Chapter 18 — Advanced Aqueous Equilibria

1

More About Chemical Equilibria: Acid– Base & Precipitation Reactions Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

Jeffrey Mack California State University, Sacramento

Stomach Acidity & Acid–Base Reactions

The Common Ion Effect • 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

The Common Ion Effect

What is the effect on the pH of a 0.25M NH 3(aq) solution when NH4Cl is added?

First let’s find the pH of a 0.25M NH3(aq) Solution:

NH3 (aq)  H2O

NH4  (aq)  OH (aq)

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!

Initial Change Equilibrium

[NH3]

[NH4+]

[OH-]

0.25 x 0.025  x

0 +x x

0 +x x

Chapter 18 — Advanced Aqueous Equilibria The Common Ion Effect

The Common Ion Effect

First let’s find the pH of a 0.25M NH3(aq) Solution: Initial Change Equilibrium

[NH3]

[NH4+]

[OH-]

0.25 x 0.025  x

0 +x x

0 +x x

K b  1.8 x 105 



2

First let’s find the pH of a 0.25M NH 3(aq) Solution: [NH4  ][OH- ] x2  [NH3 ] 0.25 - x

K b  1.8 x 105 

Assuming x is 1, indicating that the reaction is product 1 favored. K   1.8  109 rev

3

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+

Controlling pH: Buffer Solutions 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. K rev 

Kb

• An hydroxide added will immediately react with the acid so long as it is present.

1  5.6  10 4 Ka

• An hydronium ion added will immediately react wit the acid so long as it is present.

Buffer Solutions

Buffer Solutions

Problem:

Problem:

What is the pH of a buffer that has [CH 3CO2H] = 0.700 M and [CH3CO2] = 0.600 M?

What is the pH of a buffer that has [CH 3CO2H] = 0.700 M and [CH3CO2] = 0.600 M? Since the concentration of acid is greater than the base, equilibrium will move the reaction to the right.

CH3CO2H(aq)  H2O

CH3CO2 (aq)  H3O (aq)

Chapter 18 — Advanced Aqueous Equilibria

4

Buffer Solutions

Buffer Solutions

Problem:

Problem:

What is the pH of a buffer that has [CH 3CO2H] = 0.700 M and [CH3CO2] = 0.600 M?

What is the pH of a buffer that has [CH 3CO2H] = 0.700 M and [CH3CO2] = 0.600 M?

Initial Change Equilibrium

[CH3CO2H]

[CH3CO2]

[H3O+]

0.700 x 0.700  x

0.600 +x 0.600 + x

0 +x x

K a  1.8  10 5 

[H3O ]  0.600 0.700

[H3O+] = 2.1 105

Assuming that x 7 This will yield the [H3O+] and pH.

Volume of OH- added to the eq. point:

Titration of a Weak Acid with a Strong Base

HBz (aq) + OH(aq)  Bz(aq) + H2O(l) 100.0mL 

1L 0.025mol HBz 1mol OH 1L 103 mL      25mL 103 mL 1L 1molHbz 0.100molOH 1L

The new total volume of the solution is 125 mL

Titration of a Weak Acid with a Strong Base

Problem:

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)

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:

100.0 mL 

1L 0.025 mol HBz 1 mol OH 1 mol Bz      0.0025 mols Bz  103 mL 1L 1 mol Hbz 1 mol OH

The concentration of Bz- at the eq. point is: 0.0025 mols Bz  103 mL   0.020 M Bz  125 mL 1L

Titration of a Weak Acid with a Strong Base

Bz  (aq)  H2O(l)

Initial Change Equilibrium

HBz (aq)  OH (aq) K b  1.6  10 10

[Bz-]

[HBz]

[OH-]

0.020 -x 0.020 - x

0 +x x

0 +x x

Titration of a Weak Acid with a Strong Base

Problem:

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:

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:

Initial Change Equilibrium

[Bz-]

[HBz]

[OH-]

0.020 -x 0.020 - x

0 +x x

0 +x x

K b  1.6 x 1010 

x2 0.020  x

K b  1.6  10 10 

x2 0.020  x

Assuming that x 7. This is due to the production of the conjugate base of a week acid.

11 Titration of a Weak Acid with a Strong Base Conclusion: What would the pH equal at the half-way point of the titration?

Equivalence point pH = 8.25

Hint: Only ½ of the moles of weak acid have been converted to its conjugate base!

Half-way point pH = ??

Titration of a Weak Acid with a Strong Base

Titration of a Weak Acid with a Strong Base

Problem:

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!

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!  conj. Base  pH  pK a  log     Acid  

Titration of a Weak 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? At the half-way point, moles of Hbz and Bz- are equal. This is a BUFFER SOLUTION!  conj. Base  pH  pK a  log    pK a  log(1)  pK a  0  pK a   Acid  





pH   log 6.3  10 5  4.20

Acetic Acid Titrated with NaOH

Chapter 18 — Advanced Aqueous Equilibria Titration of a Weak Polyprotic Acid with a Strong Base 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 Base with a Strong Acid

12

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. K a(1)   HA  (aq)  H2O(l) H2 A(aq)  OH (aq)   The pH is relative to the amounts of conjugate acids and bases. K a( 2 )   A 2  (aq)  H2O(l) HA  (aq)  OH (aq)   At the second eq. point all of the acid has been converted to A2-, pH is determined by Kb. K Kb  w K a(2)

pH Indicators for Acid–Base Titrations

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.

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.

pH Indicators for Acid–Base Titrations

Chapter 18 — Advanced Aqueous Equilibria Natural Indicators: Red Rose Extract in Methanol Neutral pH

pH7

Solubility of Salts

13 Solubility of Salts • 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.

Lead(II) iodide

Solubility of Salts

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 Consider the solubility of a salt MX: If MX is added to water then: H2O(l) MX(s)   M (aq)  X (aq)

K sp  [M ][X ] Generally speaking, If Ksp >> 1 then MX is considered to be soluble If Ksp