Polyprotic Acids

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Polyprotic Acids Bases

• A polyprotic acid can donate more than one H
• Carbonic acid H2CO3(aq) dissolved CO2 in water
• Sulfuric acid H2SO4(aq)
• Phosphoric acid H3PO4(aq)
• A polyprotic base can accept more than one
  proton
• Carbonate ion CO32-(aq)
• Sulfate ion SO42-(aq)
• Phophate ion PO43-(aq)
• Treat each step of protonation or deprotonation
  sequentially

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• H2CO3 (aq) H2O(l) ? H3O(aq) HCO3-(aq) Ka1
  4.3 x 10-7
• HCO3-(aq) H2O(l) ? H3O(aq) CO32-(aq) Ka2
  4.8 x 10-11
• Typically
• Ka1 gtgt Ka2 gtgt Ka3 gtgt
• Harder to loose a positively charged proton from
  a negatively charged ion, because of attraction
  between opposite charges.

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• Calculate the pH of 0.010 M H2SO4(aq) at 25oC.
• Sulfuric acid is the only common polyprotic acid
  where the first deprotonation step is complete.
  The second deprotonation step is much weaker and
  adds slightly to the H3O(aq) concentration.
• For the first step assume all H2SO4(aq)
  deprotonates
• H2SO4 (aq) H2O(l) ? H3O (aq) HSO4-(aq)
• From the first step H3O(aq) 0.010 M

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• Second deprotonation
• HSO4- (aq) H2O(l) ? H3O (aq) SO42- (aq) Ka2
  0.012
• HSO4- (aq) SO42- (aq) H3O (aq)
• Initial 0.010 0 0.010
• Change -x x 0.010 x
• Equilibrium 0.010-x x 0.010 x
• Ka2 (H3O (aq))(SO42- (aq)) / (HSO4-
  (aq) )
• 0.012 (0.010x)(x) / (0.010-x)
• Solve the quadratic equation for x. Ka2 is large
  cannot assume that x ltlt 0.010
• H3O (aq) 1.4 x 10-2 M
• pH 1.9

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• Determine the pH of 0.20 M H2S(aq) at 25oC
• H2S (aq) H2O(l) ? H3O (aq) HS- (aq) Ka1
  1.3 x 10-7
• HS- (aq) H2O(l) ? H3O (aq) S2- (aq)
  Ka2 7.1 x 10-15
• For the first deprotonation step determine
  H3O(aq) using equilibrium tables. H3O(aq)
  1.6 X 10-4 M
• Can assume that x ltlt 0.20 since Ka1 is small
• Second deprotonation constant is very small, so
  ignore addition of H3O(aq) due to second step.
• pH determined by first step alone. pH 3.8

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Composition and pH

• For a solution of H2CO3(aq) at low pH the fully
  protonated species (H2CO3) dominates at high pH
  the fully deprotonated form (CO32-) dominates
  and at intermediate pH the intermediate species
  (HCO3-) dominates.
• LeChateliers principle at work
• H2CO3 (aq) H2O(l) ? H3O(aq) HCO3-(aq) Ka1
  4.3 x 10-7
• HCO3-(aq) H2O(l) ? H3O(aq) CO32-(aq) Ka2
  4.8 x 10-11

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• H2CO3(aq) H2O(l) ? H3O(aq) HCO3-(aq)
• Ka1 H3O(aq) HCO3-(aq) / H2CO3(aq)
• HCO3-(aq) H2O(l) ? H3O(aq) CO32-(aq)
• Ka2 H3O(aq) CO32-(aq) / HCO3-(aq)
• Define a(X) fraction of species X

