Aqueous Equilibria

1
Aqueous Equilibria

• Chapter 16

2
pH

• The pH of aqueous solutions not only blood
  plasma, but seawater, detergents, sap and
  reaction mixtures is controlled, by the transfer
  of protons between ions and water molecules.

3
Ions as acids and bases

• Any cation that is the conjugate acid of a weak
  base functions as an acid and lowers the pH of
  the solution. Ammonium ion the conjugate acid of
  weak base ammonia, is an acid.
• NH4 (aq) H2O(l)? H3O(aq) NH3(aq)
• Small highly charged metal cations that can act
  as Lewis acids in water, such as Al3 and Ti3
  also produce acidic solutions, even though the
  cations dont have any protons to donate. The
  protons come from the water molecules that
  hydrate the ions in solution.

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• Salts that contain the conjugate acids of weak
  bases produce acidic aqueous solutions so do
  salts that contain small, highly charged metal
  cations. Salts that contain the conjugate bases
  of weak acids produce basic aqueous solutions.

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pH of a salt solution

• Aquoeus solutions of salts with acidic cations
  have a pH lower than 7 salts with basic anions
  produce a pH higher than 7 in aqueous solutions.

6
Class Practice

• Estimate the pH of 0.15M
• Ca(CH3COO)2(aq).

7
pH of mixed solutions

• The pH of a solution of a weak acid increases
  when a salt containing its conjugate base is
  added. The pH of a solution of a weak base
  decreases when a salt containing its conjugate
  acid is added.

8
Class Practice

• Calculate the pH of a solution that is 0.500 M
  HNO2(aq) and 0.100M KNO2(aq)
• Ka4.3x10-4 for HNO2

9
Titrations

• Strong acid strong base titrations

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• Strong base and strong acid

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• In the titration of a strong acid with a strong
  base (or a strong base with a strong acid), the
  pH increases (or decreases) slowly initially,
  increases (or decreases) rapidly through pH7 at
  the stoichiometric point, and then increases (or
  decreases) slowly again.

12
Class Practice

• Suppose we carry out a titration in which 0.34M
  HCl (aq) is the titrant and the analyte initially
  consists of 25 ml of 0.25M NaOH (aq). What is the
  pH of the analyte solution after the addition of
  5 ml of titrant?

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Calculating the pH at the stochiometric point of
the titration of weak acid with strong base

• Calculate the pH of the solution resulting when 5
  ml of 0.150 M NaOH (aq) is added to 25 ml of
  0.100M HCOOH (aq) . Use Ka1.8x10-4 for HCOOH
  (aq).

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Buffer solutions

• Solutions that resist the change in pH when small
  amounts of strong acids or base are added are
  called as buffers.

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Acid Buffer

• An acid buffer solution consists of a weak acid
  and its conjugate base it has pHlt7.
• CH3COOH (aq) H2O(l)?H3O(aq) CH3CO2- (aq)
• Acid Buffer action The weak acid transfers
  protons to the OH- ions from added strong base.
• The conjugate base of the weak acid accepts
  protons from the H3O ions supplied by a strong
  acid.

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Base Buffer

• A base buffer solution consists of a weak base
  and its conjugate acid it has pH gt7.
• NH3(aq) H2O(l)?NH4 (aq) OH- (aq)
• Base Buffer action The weak base accepts protons
  from the H3O ions supplied by a strong acid.
• The conjugate acid of the weak base transfers
  protons to the OH- ions from added strong base.

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Calculate the pH of a buffer solution

• For a solution containing a weak acid HA and a
  salt that provides the conjugate base anion, A- ,
  the proton transfer equilibrium is
• HA(aq) H2O(l)?H3O (aq) A- (aq)
• KaH3O A-/HA
• Rearranging the equation
• 1/H3O 1/Ka xA-/HA

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• Taking logarithm of both sides
• log1/H3O log1/ Ka logA-/HA
• -logH3O -logKa logA-/HA
• That is pH pka logA-/HA
• pH pKa log (baseinitial/acidinitial)
• This relation is called the Henderson Hassel
  balch equation.

