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Applications of Aqueous Equilibria. Buffered Solutions. A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt. Acid/Salt Buffering Pairs.

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Applications of Aqueous Equilibria

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Applications of aqueous equilibria

Applications of Aqueous Equilibria


Applications of aqueous equilibria

Buffered Solutions

  • A solution that resists a change in pH when either hydroxide ions or protons are added.

  • Buffered solutions contain either:

    • A weak acid and its salt

    • A weak base and its salt


Applications of aqueous equilibria

Acid/Salt Buffering Pairs

The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)


Applications of aqueous equilibria

Base/Salt Buffering Pairs

The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)


Applications of aqueous equilibria

Weak Acid/Strong Base Titration

A solution that is

0.10 M CH3COOH

is titrated with

0.10 M NaOH

Endpoint is above pH 7


Applications of aqueous equilibria

Strong Acid/Strong Base Titration

Endpoint is at

pH 7

A solution that is

0.10 M HCl is titrated with

0.10 M NaOH


Applications of aqueous equilibria

Strong Acid/Strong Base Titration

A solution that is

0.10 M NaOH is titrated with

0.10 M HCl

Endpoint is at

pH 7

It is important to recognize that titration curves are not always increasing from left to right.


Applications of aqueous equilibria

Strong Acid/Weak Base Titration

A solution that is

0.10 M HCl is titrated with

0.10 M NH3

Endpoint is below

pH 7


Applications of aqueous equilibria

Titration of an Unbuffered Solution

A solution that is

0.10 M HC2H3O2

is titrated with

0.10 M NaOH


Applications of aqueous equilibria

Titration of a Buffered Solution

A solution that is

0.10 M HC2H3O2 and

0.10 M NaC2H3O2 is titrated with 0.10 M NaOH


Applications of aqueous equilibria

Comparing Results

Unbuffered

Buffered

  • In what ways are the graphs different?

  • In what ways are the graphs similar?


Applications of aqueous equilibria

Comparing Results

Buffered

Unbuffered


Buffer capacity

Buffer capacity

  • The best buffers have a ratio [A-]/[HA] = 1

  • This is most resistant to change

  • True when [A-] = [HA]

  • Make pH = pKa (since log1=0)


General equation

General equation

  • Ka = [H+] [A-][HA]

  • so [H+] = Ka [HA] [A-]

  • The [H+] depends on the ratio [HA]/[A-]

  • taking the negative log of both sides

  • pH = -log(Ka [HA]/[A-])

  • pH = -log(Ka)-log([HA]/[A-])

  • pH = pKa + log([A-]/[HA])


Applications of aqueous equilibria

This is called the Henderson-Hasselbalch Equation


Using the henderson hasselbalch equation

Using the Henderson-Hasselbalch Equation

  • pH = pKa + log([A-]/[HA])

  • pH = pKa + log(base/acid)

  • Calculate the pH of the following mixtures

  • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

  • 0.25 M NH3 and 0.40 M NH4Cl

  • (Kb = 1.8 x 10-5)


Prove they re buffers

Prove they’re buffers

  • What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the following solutions.

    • 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

    • 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)

  • What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding.

To do this I must introduce the BAAM table!


Buffer capacity1

Buffer capacity

  • The pH of a buffered solution is determined by the ratio [A-]/[HA].

  • As long as this doesn’t change much the pH won’t change much.

  • The more concentrated these two are the more H+ and OH- the solution will be able to absorb.

  • Larger concentrations bigger buffer capacity.


Buffer capacity2

Buffer Capacity

  • Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following:

  • 5.00 M HAc and 5.00 M NaAc

  • 0.050 M HAc and 0.050 M NaAc

  • Ka= 1.8x10-5


Summary

Summary

  • Strong acid and base just stoichiometry.

  • Determine Ka, use for 0 mL base

  • Weak acid before equivalence point

    • Stoichiometry first

    • Then Henderson-Hasselbach

  • Weak acid at equivalence point Kb

  • Weak base after equivalence - leftover strong base.


  • Summary1

    Summary

    • Determine Ka, use for 0 mL acid.

    • Weak base before equivalence point.

      • Stoichiometry first

      • Then Henderson-Hasselbach

  • Weak base at equivalence point Ka.

  • Weak base after equivalence - leftover strong acid.


  • Applications of aqueous equilibria

    Selection of Indicators


    Solubility equilibria

    Solubility Equilibria

    Will it all dissolve, and if not, how much?