X
a(X)
H2CO3(aq) HCO3-(aq) CO32-(aq)
The fraction of deprotonated species increases as
the pH increases
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Determine the concentration of H2CO3(aq),
HCO3-(aq), CO32-(aq), H3O(aq) present and the pH
at equilibrium in a solution that is initially
0.010 M in H2CO3. (Ka1 4.3 x 10-7, Ka2 4.8 x
10-11) Step 1 H2CO3(aq) H2O(l) ? H3O(aq)
HCO3-(aq) H2CO3(aq) H3O(aq)
HCO3-(aq) Initial 0.010 0 0 Change
-x x x Equilibrium 0.010-x
x x 4.3 x 10-7 x2 /
(0.010-x) Assume x ltlt 0.010 x 6.6 x
10-5 H2CO3(aq) 0.010 M H3O(aq) 6.6 x
10-5 M HCO3-(aq) 6.6 x 10-5 M
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HCO3-(aq) H2O(l) ? H3O(aq) CO32-(aq) Ka2
H3O(aq) CO32-(aq) / HCO3-(aq)
HCO3-(aq) H3O(aq) CO32-(aq) Initial 6.6 x
10-5 6.6 x 10-5 0 Change -y 6.6
x 10-5 y y Equilibrium 6.6 x 10-5 - y
6.6 x 10-5 y y 4.8 x 10-11 (6.6 x
10-5 y) y / (6.6 x 10-5 - y) Assume y ltlt 6.6 x
10-5 4.8 x 10-11 (6.6 x 10-5 y / (6.6 x 10-5
) y 4.8 x 10-11 At equilibrium H2CO3(aq)
0.010 M H3O(aq) 6.6 x 10-5 M HCO3-(aq)
6.6 x 10-5 M CO32-(aq) 4.8 x 10-11 M pH
4.18
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13
Buffers

• Buffer solutions resists change in pH even with
  addition of small amounts of acid or base.
• Buffer solutions are mixed solutions mixture of
  a weak acid and its conjugate base or a weak base
  and its conjugate acid.
• Human blood has a pH maintained at pH 7.4 due
  to a combination of carbonate, phosphate and
  protein buffers.
• The ocean is buffered to a pH of 8.4 by
  buffering that depends the presence of hydrogen
  carbonates and silicates.

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Buffer Action

• An acid buffer is an aqueous solution of a weak
  acid and its conjugate base.
• It buffers solutions on the acid side of neutral
  (pH lt 7).
• Example solution of CH3COOH(aq) CH3COONa(aq)
• CH3COOH / CH3COO-
• A base buffer is an aqueous solution of a weak
  base and its conjugate acid.
• It buffers solutions on the basic side of neutral
  (pH gt 7).
• Example NH3(aq) NH4Cl(aq)
• NH3/ NH4

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• Buffer solution of CH3COOH(aq) / CH3COO- (aq)
• If a small amount of strong acid is added
• H3O(aq) CH3COO-(aq) ? CH3COOH(aq) H2O(l)
• K for this reaction 1/Ka(CH3COOH(aq)) 5.5 x
  104
• The CH3COO-(aq) acts as a sink for the added
  protons, and the pH remains unchanged.
• If a small amount of strong base is added
• OH-(aq) CH3COOH(aq) ? CH3COO-(aq) H2O(l)
• K for this reaction 1/Kb(CH3COO-(aq)) 1.8 x
  109
• The CH3COOH(aq) acts as a sink for the added
  OH-, and the pH remains unchanged

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Designing a buffer

• Make a solution with particular pH so that it
  buffers about this pH.
• Consider a solution of a weak acid and its
  conjugate base
• HA(aq) H2O(l) ? H3O(aq) A-(aq)

Henderson-Hasselbach equation
Note for a solution of a weak base/conjugate
acid, use Ka of the conjugate acid
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• An optimal buffer is one in which the weak acid
  and its conjugate base have equal concentrations
• Select a weak acid that has its pKa as close as
  possible to the desired pH
• Having chosen the weak acid, use the Henderson
  Hasselbach equation to determine the ratio of
  A-(aq) and HA(aq) that will form solution
  that buffers around the desired pH

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• Calculate the pH of a buffer solution that is
  0.040 M CH3COONa (aq) and 0.080 M CH3COOH (aq).
• pKa(CH3COOH(aq)) 4.75
• CH3COOH(aq) H2O(l) ? H3O(aq) CH3COO-(aq)
• CH3COOH(aq) H3O(aq) CH3COO-(aq)
• Initial 0.080 M 0 0.040M
• Change -x x 0.040 x
• Equilibrium 0.080 - x x 0.040 x
• Ka (0.040 x) x / (0.080 - x)
• Assume x ltlt 0.040
• x 3.6 x 10-5
• pH 4.44

