Ionic Equilbria in aqueous Solutions

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Ionic Equilbria in aqueous Solutions

• Chapter 19

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The Common-Ion Effect

• Common Ion Two dissolved solutes that contain
  the same ion (cation or anion).
• The presence of a common ion suppresses the
  ionization of a weak acid or a weak base.
• Common-Ion Effect is the shift in equilibrium
  caused by the addition of a compound having an
  ion in common with the dissolved substance.

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• In 0.15M NH3, the pH is 11.21 and the percent
  dissociation is 1.1. Calculate the
  concentrations of all species present in
  solution, the pH, and
• the percent dissociation of NH3 in a solution
  that is 0.15 M in NH3 and 0.45 M in NH4Cl.

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The Common-Ion Effect

• To determine the pH, we apply the
  HendersonHasselbalch equation.
• When the concentration of HA and salt are high
  (0.1 M) we can neglect the ionization of acid
  and hydrolysis of salt.

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The Common-Ion Effect

• Calculate the pH of a 0.20 M CH3COOH solution
  with no salt added.
• Calculate the pH of a solution containing 0.20 M
  CH3COOH and 0.30 M CH3COONa.
• What is the pH of a solution containing 0.30 M
  HCOOH, before and after adding 0.52 M HCOOK?

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Buffer Solutions

• A Buffer Solution is a solution of (1) a weak
  acid or a weak base and (2) its salt both
  components must be present.
• A buffer solution has the ability to resist
  changes in pH upon the addition of small amounts
  of either acid or base.
• Buffers are very important to biological systems.

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Buffer Solutions
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Buffer Solutions

• Buffer solutions must contain relatively high
  acid and base component concentrations, the
  buffer capacity.
• Acid and base component concentrations must not
  react together.
• The simplest buffer is prepared from equal
  concentrations of acid and conjugate base.

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• Calculate the pH of a buffer system containing
  1.0 M CH3COOH and 1.0 M CH3COONa. What is the pH
  of the system after the addition of 0.10 mole of
  gaseous HCl to 1.0 L of solution?
• Calculate the pH of 0.30 M NH3/0.36 NH4Cl buffer
  system. What is the pH after the addition of 20.0
  mL of 0.050 M NaOH to 80.0 mL of the buffer
  solution?

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Buffer Solutions

• Buffer Preparation Use the HendersonHasselbalch
  equation in reverse.
• Choose weak acid with pKa close to required pH.
• Substitute into HendersonHasselbalch equation.
• Solve for the ratio of conjugate base/acid.
• This will give the mole ratio of conjugate base
  to acid. The acid should always be 1.0.

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Buffer Solutions

• Describe how you would prepare a phosphate
  buffer with a pH of about 7.40.

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Chemistry is FUN

• Maybe not a day or two before the exam
• Sometimes, not even after the graded exams are
  returned
• ACS Humor in Chemistry
• 1) How many atoms in a spoon of guacamole?
• Avocado's Number, Of course !!
• 2) What is Ba(Na)2 ?
• 3) What weapon can you make from potassium,
  nickel and iron?

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AcidBase Titrations

• Titration a procedure for determining the
  concentration of a solution using another
  solution of known concentration.
• Titrations involving only strong acids or bases
  are straightforward.
• Titrations involving weak acids or bases are
  complicated by hydrolysis of salt formed.

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AcidBase Titrations

• Strong AcidStrong Base
• The equivalence point is the point at which
  equimolar amounts of acid and base have
  reacted.

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• Weak AcidStrong Base
• The conjugate base hydrolyzes to form weak acid
  and OH.
• At equivalence point only the conjugate base is
  present.
• pH at equivalence point will always be gt7.

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AcidBase Titrations
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AcidBase Titrations

• The pH of a 25 mL 0.10 M CH3COOH sample can be
  determined after the addition of
• 0. No addition of 0.10 M NaOH.
• 1. 10.0 mL (total) of 0.10 M NaOH.
• 2. 25.0 mL (total) of 0.10 M NaOH.
• 3. 35.0 mL (total) of 0.10 M NaOH.

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AcidBase Titrations

• Exactly 100 mL of 0.10 M nitrous acid are
  titrated with a 0.10 M NaOH solution. Calculate
  the pH the pH for
• 1. The initial solution.
• 2. The point at which 80 mL of base have been
  added.
• 3. The equivalence point.
• 4. The point at which 105 mL of base have been
  added.

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AcidBase Titrations

• Strong AcidWeak Base
• The (conjugate) acid hydrolyzes to form weak
  base and H3O.
• At equivalence point only the (conjugate) acid
  is present.
• pH at equivalence point will always be lt7.

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AcidBase Titrations

• Calculate the pH when 40 mL of 0.10 M NH3 is
  titrated with a 0.10 M HCl solution.
• Before addition of HCl
• Before equivalence point
• at the equivalence point
• After the equivalence point
• Calculate the pH at the equivalence point in the
  titration of 50 mL of 0.10 M methylamine with a
  0.20 M HCl solution.