    Applications of aqueous equilibria

    • All dissolving is an equilibrium.

    • If there is not much solid it will all dissolve.

    • As more solid is added the solution will become saturated.

    • Soliddissolved

    • The solid will precipitate as fast as it dissolves .

    • Equilibrium


    Watch out

    Watch out

    • Solubility is not the same as solubility product.

    • Solubility product is an equilibrium constant.

    • it doesn’t change except with temperature.

    • Solubility is an equilibrium position for how much can dissolve.

    • A common ion can change this.


    Applications of aqueous equilibria

    Ksp Values for Some Salts at25C


    Applications of aqueous equilibria

    Solving Solubility Problems

    For the salt AgI at 25C, Ksp = 1.5 x 10-16

    AgI(s)  Ag+(aq) + I-(aq)

    O

    O

    +x

    +x

    x

    x

    1.5 x 10-16 = x2

    x = solubility of AgI in mol/L = 1.2 x 10-8 M


    Applications of aqueous equilibria

    Solving Solubility Problems

    For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5

    PbCl2(s)  Pb2+(aq) + 2Cl-(aq)

    O

    O

    +2x

    +x

    2x

    x

    1.6 x 10-5 = (x)(2x)2 = 4x3

    x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M


    Relative solubilities

    Relative solubilities

    • Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions.

    • The bigger the Ksp of those the more soluble.

    • If they fall apart into different number of pieces you have to do the math.


    The common ion effect

    The Common Ion Effect

    • When the salt with the anion of a weak acid is added to that acid,

    • It reverses the dissociation of the acid.

    • Lowers the percent dissociation of the acid.

    • The same principle applies to salts with the cation of a weak base..

    • The calculations are the same as last chapter.


    Applications of aqueous equilibria

    Solving Solubility with a Common Ion

    For the salt AgI at 25C, Ksp = 1.5 x 10-16

    What is its solubility in 0.05 M NaI?

    AgI(s)  Ag+(aq) + I-(aq)

    0.05

    O

    0.05+x

    +x

    0.05+x

    x

    1.5 x 10-16 = (x)(0.05+x)  (x)(0.05)

    x = solubility of AgI in mol/L = 3.0 x 10-15 M


    Applications of aqueous equilibria

    Precipitation and Qualitative Analysis


    Ph and solubility

    pH and solubility

    • OH- can be a common ion.

    • More soluble in acid.

    • For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.

    • CaC2O4 Ca+2 + C2O4-2

    • H+ + C2O4-2 HC2O4-

    • Reduces C2O4-2 in acidic solution.


    Precipitation

    Precipitation

    • Ion Product, Q =[M+]a[Nm-]b

    • If Q>Ksp a precipitate forms.

    • If Q<Ksp No precipitate.

    • If Q = Ksp equilibrium.

    • A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?


    Selective precipitations

    Selective Precipitations

    • Used to separate mixtures of metal ions in solutions.

    • Add anions that will only precipitate certain metals at a time.

    • Used to purify mixtures.

    • Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.


    Selective precipitation

    Selective Precipitation

    • In Basic adding OH-solution S-2 will increase so more soluble sulfides will precipitate.

    • Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3


    Selective precipitation1

    Selective precipitation

    • Follow the steps first with insoluble chlorides (Ag, Pb, Ba)

    • Then sulfides in Acid.

    • Then sulfides in base.

    • Then insoluble carbonate (Ca, Ba, Mg)

    • Alkali metals and NH4+ remain in solution.


    Applications of aqueous equilibria

    Complex Ions

    A Complex ion is a charged species composed of:

    1. A metallic cation

    2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation


    Applications of aqueous equilibria

    NH3, CN-, and H2O are Common Ligands


    The addition of each ligand has its own equilibrium

    The addition of each ligand has its own equilibrium

    • Usually the ligand is in large excess.

    • And the individual K’s will be large so we can treat them as if they go to equilibrium.

    • The complex ion will be the biggest ion in solution.


    Applications of aqueous equilibria

    Coordination Number

    • Coordination number refers to the number of ligands attached to the cation

    • 2, 4, and 6 are the most common coordination numbers


    Applications of aqueous equilibria

    Complex Ions and Solubility

    AgCl(s)  Ag+ + Cl- Ksp = 1.6 x 10-10

    Ag+ + NH3 Ag(NH3)+ K1 = 1.6 x 10-10

    Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 1.6 x 10-10

    K = KspK1K2

    AgCl + 2NH3 Ag(NH3)2+ + Cl-


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