8 pH

NaOH mol0100.0NaOH L 1

NaOH 0.500molNaOH L0200.0 HA mol0100.0HA L 1

HA 0.100molHA L1000.0

10/15/2014

6

A More Realistic Buffer Solution Problem What is the pH after adding a total of 30.00mL of 0.500M NaOH to the original (100.0mL 0.100M each) buffer

solution?

OH- (aq) + HA(aq) → A- (aq)

0.0150mol 0.0100mol 0.0100mol

0.00500mol 0.0200mol

Before:

After:

Change: -0.0100mol -0.0100mol +0.0100mol

0mol

Which base should have a greater effect on pH?

A-

A- A-

A-

Strong Base Solution

OH-

--

OH M0385.00.03000 L1000.0

OH mol00500.0

1.4150.0385 -logpOH

1.415000.41pH

12.585pH

How a buffer reactions to addition of strong base or strong acid

adding base

adding acid

Mol of conjugate base in 1mol total buffer

Note that our theoretical buffer has a Ka of 1x10-7 and thus a pKa of 7. acid] [conjugate

base] [conjugatelogppH a K

Assume we have a buffer with 1M total buffer strength. Let’s start with 0.50M of each HA and A-.

10/15/2014

7

When does a buffer stop “buffering”?

10:1

1:10

1:1 base:acid

Mol of conjugate base in 1mol total buffer

No longer buffering

A good rule of thumb: A buffer is relatively effective ± 1 pH point of the acid’s pKa. After that point the buffer is exghausted and can no longer

buffer in that direction.

Strong Acid / Strong Base Titration problems (determining the pH at different points of a titration) are very similar to buffer questions. Use change tables, and see what’s left.

Determine the pH during a titration after 1.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl.

OH- (aq) + H+ (aq) → H2O

7.50x10-5 mol 1.00x10-4 mol Before:

After:

Change: -7.50x10-5 mol -7.50x10-5 mol

0mol 2.5x10-5 mol H+

H+ Cl-

Cl-

Cl- Cl- Na+ Na+

H M0022.00.00150L L01000.0

H mol.5x102 -5

2M)-log(0.002pH 66.2pH Strong Acid Solution

Determine the pH during of10.00mL of 0.0100M HCl before

the titration. 0M)-log(0.010pH

000.2pH H+

H+

H+

H+

Cl-

Cl- Cl- Cl-

What type of solution is this?

Strong Acid Solution

10/15/2014

8

Strong Acid / Strong Base Determine the pH during a titration after 2.00mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl.

OH- (aq) + H+ (aq) → H2O

1.0x10-4 mol Before:

After:

Change:

0mol

1.0x10-4 mol

-1.0x10-4 mol

0mol

-1.0x10-4 mol Cl-

Cl-

Cl-

Cl- Na+ Na

+

Na+ Na+

00.7pH

Neutral Solution

Strong Acid / Strong Base

OH- (aq) + H+ (aq) → H2O

1.00x10-4 mol Before:

After:

Change:

2.5x10-5 mol

1.25x10-4 mol

-1.00x10-4 mol

0mol

-1.00x10-4 mol

Determine the pH during a titration after 2.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl.

---5

OH M0020.00.00250L L01000.0

OH mol.5x102

70.10M)-log(0.002pOH

70.100.14pH

30.12pH

Cl- Cl-

Cl-

Cl-

Na+ Na+

Na+

Na+ OH-

Strong Base Solution

10/15/2014

9

Acid-Base Titrations Strong Acid/Strong Base

Adding Base to Acid

Vol of NaOH added (mL)

Equivalence Point: The amount of base added equals the amount of acid present before the titration.

The acid/base reaction is complete. For strong

acid/strong base titration the pH is 7.

After

During Before

H+

H+

H+

H+

Cl-

Cl- Cl- Cl-

Before Titration

H+

H+ Cl-

Cl-

Cl- Cl-

During Titration

Na+ Na+

add NaOH

Cl-

Cl-

Cl-

Cl-

Equivalence Point

Na+ Na+

Na+ Na+

Cl- Cl-

Cl-

Cl-

Past Equivalence Point

Na+ Na+

Na+

Na+ OH- add

NaOH add

NaOH

Acid-Base Titrations Weak Acid/Strong Base

Equivalence Point: The amount of base added equals the amount of acid present before the titration.

The acid/base reaction is complete. For strong

acid/strong base titration the pH is >7.

After

During (buffer zone)

Before

Vol of NaOH added (mL)

½ volume of equivalence point pH = pKa

[HA] = [A-]

Eq. Pt.

HF

Before Titration During Titration

add NaOH

Equivalence Point Past Equivalence

Point

add NaOH

add NaOH

F- F- Na+

Na+ HF HF

HF

HF HF F-

F- Na+

Na+ F- F-

Na+

Na+ F-

F- Na+

Na+

F- F-

Na+

Na+

OH- Na+

10/15/2014

10

Acid-Base Titrations

Adding Acid to Base

Equivalence Point: The amount of acid added equals the amount of base present before the titration.

The acid/base reaction is complete. For strong

acid/strong base titration the pH is 7.