19
Class Practice

• Calculate the pH of a buffer solution that is
  0.40 M NaCH3CO2 (aq) and 0.80M CH3COOH(aq) at
  25C.
• Calculate the ratio of the molarities of CO32-
  and HCO3- ions required to achieve buffering at
  pH 9.50. The pka of H2CO3 is 10.25.

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Solubility product

• The solubility product is the equilibrium
  constant for the equilibrium between an un
  dissolved salt and its ions in a saturated
  solution.

21
Class Practice

• The molar solubility of silver chromate Ag2CrO4
  is 6.5x10-5 mol/L. determine the value of Ksp.
• The Ksp for lead(II)iodide in water is 1.4x10-8 .
  Estimate its molar solubility.

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The common ion effect

• The common-ion effect is a term used to describe
  the effect on a solution of two dissolved solutes
  that contain the same ion.The presence of a
  common ion suppresses the ionization of a weak
  acid or a weak base.

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• If both sodium acetate and acetic acid are
  dissolved in the same solution they both
  dissociate and ionize to produce acetate ions.
  Sodium acetate is a strong electrolyte so it
  dissociates completely in solution. Acetic acid
  is a weak acid so it only ionizes slightly.
• NaC2H3O2(s) ? Na(aq) C2H3O2-(aq) HC2H3O2(l) ?
  H(aq) C2H3O2-(aq)

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• According to Le Chatelier's principle, the
  addition of acetate ions from sodium acetate will
  suppress the ionization of acetic acid and shift
  its equilibrium to the left. This will decrease
  the hydrogen ion concentration and thus the
  common-ion solution will be less acidic than a
  solution containing only acetic acid.

25
Class Practice

• What is the approximate molar solubility of
  silver chloride in 0.10M NaCl (aq)?

26
Predicting precipitation

• In the Ksp expression the right hand side of the
  expression is known as the "ion product". At
  saturation when the ions in solution are in
  equilibrium with the solid slightly soluble salt,
  the ion product is equal to a fixed value called
  the solubility product constant
• Ksp ion product

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• Two further cases exist
• ion product lt Ksp
• ion product gt Ksp
• If the ion product lt Ksp then no precipitation
  will occur even though the salt may be insoluble
  according to the solubility rules. If, on the
  other hand, the ion product gt Ksp the ion
  concentration will be large enough for
  precipitation to occur

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• The concentration of Calcium ion in blood plasma
  is 0.0025 M. If the concentration of Oxalate ion
  is 1 X 10-8 M. Will Calcium Oxalate, CaC2O4
  precipitate? Ksp 2.3 X 10-9.

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• Write the slightly soluble salt equilibria.
  CaC2O4 Ca2 C2O4-2
• Write the ion product ion product Ca2
  C2O4-2
• Identify the given ion concentrations
• Ca2 .0025 2.5 X 10-3 M
• C2O4-2 1 X 10-8 M
• Plug in the given ion concentrations into the ion
  product equation

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• ion product Ca2 C2O4-2 (2.5 X 10-3) ( 1
  X 10-8)
• ion product 2.5 X 10-11
• Compare the ion product value with Ksp value and
  make a conclusion. Since the ion product value
  (2.5 X 10-11) is less than the Ksp (2.3 X 10-9)
  we have to conclude that no precipitation will
  take place.

31
Class Practice

• Lead II Chromate, PbCrO4, is used in yellow paint
  pigment("chrome yellow"). When the solution is
  5.0 X 10-4 M in Pb2 ion and 5.0 X 10-5 M in
  Chromate ion. Would you expect some of the Lead
  Chromate to precipitate. Ksp PbCrO4 1.8 X
  10-14.

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Homework

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