Chemistry 100 - Chapter 17

Applications of Aqueous Equilibria

The Common Ion Effect

Add sodium formate (HCOONa) to a solution of formic acid (HCOOH) that has already established equilibrium?

HCOOH (aq) ⇄ H+ (aq) + HCOO- (aq)weak acid

HCOONa (aq) HCOO- (aq) + Na+ (aq) strong electrolyte

Another Example of the Common Ion Effect

What would happen if we added HCOOH to a solution of a strong acid?

HCl (aq) H+ (aq) + Cl- (aq) strong acid

HCOOH (aq) ⇄ H+ (aq) + HCOO- (aq) The ionization of the weak acid

would be supressed in the presence of the strong acid!

By Le Chatelier’s Principle, the value of the weak acid is decreased!

HCOOHHCOOH

logKlog a

How would we calculate the pH of these solutions?

HCOOH]HCOO][H[

Ka

]HCOOHlog[]HCOOlog[]Hlog[pKa

note pH = -log [H+]

Define pKa = -log (Ka )

The Buffer Equation

]HCOOH[]HCOO[

logpHpKa

• Substituting and rearranging

]HCOOH[]HCOO[

logpKpH a

The Generalized Buffer Equation

The solution pH is determined by the ratio of the weak acid to the conjugate

base at equilibrium. Henderson-Hasselbalch equation

]acid weak[]base .conj[

logpKpH a

The Definition of a Buffer

Buffer a reasonably concentrated solution of a weak acid and its conjugate base.

Buffer solutions resists pH changes when additional strong acid or strong base are added to the solutions.

Note: The Henderson-Hasselbalch equation is really only valid for pH ranges near the pKa of the weak acid!

How Do We Use the Buffer (the H-H) Equation?

The pH of the buffer is determined by the concentration ratio of weak acid to conjugate base at equilibrium.

How different are the equilibrium concentrations of weak acid/conjugate base from the initial concentrations?

Buffer CH3COONa (aq) and CH3COOH (aq))

CH3COOH (aq) ⇄ CH3COO- (aq) + H+ (aq)

The Equilibrium Data Table

[CH3COOH]

[H+]

[CH3COO-]

Start A 0 B

Change -x + x +x

m (A-x) (x) (B+x)

According to the Henderson-Hasselbalch Equation, the pH of the solution is calculated as follows

]xA[]xB[

logpKxlogpH a

What if the original concentrations of acid and base ([A] and [B], respectively) are much larger than x (i.e., the value of the weak acid is very small)?

The pH of the solution will be almost entirely due to the original concentrations of acid and base!!

]A[]B[

logpKpH a

The pH of the solution changes very little after adding strong acid or base (it is

buffered)

Examples of Buffer Calculations

How do we calculate the pH of a buffer solution?

How would we prepare a buffer solution of a specified pH?

The pH of a Buffer Solution

Major task obtain the ratio of the concentrations

of conjugate base to weak acid! Using the Ka of the appropriate acid,

the pH of the solution is obtained from the Henderson-Hasselbalch equation.

Preparing a Buffer Solution of a Specific pH

The first step in the process is to choose a suitable weak acid.

The H-H equation is really only valid in a pH range near the pKa of the weak acid;

Second Step Calculate the required ratio of conjugate base to weak acid from the Henderson-Hasselbalch Equation

From the previous problem, there are a number of concentrations where the ratio of the conjugate base to the weak acid will be acceptable

[CH3CH2COONa] = 0.13 M; [CH3CH2COOH] = 0.10 M [CH3CH2COONa] = 0.065 M;[CH3CH2COOH] =

0.050 M [CH3CH2COONa] = 0.39 M; [CH3CH2COOH] = 0.30 M[CH3CH2COONa] = 0.65 M; [CH3CH2COOH] = 0.50 M

[CH3CH2COONa] = 1.3 M; [CH3CH2COOH] = 1.0 M

Buffer Capacity

Buffer Capacity refers to the amount of strong acid/base that can be added to the buffer solution

Buffer capacity is directly related to the concentrations of the weak acid and conjugate base in the buffer solution

Choosing Concentrations for a High Buffer Capacity

[CH3CH2COONa] = 0.13 M; [CH3CH2COOH] = 0.10 M

NOT ACCEPTABLE[CH3CH2COONa] = 0.065 M; [CH3CH2COOH] =

0.050 MNOT ACCEPTABLE

[CH3CH2COONa] = 0.39 M; [CH3CH2COOH] = 0.30 MREASONABLE

[CH3CH2COONa] = 0.65 M; [CH3CH2COOH] = 0.50 MGOOD CHOICE

[CH3CH2COONa] = 1.3 M; [CH3CH2COOH] = 1.0 MGOOD CHOICE

Adding Acid to Buffer Solutions

What happens when we add strong acid solutions to the buffer?

