Acids and Bases

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Acids and Bases
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2003 AP

• 1 A-E

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Write the Equilibrium, Kb, for the reaction
represented
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Writing an equilibrium expression for 1a

• Using the law of mass action given the chemical
  equilibrium equation
• Concentration Products over Reactants raised to
  their stoichiometric coefficients excluding pure
  liquids and solids!

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Writing an equilibrium expression for 1a

• In this case, equilibrium expression consists of
  the products of the concentrations of Conjugate
  acid of Aniline and Hydroxide Ion over the
  concentration of Aniline.

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A Sample of aniline is dissolved in water to
produce 25mL of a 0.10M. The pH of the solution
is 8.82. Calculate the Equilibrium Constant
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Calculate the Kb for the reaction

• Use the equilibrium expression above to determine
  the Kb.
• The concentration of aniline is 0.1M
• The Hydroxide and Conjugate acid concentration
  can both be determined after calculation of the
  pOH (14-8.82).
• Take the pOH and raise it to the 10(-pOH)

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Calculate the Kb for the reaction

• This will yield the concentration for OH- and
  Aniline Conjugate because they are produce in the
  same proportion 11.
• Multiply these concentrations and divide them by
  the initial concentration of Aniline.

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Calculate Kb

• Assume the initial concentration of OH- is
  negligible.
• The reason x is not subtracted from the initial
  concentration of aniline is because x is so small
  that it is considered negligible.

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The solution prepared in part b is titrated with
a 0.10M. Calculate the pH of the solution when
5.0mL of the acid has been added.
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Calculate pH after 5mL of HCl is added

• Set up a net chemical equation for the reaction
  of Aniline and H (Strong Acid)
• Since the initial concentrations of OH- and
  conjugate acid are small they are not taken into
  account.
• Multiply the Molarity of Aniline by the volume in
  liters (same with the HCl)
• From here you have the moles of both Aniline and
  HCl

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Calculate pH after 5mL of HCl is added

• The H will combine with Aniline to form a
  conjugate acid and goes to completion.
• Subtract the number of moles of H from the moles
  of Aniline. This gives a new number of moles of
  Aniline.
• Since all of the H moles are consumed, it is
  equal to the number of moles of the Conjugate
  acid.

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Calculate pH after 5mL of HCl is added

• To determine the pH, we will use a variation of
  the Henderson-Hassalbauch. Below
• From here we take the pKb calculated above and
  the mole ratio of the acid over the base.
• The pOH can be determined
• To get the pOH we subtract the pOH from 14 to get
  the pH.

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Calculate the pH at the equivalence point.
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pH at the equivalence point

• We already know the moles of the Aniline.
• The moles of Aniline must equal the moles of HCl
  for the equivalence point to be reached. (Ratio
  11)
• From here we have the only Aniline Conjugate Acid
  moles.
• We must determine the concentration of conjugate
  acid and the Ka of the conjugate

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pH at the equivalence point

• 25mL initially of aniline
• To determine the volume of HCl take the moles of
  HCl and divide it by the concentration which
  yields the volume required.
• Convert all volumes to liters and add the initial
  volume of aniline with the volume of HCl added.
• Take the moles of conjugate acid and divide it by
  this new volume

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pH at the equivalence point

• To determine the Ka dived the Kb into (1.0 x
  10-14).

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pH at the equivalence point

• Write the Chemical reaction for the behavior of
  Aniline conjugate in water and make an
  equilibrium expression.
• x is not subtracted from the initial
  concentration because it is considered
  negligible.
• Multiply the concentration of Conjugate aniline
  concentration by the Ka. Then take the square
  root of the product.

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pH at the equivalence point

• Then take log (x) of the answer
• This will give you the pH at the equivalence
  point
• pH 2.97

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Which of the following indicators listed is most
suitable for this titration.
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Selecting an Indicator

• Based on the calculations above the pH at the end
  point is 2.97
• Erythrosine is optimal because based on the color
  change in an acidic pH
• It is a weak based titration with a strong acid.

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2002
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Calculate the value H in an HOBr solution that
has a pH of 4.95
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Calculate H

• Given the pH is 4.95
• Use the exponent 10 raised the pH
• This will yield the H

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Write the equilibrium constant expression for the
ionization of HOBr in an HOBr solution with a
H 1.8 x 10(-5)
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Equilibrium constant expression

• Given the chemical equilibrium reaction and the
  concentration of H
• H OBr- This is true because as HOBr
  dissociates the products form in same proportions
  
• Exclude x because it is negligible.
• Use the Law of Mass action Products over
  Reactants raised to their stoichiometric
  coefficients.

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• Use the equilibrium constant.
• The product of the concentration divided by the
  Ka yields the concentration of HOBr.