Strong Acid/Strong Base

After

During

Before

Na+ Na+

Na+ Na+ OH- OH- OH-

OH- add HCl

Na+ Na+

Na+ Na+ OH- OH-

Cl-

Cl-

add HCl

Cl-

Cl-

Cl-

Cl-

Equivalence Point

Na+ Na+

Na+ Na+

Before Titration During Titration

add HCl

Cl-

Cl-

Cl-

Cl- Na+ Na

+

Na+ Na+ Cl-

Past Equivalence Point

H+

Strong Acid/Weak Base

Acid-Base Titrations

After

During (buffer zone)

Before

Equivalence Point: The amount of acid added equals the amount of base present before the titration.

The acid/base reaction is complete. For strong

acid/weak base titration the pH is

10/15/2014

11

Strong Acid / Weak Base Titrations

Determine the pH during a titration after 1.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF.

Determine the pH during of10.00mL of 0.0100M HF before the titration. Ka = 6.3x10-4

Determine the pH during a titration after 2.00mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF.

Determine the pH during a titration after 2.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF.

Polyprotic acid Titrations

Since there are multiple protons to remove from a polyprotic acid, there are multiple equivalence points – one for each proton.

The the pH of the equivalence point of an amphoteric species (on that can either

donate or accept an proton)

2

p p species amphotericof pH 1nn

aa

KK

H3PO4 H2PO4- HPO4

2- PO43-

⇌ ⇌ ⇌

pKa1 = 2.12 pKa2 = 7.20 pKa3 = 12.38

- 2- 3-

10/15/2014

12

Solubility Product – Ksp In CHM151, we talked about ionic compounds that were “insoluble” in water. In reality, those

compounds do dissolve, just a very. very small amount.

For example, based on our rules, AgBr is insoluble. How some very small amount does dissociate into ions. The represent this we have another K expression called Ksp.

The solubility of AgBr is represented by the equation:

Since solids do not appear in any K expression…

Ksp = [Ag+][Br-] and Ksp = 5.4x10

-13

Ag+ Br-

So, how do we determine the concentration of Ag+ and Br- in the solution?

0M

+x

x

0M

+x

x

Initial Change Equilibrium

AgBr(s) ⇌ Ag+(aq) + Br-(aq)

132 5.4x10x

132 5.4x10x

M7.3x10x 7

[Ag+] = [Br-] = 7.3x10-7M This is called the “molar

solubility”

Solubility Product – Ksp (more complicated example)

Mg2+

OH-

OH-

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

How do we deal with sparingly soluble salts that do not have cations and anions in a 1:1 ratio?

Ksp = [Mg2+][OH- ]2

0M

+x

x

0M

+2x

2x

Initial Change Equilibrium

(x)(2x)2 =1.2x10-11

(x)(4x2) = 4x3 = 1.2x10-11

Ksp = 1.2x10-11

3

11

4

1.2x10x

x = 1.44x10-4M

[Mg2+] = x = 1.4x10-4 M

[OH-] = 2x = (2)1.44x10-4 M = 2.9x10-4M

10/15/2014

13

Solubility Product – Ksp

Ag+ Br-

Ag+ Br-

Add solid AgBr

Remember LeChatelier’s Principle holds that the addition of a solid should not shift equilibrium. Does this hold true? YES! Assuming there is already some solid in the solution, the concentration of ions does not change.

Ag+ Ag+

Ag+ Add water

Initially, as the water is added, the concentration of ions is decreased. Based on LeChatelier’s Principle the reaction should shift toward the product’s side. Thus, in the end, more AgBr is dissolved and the concentration of ions is the same.

Br-

Br- Br-

In all fours cases, the [Ag+] 7.3x10-7 M and [Br-] is 7.3x10-7 M

Factors that effect solubility of sparingly soluble salts

Common-ions

pH (acid/base reactions

Acids will increase the solubility of a sparingly soluble salts with a basic ion (OH-, F- ,etc.) through an acid/base reaction.

Fe2+ OH-

OH-

Add HCl

Fe2+ Cl-

Cl- Fe2+ Cl-

Cl-

OH- Fe2+ OH-

Equilibrium reestablished

Common ions will decrease solubility of a sparingly soluble salt. Same as common ion effect.

Formation of Complex ions (Lewis acid/base reactions

Cu2+

Add a lot of NH3

OH-

OH-

10/15/2014

14

Common ion of a sparingly soluble salt

Let’s say we have 0.50L of 0.50M NaOH. Into this solution we drop in a large chunk of Mg(OH)2. Determine the concentrations of all ions in solution at equilibrium.

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

0M

+x

x

0.50M

+2x

0.50+2x

Initial Change Equilibrium

Ksp = [Mg2+][OH- ]2

Ksp = 1.2x10-11

1.2x10-11 = (x)(0.50)2 = 0.25x

1111

4.8x100.25

1.2x10x

[Mg2+] = x = 4.8x10-11 M

[OH-] = 0.50+2x = 0.50 +(2)4.8x10-11 ≈ 0.50M OH-

Na+

OH- Na+

Add Mg(OH)2