[H+] increased and the basic part of the buffer goes to work

H+ (aq) + CH3COO- (aq) CH3COOH (aq) This is the reverse of the usual acid

dissociation equilibrium, hence, the reaction essentially goes to completion

Adding Base to Buffer Solutions

What happens when we add the base to the buffer?

The [OH-] increases and the acid part of the buffer goes to work

OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

This is the reverse of the usual base dissociation equilibrium, hence, the reaction essentially goes to completion

Acid-base Titration Curves

Chapter 4 – acid-base titrations A known (standard) basic solution is

slowly added to an unknown acid solution

What if we monitored the pH of the solution as a function of added titrant acid base titration curve is generated

Three cases to consider

Strong Acid/strong Base

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

Net ionicH+ (aq) + OH- (aq) H2O (l)

When n(H+) = n(OH-), we are at the equivalence point of the titration

Product of reaction is a strong acid/strong base salt. The pH at the equivalence point is 7.00.

The Titration Curve

Weak Base /Strong Acid

NH3 (aq) + HCl (aq) NH4Cl (aq)

Net ionicNH3 (aq) + H+ (aq) NH4

+ (aq)

Equivalence point, n(NH3) = n(H+). pH of the solution < 7.00. Determined by the

ionization of the conjugate acid

NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)

The Titration Curve

Eq. point

pH

V (strong acid) / mL

11.00

9.00

7.00

5.00

3.00

Weak Acid/Strong base.

NaOH (aq) + CH3COOH (aq) CH3COONa (aq) + H2O (l)

Net ionicOH- (aq) + CH3COOH (aq) CH3COO- (aq) +

H2O (l)

The Titration Curve

Equivalence point when n(CH3COOH) = n(OH-),

At the equivalence point pH of the solution is determined by the

ionization of the conjugate base of the weak acid

CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)

Therefore, the pH at the equivalence point is > 7.00!

Comparison Between Strong/Weak Acid Titrations

Indicators

Indicators are used to detect the endpoint of the acid-base titration.

Indicators are weak acids. Their ionization can be represented by the following reaction.

HIn (aq) ⇌ H+ (aq) + In- (aq)

Usually coloured

Also usually coloured, but the colour is different than for the acid form of the indicator.

Choose the indicator whose transition range (i.e., the pH range where it changes colour) matches the steep part of the titration curve

Note: we can use the following ratios as a guide.

[HIn] / [In-] > 10 acid colour dominates. [In-] / [HIn] > 10 base colour dominates.

Strong Acid/Strong Base steep part of titration curve pH 4-10. A number of indicators change colour in this range

Weak Acid/Strong Base steep part of titration curve pH >7.0. The indicator colour change must occur in this range

Strong Acid/Weak Base steep part of titration curve pH <7.0. The indicator colour change must occur in this range

Indicators in Titrations

Solubility Equilibria

Examine the following systemsAgCl (s) ⇄ Ag+ (aq) + Cl- (aq)

BaF2 (s) ⇄ Ba2+ (aq) + 2 F- (aq) Using the principles of chemical

equilibrium, we write the equilibrium constant expressions as follows

6222eqsp 10x0.1F BaBaFKK

10

eqsp

eq

10x8.1Cl AgAgClKK

constant AgCl note

AgCl

Cl AgK

The Definition of the Ksp

Ksp the solubility product constant. The product of the molar

concentrations of the dissolved ions in equilibrium with the undissolved solid at a particular temperature.

Don’t confuse the solubility of the solid with the Ksp. These quantities

are related, but they are not the same.

Calculate the solubility of a solid in the presence of a common ion.

Examples of Ksp Calculations

Calculate the solubility of a sparingly soluble solid in water.

Calculate the solubility of a solid as a function of the pH of the solution.

Solubility of Sparingly Soluble Solids in Water

AgCl (s) ⇄ Ag+ (aq) + Cl- (aq) We approach this using the

principles of chemical equilibrium. equilibrium data table, establish and

solve for our unknown quantity!

The Common Ion Effect

What about the solubility of AgCl in solution containing NaCl (aq)?

AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)NaCl (aq) Na+ (aq) + Cl- (aq)

AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)

Equilibrium is displaced to the left by LeChatelier’s principle (an example of the common ion effect).