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Calculate the volume of 0.115M Ba(OH)2 needed to
reach equivalence when titrated into 65mL sample
of 0.146M of HOBr
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Calculate Volume Needed to Reach Equivalent Point

• Convert the Volume of HOBr to liters
• Multiply the volume in Liters by the Molarity of
  HOBr in Solution
• This will give you the moles of the HOBr in
  solution
• The titrant used is Ba(OH)2 (Strong Base) so the
  concentration of OH- must be doubled therefore
  you must multiply the concentration of Ba(OH)2 to
  get the concentration of OH-

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Calculate Volume Needed to Reach Equivalent Point

• Take the Moles of HOBr (HOBr OH-) that was
  calculated and divide it by the molarity of OH-
  (Ba(OH)2 x 2).
• This will give you the volume in liters necessary
  to added to reach eq point

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Indicate whether the pH at equivalence point is
less than 7, 7, or greater than seven. Explain
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What will the pH at the eq point be?

• You can go through the calculations to determine
  the pH specifically at the eq point.
• The basic rule of thumb is that if you are
  titrating a weak acid with a strong base then the
  pH at the end point will be greater than 7.

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Calculate the number of the moles of NaOBr(s)
that would have to be added to 125mL of 0.160M
HOBr to produce a buffer solution with a H of
5.00 x 10(-9)
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Number of Moles needed to produce certain
Concentration

• The Henderson-Hasselbauch equation can be used
  for this scenario (See equation Below)
• Given the H take the log(H) and determine
  the pH for the equation

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Number of Moles needed to produce certain
Concentration

• You have already been given the Ka so plug that
  in the equation as well
• Separate the ratio of concentrations into log(B)
  log (A) (Refer to Log Rules)

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Number of Moles needed to produce certain
Concentration

• Solve for the Log (Base) and then eliminate the
  log function by raising the solution using the
  base of 10.
• This will yield your concentration of Base

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Number of Moles needed to produce certain
Concentration

• Multiply the concentration by the volume of
  solution in liters
• This will yield the number of NaOBr moles

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HOBr is a weaker acid than HBrO3. Account for
this fact.
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Strong Acid/Weak Acid

• The number of oxygen atoms will effect the
  strength of the acid. HOBr has a single oxygen
  atom while HBrO3 contains three.
• Oxygen is a very electro negative atom and will
  draw electrons away from the H-Br bond thus
  weakening the bond making it easier to dissociate
  when in dissolved in water. The more oxygen atoms
  the weaker the H-Br bond.
• Charge of X
• The charge on the X which in this case is the Br
  atom has a different charge in HOBr (1) than in
  HBrO3 (5)
• Strength of O-H Bond
• Strength of X-O Bond

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Arrhenius Acid Base Concept

• Arrhenius Acid/Base Concept
• Acids produce Hydrogen ions (H) within an
  aqueous solution
• Bases produce Hydroxide Ions (OH-) in solution
• Definition is limited because it applies only to
  acids and bases that can dissociate OH- and H
  ions
• Examples
• NaOH will dissociate into Na and OH-
• HCl will dissociate into H and Cl-

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Bronsted-Lowry Model

• The model definition of Acid/Base
• Bronsted Acid A proton donor
• Bronsted Base A proton acceptor
• The definition applies to many more molecules
  that may exhibit Acid/Base qualities but do not
  directly produce OH- or H ions.
• Every Acid and Base has a conjugate Acid or Base
• Water can act as an acid and a base
• H3O (Hydronium ion) (acid) and OH- (Base)

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Bronsted-Lowry Model

• Example of Bronsted Base
• NH3(aq) H2O(l) ? NH4(aq) OH-(aq)
• Example of Bronsted Acid
• HC2H3O2(aq) H2O(l) ? H3O(aq) C2H3O2-(aq)

Base
Acid Conjugate Acid
Conjugate Base
Weak Acid Base
Conjugate Acid Conjugate Base
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Lewis Acid/Base Definition

• Lewis Acid/Base Definition
• Lewis Acid Electron Pair Acceptor
• Lewis Base Electron Pair Donor
• Encompasses an even wider variety of molecules
  (Bronsted and Arrhenius) even ones that do not
  donate protons or produce OH- ions.
• Must be aware of the Lewis structure of a
  particular molecule to determine whether it is a
  Lewis Acid or Base.

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Lewis Acid/Base Definition

• Examples
• BF3(g) NH3(g) ? F3BNH3(g)
• BF3 is the Lewis Acid because it has no free
  unpaired electrons with only has 6 electrons
  around the central atom (Boron will require one
  more pair of electrons to complete the valence
  shell)
• NH3 is the Lewis Base because the molecule has a
  completed octet valence shell with free unpaired
  electrons on the central atom. The BF3 will
  accept these unpaired electrons and form a
  covalent bond.

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Lewis Acid/Base Definition

• Example
• Ni2(aq) 6NH3(aq) ? Ni(NH3)62(aq)
• Ni2 is the Lewis Acid because it is a cation
  which will attract negatively charged electrons
  to toward itself.
• The NH3 is the Lewis Base because it provides the
  free unpaired electrons for the Ni2

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Lewis Acid/Base Definition

• Example
• Even earlier definitions are encompassed in the
  Lewis Acid Model
• H H2O ? H3O

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