More Acid and Base Chemistry - Glendale Community College

10/15/2014

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:

6.02x105 x100 0.0489% 0.123

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?

HClO(aq)+ H2O(l) H3O+(aq) + ClO-(aq)

Ka

[H3O ][Cl O ] [HCl O]

2.95x108 (x)(0.100 x) 0.123 - x

Assume these "x's" will be very small

I nitial

0.123M

C hange Equilibirum

-x 0.123M - x

2.95x108 (x)(0.100) 0.123

3.63x109 0.100x

x 3.63x108

3.63x108 x100 0.0000295% 0.123

0M

0.100M

+x

+x

Hypoclorite ion already in solution before acid added

x

0.100+x

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.

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Common-ion effect

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

6.02x105 x100 0.0489% 0.123

Ka

[H3O ][Cl O [HCl O]

]

HClO(aq)+ H2O(l) H3O+(aq) + ClO-(aq)

I nitial

0.123M

0.100M

0M

C hange

-x

+x

+x

Equilibirum 0.123M - x

0.100+x

x

Hydromiun ion already in solution before acid added

2.95x108 (0.100 x)(x) 0.123 - x

Assume these "x's" will be very small

2.95x108 (0.100)(x) 0.123

3.63x109 0.100x

x 3.63x108

3.63x108 x100 0.0000295% 0.123

Both products of dissociation of HClO, ClOand 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.

Plain ole water

pH = -7

pH = 7

Add HCl

H+

Cl-

pH = 0

HCl adds H3O+ to the solution with its reaction with water.

HCl + H2O H3O+ + Cl-

Through the reaction, 1 mol of H3O+ is added to the water.

NH3 /NH4+ buffer solution

NH4+ NH3

NH3 NH4+

NNHH4+3NNHH4+3

pH = 9.25

Add HCl

NH4+ NH4+ Cl-

NH3 NH4+

NNHH4+3NNHH4+3

pH = 9.03

pH = -0.22

HCl reacts with NH3 to give more NH4+ ions. H3O+ can only be added to the

solution through the NH3/NH4+ equilibrium

HCl + NH3 NH4+ + Cl-

Through the reaction, only 3.7x10-10 mol of H3O+ is added to the water.

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Calculating pH of buffer solutions

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.

Ka

[H3O ][NH3] [NH4 ]

5.8x1010 (x)(3.0 x) 5.0 - x

5.8x1010 (x)(3.0) 5.0

x 9.7x1010

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

NH4+(aq)+ H2O(l) H3O+(aq) + NH3(aq)

Initial

5.0M

0M

Change

-x

+x

Equilibrium 5.0M - x

x

3.0M +x 3.0M+x

Because the [NH3] +

[NH4+] = 8.0M, this is

an 8.0 M buffer

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

Henderson ? Hasslebalch Formula

pH

p

Ka

l

o

g[conjuga teba s e ] [conjuga tea ci d]

?Change in acid/base concentrations by dissociation negligible.

[acid/base]initial [acid/base]equilibrium

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

pH 9.24 log 3.0M 5.0M

pH=9.02

Adding Strong Acid/Base to a Buffer Solution

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."

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).

Note that the HCl (acid) will react with the NH3 (base) in the solution. The strong acid or base will always react completely. Remember, 100% dissociation.

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

H+ (aq) + NH3(aq) NH4+(aq) Before: 1.0mole 3.0moles 5.0moles Change: -1.0mole -1.0mole +1.0mole After: 0mole 2.0moles 6.0moles

pH

p

Ka

l

o

g[conjuga teba s e ] [conjuga tea ci d]

pH 9.24 log 2.0mol 6.0mol

pH=8.76

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,

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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.

pH

p

Ka

l

o

g[conjuga teba s e ] [conjuga tea ci d]

pH 4.19 log 0.100M 0.100M

pH 4.19

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

OH- (aq) + HA(aq) A- (aq)

Before: 0.00500mol

Change: -0.00500mol

After:

0mol

0.0100mol 0.0100mol -0.00500mol +0.00500mol 0.00500mol 0.0150mol

A-

A-

A- HA

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

Thus, H-H can be used

Buffer Solution

pH 4.19 log 0.0150mol pH 4.67 0.00500mol

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A-

A-

A- HA

A- A-

A-

A-

A- OH- AHA H+ HA

Cl- Na+

Na+

Cl-

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. 0.0200L NaOH 0.500mol NaOH 0.0100mol NaOH

1 L NaOH

0.1000L HA 0.100mol HA 0.0100mol HA 1 L HA

OH- (aq) + HA(aq) A- (aq)

Before: 0.0100mol

Change: -0.0100mol

After:

0mol

0.0100mol -0.0100mol

0mol

0.0100mol +0.0100mol 0.0200mol

A- A-

A-

A-

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

0.0200mol A- 0.167M A0.02000L 0.1000L

F- (aq)+ H2O(l) OH-(aq) + HF(aq)

Initial

0.167M

0M

0M

Change

-x

+x

+x

Equilibrium 00.167M - x

x

x

Weak Base Solution

x2 1.6x1010 0.167

x 5.2x106

[OH- ] x 5.2x106 pH 14 pOH 14 ( log 5.2x10-6)

pH 8.72

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