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Suppose that a solution is made by dissolving 1.2
g NaOH(s) (0.030 moles) in 500 mL of the buffer
solution in the previous problem. Calculate the
pH of the resulting solution and the change in
pH. Assume the volume of the solution to be
constant. The OH-(aq) will react with
CH3COOH(aq) CH3COOH(aq) OH-(aq) ? CH3COO-(aq)
H2O(l) Moles of OH- added 0.030 moles Moles of
CH3COOH(aq) (0.500 L) (0.080 M) 0.040
mol Amount of unreacted CH3COOH(aq) 0.010
moles Molarity of CH3COOH(aq) 0.020 M
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Moles of CH3COO-(aq) initial amount amount
formed by reaction of OH- (aq) and CH3COOH(aq)
(0.040M x 0.500 L) (0.030 moles) 0.050
moles Molarity of CH3COO-(aq) 0.10
M CH3COOH(aq) H2O(l) ? CH3COO-(aq)
H3O(aq) pH pKa log (CH3COO-(aq) /
CH3COOH(aq) ) 5.45 If the solution had
contained HCl at pH 4.4, addition of the NaOH
would have raised the pH to 12.8
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Titrations

• Strong Acid - Strong Base
• H3O(aq) OH-(aq) ? 2H2O(l)
• pH changes slowly initially, changes rapidly
  through pH 7 (equivalence point) and then
  changes slowly again
• If the analyte is a strong acid, pH increases as
  base is added

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If the analyte is a strong base, pH decreases as
acid is added
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Analyte 25.00 mL of 0.250 M NaOH(aq) Titrant
0.340 M HCl(aq) Determine the pH of the solution
when 5.00mL of titrant added Answer pH
13.18 Determine the amount of titrant that must
be added to reach the equivalence point? Answer
18.4 mL Determine the pH of the solution after
the addition of 20.4 mL of titrant. Answer pH
1.82
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Strong Acid-Weak Base and Weak Acid - Strong Base
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CH3COOH(aq) OH-(aq) -gt CH3COO-(aq) H2O(l)
Slow change in pH before equivalence point
solution is a buffer CH3COOH(aq)/CH3COO-(aq) At
halfway point HA A- pH pKa At
equivalence, pH determined by CH3COO-(aq)
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• Changes in pH during a titration of a weak
  acid/base with a strong base/acid
• Halfway to the stoichiometric point, the pH pKa
  of the acid
• The pH is greater than 7 at the equivalence point
  of the titration of a weak acid and strong base
• The pH is less that 7 at the equivalence point of
  the titration of a weak base and strong acid
• Beyond the equivalence point, the excess strong
  acid or base will determine the pH of the solution

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Titration of 100.0 mL of 0.1000 M CH3COOH(aq)
with 0.1000 M NaOH Before addition of NaOH pH
determined by CH3COOH(aq) CH3COOH(aq) H2O(l) ?
H3O(aq) CH3COO-(aq) Answer pH 2.88 Before
the equivalence point determine pH for a
buffer Addition of 30.00 mL of NaOH(aq) The
OH-(aq) reacts with the CH3COOH(aq). Determine
concentration of CH3COOH(aq) and CH3COO- (aq) in
solution after addition of the base. Answer pH
4.38 At half equivalence CH3COOH(aq)
CH3COO-(aq) pH pKa
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At equivalence enough OH-(aq) added to react
with all CH3COOH(aq). For this problem,
equivalence is reached when 100.0mL of OH- is
added i.e. 0.01000 moles of OH-(aq) added
Solution contains 0.01000 moles CH3COO-(aq) in
200.0 mL solution CH3COO-(aq) 0.05000 M pH
determined by CH3COO-(aq) H2O(l) ? CH3COOH(aq)
OH- (aq) pH 8.72 (note greater than
7.0) Beyond equivalence pH determined by excess
OH-(aq)
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• Estimate the pH at the equivalence point of the
  titration of 25.00 mL of 0.100 M HCOOH(aq) with
  0.150 M NaOH(aq)
• (Ka(HCOOH) 1.8 x 10-4)
• At the equivalence point, enough NaOH(aq) has
  been added to react with all the HCOOH(aq)
  forming HCOO-(aq)
• The reaction
• HCOO-(aq) H2O(l) ? HCOOH(aq) OH-
• determines the pH at equivalence
• Answer 8.26