Solubility and pH

What happens when we try to dissolve a solid like Mn(OH)2 in solutions of varying pH?

We first calculate the pH of the saturated solution of the Mn(OH)2.

Suppose that we try to dissolve Mn(OH)2 in an acidic solution

Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)

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

Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)

Equilibrium is displaced to the right by LeChatelier’s principle.

Solubility of CaF2 vs. pH

Increase the [OH-] in the solution looking at an example of the common ion effect

Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)NaOH (aq) Na+ (aq) + OH- (aq)

Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)

Equilibrium is displaced to the left by LeChatelier’s principle.

Any solid that produces a moderately basic ion on dissociation (e.g., CaF2, MgCO3). solubility will decrease as the pH is

increased (i.e., the [OH-] in the solution is increased).

solubility will increase as the pH is decreased (i.e., the [H+] in the solution is increased).

What about solids whose anions do not exhibit basic tendencies? PbCl2, the Cl- has no tendency to react

with added acid. solubility of PbCl2, does not depend on the

solution pH!

Predicting Precipitation: the Qsp Value

Let’s examine the following equilibrium system.

AgCl (s) ⇄Ag+ (aq) + Cl- (aq) Let’s say that we two solutions so that

they would have the following concentrations.

[NaCl]o = 1.0x 10-5 M [Cl-]o = 1.0 x 10-5 M

[AgNO3] = 1.0 x 10-6 M [Ag+]o = 1.0 x 10-6 M Would we be able to predict whether or

not a precipitate will occur?

The QspValue

• Define the solubility product quotient Qsp.

We now examine the magnitude of the solubility product quotient (Qsp) with

respect to the Ksp. Qsp < Ksp no precipitate will formQsp > Ksp a precipitate will form

Qsp = Ksp saturated solution.

1156oosp 10x0.110x0.1 10x0.1]Cl[]Ag[Q

The Formation of Metal Complexes

We see that a number of metal anions can act as Lewis acids; therefore, these ions can react strongly with Lewis bases and form complex ions.

AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)Ag+ (aq) + 2 NH3 (aq) ⇄ Ag(NH3)2

+ (aq) For the NH3 (aq) to increase the solubility

of the metal salt, the NH3 (aq) must be a stronger Lewis base than the water molecules that it displaces.

Solubility of many metal containing compounds increases markedly in the presence of suitable complexing species NH3 (aq) OH- (aq) CN- (aq)

The species Ag(NH3)2+ (aq) is known

as a complex ion. The equilibrium constant for the second reaction, Kf

72

3

23f 10x7.1

NH Ag

NHAgK

is known as the formation constant for the complex ion.

The Magnitudes of Kf values

The complexation reaction effectively removes all the Ag+ (aq) from the solution.

For the original equilibrium systemAgCl (s) ⇄ Ag+ (aq) + Cl- (aq)

the equilibrium position is strongly displaced to the right by LeChatelier’s principle. The solubility of AgCl is increased significantly in the presence of

the complexing agent!

• Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS (Ksp = 610-37) is less soluble than ZnS (Ksp = 210-25), CuS will be removed from solution before ZnS.

• As H2S is added to the green solution, black CuS forms in a colorless solution of Zn2+(aq).

• When more H2S is added, a second precipitate of white ZnS forms.

Precipitation and Separation of Ions

• Ions can be separated from each other based on their salt solubilities.

• Example: if HCl is added to a solution containing Ag+ and Cu2+, the silver precipitates (Ksp for AgCl is 1.8 10-10) while the Cu2+ remains in solution.

• Removal of one metal ion from a solution is called selective precipitation.

Precipitation and Separation of Ions

• Qualitative analysis is designed to detect the presence of metal ions.

• Quantitative analysis is designed to determine how much metal ion is present.

• We can separate a complicated mixture of ions into five groups:– Add 6 M HCl to precipitate insoluble chlorides

(AgCl, Hg2Cl2, and PbCl2).

– To the remaining mix of cations, add H2S in 0.2 M HCl to remove acid insoluble sulfides (e.g. CuS, Bi2S3, CdS, PbS, HgS, etc.).

– To the remaining mix, add (NH4)2S at pH 8 to remove base insoluble sulfides and hydroxides (e.g. Al(OH)3, Fe(OH)3, ZnS, NiS, CoS, etc.).

Qualitative Analysis for Metallic Elements

– To the remaining mixture add (NH4)2HPO4 to remove insoluble phosphates (Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4).

– The final mixture contains alkali metal ions and NH4

+.

Qualitative Analysis for Metallic Elements