Chemistry 100 - Chapter 17

Applications of Aqueous Equilbria

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 + strong electrolyte (aq)

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  the weak acid is decreased! value of

 How would we calculate the pH of these solutions?

K a  [ H  ][ HCOO   HCOOH  ]  log K a   log     HCOO   HCOOH    

Define pK a = -log (K a ) pK a   log[ H  ]  log[ HCOO  ]  log[ HCOOH ] note pH = -log [H + ]

The Buffer Equation

• Substituting and rearranging pK a  pH  log    [ [ HCOO  ] HCOOH ]    pH  pK a  log    [ [ HCOO  ] HCOOH ]   

The Generalized Buffer Equation

pH  pK a  log [ conj .

base ] [ weak acid ] 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

   Buffer solution of a weak acid and its conjugate base.

 a reasonably concentrated 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  (aq)) CH 3 COONa (aq) and CH CH 3 COOH (aq) ⇄ CH 3 3 COOH COO- (aq) + H + (aq) The Equilibrium Data Table Start Change  m [CH 3 COOH] A -x (A-x) [H + ] 0 + x (x) [CH 3 COO B +x (B+x) ]

 According to the Henderson-Hasselbalch Equation, the pH of the solution is calculated as follows  pH   log    pK a  log [ B [ A   x ] x ] 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!!

pH  pK a  log [ B ] [ 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 K a 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 pK a 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 [CH 3 CH 2 COONa] = 0.13 M; [CH 3 CH 2 COOH] = 0.10 M [CH 3 CH 2 COONa] = 0.065 M;[CH 3 CH 2 COOH] = 0.050 M [CH 3 CH 2 COONa] = 0.39 M; [CH 3 CH 2 COOH] = 0.30 M [CH 3 CH 2 COONa] = 0.65 M; [CH 3 CH 2 COOH] = 0.50 M [CH 3 CH 2 COONa] = 1.3 M; [CH 3 CH 2 COOH] = 1.0 M

Buffer Capacity

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

[CH 3 CH 2 COONa] = 0.13 M; [CH 3 CH 2 COOH] = 0.10 M [CH 3 CH 2 NOT ACCEPTABLE COONa] = 0.065 M; 0.050 M [CH 3 CH 2 COOH] = NOT ACCEPTABLE [CH 3 CH 2 COONa] = 0.39 M; [CH 3 CH 2 COOH] = 0.30 M REASONABLE [CH 3 CH 2 COONa] = 0.65 M; [CH 3 CH 2 COOH] = 0.50 M GOOD CHOICE [CH 3 CH 2 COONa] = 1.3 M; [CH 3 CH 2 COOH] = 1.0 M GOOD 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) + CH 3 COO (aq)  CH 3 COOH (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) + CH 3 COOH (aq)  H 2 O (l) CH 3 COO (aq) + 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) H +  NaCl (aq) + H 2 O (l) Net ionic (aq) + OH (aq)  H 2 O (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

NH 3 NH 3 (aq) + HCl (aq)  Net ionic (aq) + H + (aq)  NH 4 Cl (aq) NH 4 + (aq)   Equivalence point, n(NH 3 ) = n(H + ). pH of the solution < 7.00. Determined by the ionization of the conjugate acid NH 4 + (aq) ⇌ NH 3 (aq) + H + (aq)

The Titration Curve

pH 11.00

9.00

7.00

5.00

3.00

Eq. point V (strong acid) / mL

Weak Acid/Strong base.

NaOH (aq) + CH 3 COOH (aq)  (aq) + H 2 O (l) Net ionic OH (aq) + CH 3 COOH (aq)  + H 2 O (l) CH CH 3 3 COONa COO (aq)

The Titration Curve

   Equivalence point when n(CH 3 COOH) = n(OH ), At the equivalence point ◦ pH of the solution is determined by the ionization of the conjugate base of the weak acid CH 3 COO (aq) + H 2 O (l) ⇌ CH 3 COOH (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  [HIn] > 10  acid colour dominates. [In base colour dominates.

