AP Chapter 17

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AP Chapter 17

• Ionic Equilibria of Weak Electrolytes

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Review Test

• 20 AP Multiple Choice Questions

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pH and pOH scales
basic
acidic
pH pOH 14.00
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H H3O
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pH -logH

• The pH scale is a logarithmic scale.
• This means that in order to change the pH by one
  unit there must be a tenfold change in the H.
• Find the pH of a solution with H 0.010M
• Find the pH of a solution with H 0.10M

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Example 17.1 Page 521

• What is the pH of a solution of HCl with a
  concentration of 1.2 x 10-3M?

pH 2.92
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Example 17.2 Page 521

• Calculate the pH of
• 0.10 M solution of HNO3
• 0.10 M solution of CH3CO2H (1.3 ionized)

pH 1.00
pH 2.89
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Example 17.4 Page 522

• Calculate the H3O of a solution with a pH of
  9.0

H3O 1.0 x 10-9
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Example 17.6 Page 523

• Calculate the pH and pOH of a 0.0125 M KOH
  solution.

pOH 1.903 pH 12.097
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Calculate the OH- of a solution with a pH of
5.56.
OH- 3.63 x 10-9
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What are the ion concentrations in a 0.10M HCl
solution?
0.10M HCl 0.10M H 0.10M Cl-
Strong electrolytes dissociate completely
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What are the ion concentrations in a 0.15 M
K2SO4 solution?
0.15M K2SO4 0.30M K 0.15M SO42-
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At equilibrium, a solution of acetic acid,
CH3CO2H 0.0788M and H3O CH3CO2-
0.0012M. What is the Ka of acetic acid?
Example 17.8 page 527
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Example 17.9 page 527
The pH of a 0.0516 M solution of nitrous acid,
HNO2, is 2.34. What is the Ka?
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Additional Ka values
Table contains more Ka values than are pictured
here
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Calculate the H3O, CH3CO2-, and CH3CO2H
in a 0.100 M solution of acetic acid. What is
the Ka 1.8 x 10-5.
Example 17.10 page 528
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What else can you determine in the previous
example?

• pH, pOH, OH-
• percent ionization

Calculate this value.
Percent ionization 1.3
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What is the percent ionization of a 0.25 M
solution of trimethylamine, (CH3)3N, a weak base
with a Kb 7.4 x 10-5
1.7
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What is the pH of the trimethylamine solution?

• OH- 4.3 x 10-3
• pOH -log4.3 x 10-3 2.37
• pH 14 2.37 11.63
• OH- 4.3 x 10-3
• H 1 x 10-14 4.3 x 10-3 2.3 x 10-12
• pH -log2.3 x 10-12 11.63

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Appendix G has additional values
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Diprotic and Triprotic Acids

• A diprotic acid ionizes in two steps because it
  has two ionizable hydrogens.
• A triprotic acid ionizes in three steps because
  it has three ionizable hydrogens.

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The Stepwise Dissociation of Phosphoric Acid. A
triprotic acid.
H3PO4 (aq) H2O(l)
H2PO4-(aq) H3O(aq)
H2PO4-(aq) H2O(l)
HPO42-(aq) H3O(aq)
HPO42-(aq) H2O(l) PO43-(aq)
H3O(aq)
H3PO4 (aq) 3 H2O(l) PO43-(aq)
3 H3O(aq)
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Each ionization step occurs to a lesser extent
than one preceding it.
H3PO4 (aq) H2O(l)
H2PO4-(aq) H3O(aq)
Ka 7.5 x 10-3
H2PO4-(aq) H2O(l)
HPO42-(aq) H3O(aq)
Ka 6.3 x 10-8
HPO42-(aq) H2O(l)
PO43-(aq) H3O(aq)
Ka 3.6 x 10-13
H3PO4 (aq) 3 H2O(l)
PO43-(aq) 3 H3O(aq)
Ka ?
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0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate H3O.
Example 17.19 page 548
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0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate H3O.
We can work the problem using either Ka or Kb
Example 17.19 page 548
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• Compare this answer to H3O from problem 17.10
  pp. 528.

