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Chemistry 100 - Chapter 17 . Applications of Aqueous Equilibria. The Common Ion Effect. Add sodium format e (HCOONa) to a solution of formic acid (HCOOH) that has already established equilibrium? HCOOH (aq) ⇄ H + (aq) + HCOO - (aq) weak acid

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chemistry 100 chapter 17

Chemistry 100 - Chapter 17

Applications of Aqueous Equilibria

the common ion effect
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
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!

slide5

Define pKa = -log (Ka )

note pH = -log [H+]

the buffer equation
The Buffer Equation
  • Substituting and rearranging
the generalized buffer equation
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

the definition of a buffer
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
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?

slide10

Buffer  CH3COONa (aq) and CH3COOH (aq))

CH3COOH (aq) ⇄ CH3COO- (aq) + H+ (aq)

The Equilibrium Data Table

slide11

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)?

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

slide12

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

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

examples of buffer calculations
Examples of Buffer Calculations
  • How would we prepare a buffer solution of a specified pH?

How do we calculate the pH of a buffer solution?

the ph of a buffer solution
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
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

slide16

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

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
Choosing Concentrations for a High Buffer Capacity

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

NOT ACCEPTABLE

[CH3CH2COONa] = 0.065 M; [CH3CH2COOH] = 0.050 M

NOT ACCEPTABLE

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

REASONABLE

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

GOOD CHOICE

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

GOOD CHOICE

adding acid to buffer solutions
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
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
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
Strong Acid/strong Base
  • 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.

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

Net ionic

H+ (aq) + OH- (aq)  H2O (l)

weak base strong acid
Weak Base /Strong Acid
  • 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)

NH3 (aq) + HCl (aq)  NH4Cl (aq)

Net ionic

NH3 (aq) + H+ (aq)  NH4+ (aq)

the titration curve1
The Titration Curve

11.00

pH

9.00

7.00

Eq. point

5.00

3.00

V (strong acid) / mL

weak acid strong base
Weak Acid/Strong base.

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

Net ionic

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

slide28

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!
indicators
Indicators

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

Usually coloured

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)

slide31

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.

slide32

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

solubility equilibria
Solubility Equilibria

Examine the following systems

AgCl (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

the definition of the k sp
The Definition of the Ksp

Don’t confuse the solubility of the solid with the Ksp. These quantities are related, but they are not the same.

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.

examples of k sp calculations
Examples of Ksp Calculations
  • Calculate the solubility of a solid in the presence of a common ion.
  • Calculate the solubility of a solid as a function of the pH of the solution.

Calculate the solubility of a sparingly soluble solid in water.

solubility of sparingly soluble solids in water
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 effect1
The Common Ion Effect

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

solubility and ph
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.

slide41

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

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)

slide43

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

  • 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)

slide44

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).
slide45

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 q sp value
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 q sp value
The QspValue

We now examine the magnitude of the solubility product quotient (Qsp) with respect to the Ksp.

Qsp < Ksp no precipitate will form

Qsp > Ksp a precipitate will form

Qsp = Ksp saturated solution.

  • Define the solubility product quotient  Qsp.
the formation of metal complexes
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.

slide49

Solubility of many metal containing compounds increases markedly in the presence of suitable complexing species

    • NH3 (aq)
    • OH- (aq)
    • CN-(aq)
slide50

is known as the formation constant for the complex ion.

The species Ag(NH3)2+ (aq) is known as a complex ion. The equilibrium constant for the second reaction, Kf

slide52

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!

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

For the original equilibrium system

AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)

slide53

Precipitation and Separation of Ions

  • Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS (Ksp = 610-37)is less soluble than ZnS (Ksp = 210-25), 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.
slide54

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.
slide55

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

  • Quantitative analysis is designed to determine how much metal ion is present.
slide56

Qualitative Analysis for Metallic Elements

  • 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.).
slide57

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+.