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AcidBase Titrations

• Polyprotic Acids

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Solubility Equilibria

• Solubility Product is the product of the molar
  concentrations of constituent ions and provides a
  measure of a compounds solubility.
• MX2(s) ? M2(aq) 2 X(aq)
• Ksp M2X2

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Solubility Equilibria

• Al(OH)3 1.8 x 1033
• BaCO3 8.1 x 109
• BaF2 1.7 x 106
• BaSO4 1.1 x 1010
• Bi2S3 1.6 x 1072
• CdS 8.0 x 1028
• CaCO3 8.7 x 109
• CaF2 4.0 x 1011
• Ca(OH)2 8.0 x 106
• Ca3(PO4)2 1.2 x 1026
• Cr(OH)3 3.0 x 1029
• CoS 4.0 x 1021
• CuBr 4.2 x 108

MnS 3.0 x 1014 Hg2Cl2 3.5 x 1018 HgS 4.0 x
1054 NiS 1.4 x 1024 AgBr 7.7 x 1013 Ag2CO3
8.1 x 1012 AgCl 1.6 x 1010 Ag2SO4 1.4 x
105 Ag2S 6.0 x 1051 SrCO3 1.6 x 109 SrSO4
3.8 x 107 SnS 1.0 x 1026 Zn(OH)2 1.8 x
1014 ZnS 3.0 x 1023
CuI 5.1 x 1012 Cu(OH)2 2.2 x 1020 CuS 6.0 x
1037 Fe(OH)2 1.6 x 1014 Fe(OH)3 1.1 x
1036 FeS 6.0 x 1019 PbCO3 3.3 x 1014 PbCl2 2.4
x 104 PbCrO4 2.0 x 1014 PbF2 4.1 x
108 PbI2 1.4 x 108 PbS 3.4 x 1028 MgCO3 4.0 x
105 Mg(OH)2 1.2 x 1011
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Solubility Equilibria

• The solubility of calcium sulfate (CaSO4) is
  found experimentally to be 0.67 g/L. Calculate
  the value of Ksp for calcium sulfate.
• The solubility of lead chromate (PbCrO4) is
  4.5 x 105 g/L. Calculate the solubility
  product of this compound.
• Calculate the solubility of copper(II) hydroxide,
  Cu(OH)2, in g/L.

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The Common-Ion Effect and Solubility
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The Common-Ion Effect and Solubility

• The solubility product (Ksp) is an equilibrium
  constant precipitation will occur when the ion
  product exceeds the Ksp for a compound.
• If AgNO3 is added to saturated AgCl, the increase
  in Ag will cause AgCl to precipitate.
• Q Ag0 Cl0 gt Ksp

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The Common-Ion Effect and Solubility
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The Common-Ion Effect and Solubility
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The Common-Ion Effect and Solubility

• Calculate the solubility of silver chloride (in
  g/L) in a 6.5 x 103 M silver chloride solution.
• Calculate the solubility of AgBr (in g/L) in(a)
  pure water(b) 0.0010 M NaBr

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Complex Ion Equilibria and Solubility

• A complex ion is an ion containing a central
  metal cation bonded to one or more molecules or
  ions.
• The formation constant (Kf) is the equilibrium
  constant for the complex ion formation.

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Complex Ion Equilibria and Solubility
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Complex Ion Equilibria and Solubility
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Complex Ion Equilibria and Solubility

• ION Kf
• Ag(NH3)2 1.5 x 107
• Ag(CN)2 1.0 x 1021
• Cu(CN)42 1.0 x 1025
• Cu(NH3)42 5.0 x 1013
• Cd(CN)42 7.1 x 1016
• CdI42 2.0 x 106

ION Kf HgCl42 1.7 x 1016 HgI42 3.0 x
1030 Hg(CN)42 2.5 x 1041 Co(NH3)63 5.0 x
1031 Zn(NH3)42 2.9 x 109
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Complex Ion Equilibria and Solubility

• A 0.20 mole quantity of CuSO4 is added to a liter
  of 1.20 M NH3 solution. What is the concentration
  of Cu2 ions at equilibrium?
• If 2.50 g of CuSO4 are dissolved in 9.0 x 102 mL
  of 0.30 M NH3, what are the concentrations of
  Cu2, Cu(NH3)42, and NH3 at equilibrium?

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Complex Ion Equilibria and Solubility

• Calculate the molar solubility of AgCl in a 1.0 M
  NH3 solution.
• Calculate the molar solubility of AgBr in a 1.0 M
  NH3 solution.

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Solubility Equilibria

• Ion Product (Q) Solubility equivalent of the
  reaction quotient. It is used to determine
  whether a precipitate will form
• Q lt Ksp UnsaturatedQ Ksp SaturatedQ gt
  Ksp Supersaturated precipitate forms.

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Solubility Equilibria

• Exactly 200 mL of 0.0040 M BaCl2 are added to
  exactly 600 mL of 0.0080 M K2SO4. Will a
  precipitate form?
• If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
  0.100 M CaCl2, will precipitation occur?

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