• How does this illustrates LeChatliers Principle.

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Example 17.10 page 528
Calculate the H3O, CH3CO2-, and CH3CO2H
in a 0.100 M solution of acetic acid. What is
the Ka 1.8 x 10-5.
CH3CO2H H2O ? CH3CO2- H3O H3O 1.3 x 10-3
0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate H3O.
H3O 3.6 x 10-6
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10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
mixed. Calculate the OH-.
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10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
mixed. Calculate the OH-.
Example 17.20 page 549
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Common Ion Problems.What is the common ion in
each example?

• Example 17.20 page 549
• 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
  mixed. Calculate the OH-.
• Example 17.19 page 548
• 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
  dissolved in 1.0L of solution. Calculate H3O.
• Buffers are also examples of the common ion
  effect since they are mixtures where both
  substances produce the same ion.
• Both of the solution mixtures above are buffers.

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Buffers

• A buffer solution is an aqueous solution
  consisting of a mixture of a weak acid and its
  conjugate base or a weak base and its conjugate
  acid.
• Buffer solutions have the property that the pH of
  the solution changes very little when a small
  amount of acid or base is added to it.

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Demonstration Buffered vs. Non-buffered
solutions
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Why are these solutions buffers?

• Example 17.20 page 549
• 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
  mixed. Calculate the OH-.
• Example 17.19 page 548
• 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
  dissolved in 1.0L of solution. Calculate H3O.

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How buffers work

• Weak acids and weak bases tend to remain in high
  concentrations when added to water because by
  definition they do not ionize much in water since
  they are weak.
• However, they are very likely to react with any
  added strong base or strong acid.

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Why are weak acids/bases used to create buffers?
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Remember this problem

• Calculate the pH of
• 0.10 M solution of HNO3
• 0.10 M solution of CH3CO2H (1.3 ionized)
• What would the CH3CO2H have to be for it to
  have the same pH as the HNO3 assuming the 1.3
  ionization factor does not change?
• 0.10M CH3CO2H(0.013)
• CH3CO2H 7.7 M

pH 1.00
pH 2.89
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Why are weak acids/bases used to create buffers?

• Thats right 7.7 M CH3CO2H vs. 0.10M have the
  same pH.
• In other words the CH3CO2H is 77 times greater
  HNO3.
• It takes much more base to change the pH of a
  weak acid solution as compared to a strong acid
  solution of the same pH because there are many
  more moles of acid available to neutralize base
  in the weak acid as compared to the strong acid.

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Weak Acid (HA) and its conjugate base (A-) buffer
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Adding a strong base to the buffer

• If a strong base is added to a buffer, the weak
  acid will give up its H in order to transform
  the base (OH-) into water (H2O) and the conjugate
  base
• HA OH- ? A- H2O.
• Since the added OH- is consumed by this reaction,
  the pH will change only slightly.

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Adding a strong acid to the buffer

• If a strong acid is added to a buffer, the weak
  base will react with the H from the strong acid
  to form the weak acid, HA
• H A- ? HA
• The H gets absorbed by the A- instead of
  reacting with water to form H3O (H), so the pH
  changes only slightly.

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An effective buffer requires relatively equal
amounts of weak acid and conjugate base.
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Basic Buffers

• Note that the same ideas hold true for weak
  bases, (B), and their conjugate acids (BH).

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For the most effective buffers Ka H3O
HA H2O ? A- H3O
Consider acetic acid
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You want to make the most effective buffer you
can using acetic acid. What would the H3O be?
Calculate the pH of this buffer.
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I want to make a buffer with a pH of 3.14. Which
acid should I use?
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Did you choose well?

• Or will you be flung into the Gorge of Eternal
  Peril.

