Upload
devika
View
39
Download
0
Embed Size (px)
DESCRIPTION
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 - PowerPoint PPT Presentation
Citation preview
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!
HCOOHHCOOHlogKlog 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 ionic
H+ (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 ionic
NH3 (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 colouredAlso 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 noteAgCl
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 AgNHAgK
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