] /

   Strong Acid/Strong Base  titration curve pH 4-10. A number of indicators change colour in this range steep part of 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 systems AgCl (s) ⇄ Ag + (aq) + Cl BaF 2 (s) ⇄ Ba 2+ (aq) + 2 F (aq) (aq) Using the principles of chemical equilibrium, we write the equilibrium constant expressions as follows

K eq note  K  sp 

  

 AgCl  AgCl K  eq   Cl    constant AgCl   Ag  1 .

8 x 10  10 K sp  K eq  BaF 2  

  

2  1 .

0 x 10  6

The Definition of the K

sp   K sp  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 K sp . These quantities are related, but they are not the same.

Examples of K

sp

Calculations

 Calculate the solubility of a sparingly soluble solid in water.

 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.

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) ⇄ Mn 2+ (aq) + 2 OH (aq) H + Mn(OH) (aq) + OH 2 (s) ⇄ Mn (aq) 2+  H 2 O (l) (aq) + 2 OH (aq) Equilibrium is displaced to the right by LeChatelier’s principle.

Solubility of CaF

2

vs. pH

 Increase the [OH ] in the solution ◦ looking at an example of the common ion effect Mn(OH) 2 (s) ⇄ NaOH (aq)  Mn 2+ Na + (aq) + 2 OH (aq) + OH (aq) (aq) Mn(OH) 2 (s) ⇄ Mn 2+ (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., CaF 2 , MgCO 3 ).

◦ 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? ◦ PbCl 2 , the Cl has no tendency to react with added acid. ◦ solubility of PbCl 2, does not depend on the solution pH!

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 NH For the NH displaces. 3 3 (aq) the metal salt, the NH 3 ⇄ Ag(NH 3 ) 2 + (aq) (aq) to increase the solubility of (aq) must be a stronger Lewis base than the water molecules that it

 Solubility of many metal containing compounds increases markedly in the presence of suitable complexing species ◦ ◦ NH 3 OH (aq) (aq) ◦ CN (aq)

 The species Ag(NH 3 ) 2 complex ion. The equilibrium constant for the second reaction, K + f (aq) is known as a K f   Ag  NH 3    NH 3  2  2   1 .

7 x 10 7 is known as the formation constant for the complex ion .

The Magnitudes of K

f

values

  The complexation reaction effectively removes all the Ag+ (aq) from the solution. For the original equilibrium system AgCl (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!

Predicting Precipitation: the Q

sp

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  [AgNO 3 ] = 1.0 x 10 -6 M  [Cl ] o [Ag + ] o = 1.0 x 10 -5 = 1.0 x 10 -6 M M Would we be able to predict whether or not a precipitate will occur?

The Q

sp

Value

• Q sp Define the solubility product quotient   [ Ag  ] o [ Cl  ] o  Q sp .

 1 .

0 x 10  6  1 .

0 x 10  5   1 .

0 x 10  11 We now examine the magnitude of the solubility product quotient (Q sp ) with respect to the K sp . Q sp Q sp Q < K sp > K sp sp = K sp no precipitate will form a precipitate will form saturated solution.

Precipitation and Separation of Ions

• Consider a mixture of Zn 2+ (

aq

) and Cu 2+ (

aq

).

( CuS (

K sp K sp

= 6  10 -37 ) is less soluble than ZnS = 2  10 -25 ), CuS will be removed from solution before ZnS.

• As H 2 S is added to the green solution, black CuS forms in a colorless solution of Zn 2+ (

aq

).

• When more H 2 S 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 precipitates ( +

K

and Cu 2+ , the silver

sp

for AgCl is 1.8  10 -10 ) while the Cu 2+ remains in solution.

• Removal of one metal ion from a solution is called

selective precipitation

.

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

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

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, Hg 2 Cl 2 , and PbCl 2 ).

– To the remaining mix of cations, add H 2 S in 0.2

M

HCl to remove acid insoluble sulfides (e.g. CuS, Bi 2 S 3 , CdS, PbS, HgS, etc.).

– To the remaining mix, add (NH 4 ) 2 S 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 (NH 4 ) 2 HPO 4 remove insoluble phosphates (Ba 3 (PO 4 ) to 2 , Ca 3 (PO 4 ) 2 , MgNH 4 PO 4 ).

– The final mixture contains alkali metal ions and NH 4 + .