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The Bridge of Death 310
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I want to make a buffer with a pH of 3.14. Which
acid should I use?
Ka H3O H3O antilog (-pH) Ka antilog
(-3.14) 7.2 x 10-4
HF
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I want to make a buffer with a pH of 3.75. Which
acid should I use?
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I want to make a buffer with a pH of 3.75. Which
acid should I use?
Ka H3O H3O antilog (-pH) Ka antilog
(-3.14) 7.2 x 10-4
Formic Acid (HCO2H)
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For the most effective buffers Kb OH-
Consider ammonia
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For the most effective buffers Kb OH-
B H2O ? BH OH-
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I want to make a buffer with a pH of 10.64.
Which base should I use?
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I want to make a buffer with a pH of 10.64.
Which base should I use?
Kb OH- Kb antilog (-pOH) Kb antilog
(-3.36) 4.4 x 10-4
Methylamine (CH3NH2)
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Effective Buffers

• The previous formulas only apply when the
  concentrations of weak acid and conjugate base or
  weak base and conjugate acid are equal.
• However it is important to note that you do not
  have to have equal concentrations to have a
  buffer.

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Henderson Hasselbach Equation
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0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate H3O.
Example 17.19 page 548
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Henderson Hasselbach Equation
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10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
mixed. Calculate the OH-.
Example 17.20 page 549
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I want to make a the most effective buffer
possible using acetic acid. What chemicals
should I get from the stock room?

• Describe how you would make the buffer?
• What would the pH of the buffer be?

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I have no acetate salts. Can I still make the
buffer using acetic acid? Explain.
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I want to make a the most effective buffer
possible using ammonia. What chemicals should I
get from the stock room?

• Describe how you would make the buffer?
• What would the pH of the buffer be?

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I have no ammonium salts. Can I still make a
buffer using ammonia? Explain.
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Making Effective Buffers

1. Equal moles of a weak acid and a salt of its
   conjugate base.
2. Equal moles of a weak base and a salt of its
   conjugate acid.
3. A weak acid and half as many moles of strong
   base.
4. A weak base and half as many moles of strong
   acid.
5. Use a pH meter

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pH meter
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Calculate the pH of a buffer that is 0.10 M
acetic acid and 0.10 M sodium acetate.
Example 17.21 page 551

• pH 4.74

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Buffer Capacity

• Buffer capacity the amount of an acid or base
  that can be added to a volume of a buffer
  solution before its pH changes significantly.
• Buffer capacity depends on the amount (moles) of
  the conjugate pair used to make the buffer.
• Buffer solutions have essentially lost their
  buffering capabilities when one component of the
  conjugate pair is about 10 or less of the other.

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Page 555
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Buffers

• Which buffer has the greater capacity?
• Which buffer is more effective?
• Buffer A
• 100mL of 0.1M CH3CO2H with 100 mL of 0.1M
  NaCH3CO2.
• Buffer B
• 100mL of 1.0M CH3CO2H with 100 mL of 1.0M
  NaCH3CO2.

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Calculate the OH- of a 0.050 M solution of
NaCH3CO2

• OH- 5.3 x 10-6

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Calculate the pH of a 0.050 M solution of NaCH3CO2

• pH 8.72

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Calculate the percent reaction of a 0.050 M
solution of NaCH3CO2

• 0.011

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Calculate the pH of a 0.10 M solution of AlCl3
(Ka 1.4 x 10-5)

• pH 2.93

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Titration
titrant
analyte
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Titration Curves

• A titration curve is a plot of the pH against the
  volume of acid or base added in a titration.
• The equivalence point or endpoint for a titration
  is the point at which exactly enough of the
  titrant has been added to completely react with
  the analyte.
• In other words, at the equivalence point, the
  number of moles of titrant added corresponds
  exactly to the number of moles of substance being
  titrated, the analyte, according to the reaction
  stoichiometry.

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Table 17.8 Page 568
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How do we choose an appropriate indicator for a
titration?
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Indicators

• An acid base indicator is a substance that
  indicates how acidic or basic a solution is using
  certain color changes.
• Indicators are normally weak acids that have a
  different color than their conjugate base.
• HA H2O ? H3O A-
• Indicators can also be weak bases that have a
  different color than their conjugate acid.

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Indicators

• Remember that the equivalence point of a
  titration is where you have mixed the two
  substances in exact stoichiometric proportions.
• You obviously need to choose an indicator which
  changes color as close as possible to that
  equivalence point.
• The indicator should have a pKa value near the pH
  of the titration's endpoint. 
• That varies from titration to titration.

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pKa and pH range of Acid-Base Indicators
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Which indicator(s) would be appropriate for the
titrations below?
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Indicators
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Choosing an Indicator

• Remember that an indicator is normally a weak
  acid.
• HA H2O ? H3O A-
• Write the equilibrium expression for the reaction
  of the indicator.

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Choosing an Indicator

• HA H2O ? A- H3O

How would the HA compare to the A- halfway to
the equivalence point of the reaction of HA with
water?
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Choosing an Indicator

• HA H2O ? A- H3O

HA A- halfway to the equivalence point.
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Choosing an Indicator

• HA H2O ? A- H3O

• Therefore halfway to the equivalence point for
  the reaction of any indicator is an excellent
  point to pick the indicator for our acid-base
  titration.
• Why?

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Choosing an Indicator

• HA H2O ? A- H3O

• HA A- halfway to the equivalence point.
• This is the point that the indicator changes
  color.

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Choosing an Indicator

• HA H2O ? A- H3O

• HA A- halfway to the equivalence point.
• Less than halfway to the equivalence point of
  this reaction HA gt A- and therefore the color
  is still red.
• More than halfway to the equivalence point of
  this reaction A- gt HA and therefore the color
  is still blue.

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Choosing an Indicator

• HA A- halfway to the equivalence point.

• At half way to the equivalence point Ka H3O
• At half way to the equivalence point pKa pH

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Choosing an Indicator

• At half way to the equivalence point Ka H3O
• At half way to the equivalence point pKa pH
• Therefore we want to choose an indicator that has
  a pKa pH at the equivalence point of the
  titration we are performing.
• This is because this is the point at which the
  indicator will go through its most significant
  color change.

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Now that we have chosen an appropriate indicator
for a titration. Lets consider in more detail
what happens during the titration of different
types of acids.
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What substances are involved in each of the
titrations?
(b)
(a)
(d)
(c)
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What species are present at half way to the
equivalence point?
NaOH HCl ? NaCl H2O

• At half way to the equivalence point the main
  species present are H (hydronium), Cl- and Na.
• The solution is still acidic.

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What species are present at half way to the
equivalence point?
NaOH CH3CO2H ? NaCH3CO2 H2O

• At half way to the equivalence point the main
  species present are CH3CO2H, Na, and CH3CO2- and
  to a lesser extent H (hydronium).
• The solution is still acidic.

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The titration curve is produced when a 10.0 ml
sample of HCl is titrated with 0.100M NaOH. What
is the concentration of the HCl solution.
0.25M HCl
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The titration curve is produced when a 10.0 ml
sample of CH3CO2H is titrated with 0.100M NaOH.
What is the concentration of the CH3CO2H solution.
0.25M CH3CO2H
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See next slide for a clue.
Determine the Ka of the weak acid using the
titration curve.
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a
106

Determine the Ka of the weak acid using the
titration curve.
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Diprotic Acid

• When titrating a diprotic acid with a strong base
  it is essentially like doing two titrations at
  once.

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Titration curve of a diprotic acid
HSO4- ? H SO42-
H2SO4 ? H HSO4-
The titration curve shown above is for a diprotic
acid such as H2SO4. This proves that polyprotic
acids lose their protons in a stepwise manner.
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How would you determine the Ka for each acid
(H2SO4 and HSO4-)?
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How would you determine the Ka for each acid
(H2SO4 and HSO4-)?
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Why do you have to learn all this very
challenging information?
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This stuff is hard. Why do you have to learn it?
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Why? Because I had to suffer through this crap
and now its payback time!