Titration

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Titration
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Imagine we have 25 ml of 0.250M NaOH solution

• pH of this solution
• pOH - log (0.250 M) 0.6 so pH 13.4
  (point A)
• Calculation of pH after the addition of 5 mL of
  0.340 M HCl solution
• number of moles of OH- 25/1000 . 0.250
  6.25 . 10-3
• number of moles of H added 5/1000 . 0.340
  1.7 . 10-3
• number of OH- moles remained 4.55 . 10-3
• OH- in 30 ml 4.55 . 10-3 . 1000 /30 ml
  0.152 M
• pOH - log (0.152 M) 0.820 so pH 13.18
  (point B)
• pH after the addition of 10 ml of acid 12.91
  (point C)

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• Volume of 0.340 HCl for neutralization of 25 ml
  of 0.25M NaOH
• V / 1000 . 0.340 M 6.25. 10-3 so V
  18.4 ml, pH 7.00 (point S)
• Calculation of pH after the addition of 19.4 ml
  HCl
• number of moles of H added 19.4/1000 .
  0.34 6.60 . 10-3
• number of moles of excess H 6.60 -
  6.25 . 10-3 0.35 . 10-3
• H in 44.4 ml ( 19.4 25.0) 0.35. 10-3
  . 1000 /44.4 ml 7.9.10-3 M
• pH - log (7.9 . 10-3 M) 2.10 (point D).

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• A plot of the pH of the NaOH solution,
  against the volume of the added HCl is called a
  pH curve. In point S the number of moles of acid
  added are equal to the number of moles of NaOH.
• Sometimes we need to know how much acid or base
  is present in the sample. With adding a solution
  of accurate concentration (titrant), from a buret
  to a flask containing the sample (analyte) and
  accurate detection of the S point (stoichiometric
  point, equivalent point) we can do that.

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Imagine we have 25.00 ml of 0.100M HCOOH solution

• pH of this solution
• HCOOH H2O HCOO- H3O
• 0.100 - x x
  x
• Ka x2 / 0.100 - x 1.8 . 10-4 x
  H2O 4.2 . 10-3 (approxim.)
• pH - log (4.2 . 10-3 ) 2.4 (point A)
• Calculation of pH after the addition of 5.00 mL
  of 0.150 NaOH solution
• number of moles of HCOOH 25.00 /1000 .
  0.100 2.5 . 10-3
• number of moles of OH- added 5.00 /1000 .
  0.150 0.75 . 10-3
• number of HCOO- formed 0.75 . 10-3
• HCOO- in 30 ml 0.75 . 10-3 . 1000 /30
  ml 0.0250 M
• number of moles of HCOOH remained 1.75 .
  10-3
• HCOOH in 30 ml 1.75 . 10-3 . 1000 /30
  ml 0.0583 M

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• HCOOH H2O HCOO- H3O
• 0.0583 - x 0.0250 x
  x
• Ka 0.0250 . x / 0.0583 1.8 . 10-4
  x 4.2 . 10-4 (approxim.)
• pH - log (4.2 . 10-4 ) 3.38 (point B)
• pH after the addition of 10 ml of NaOH 3.92
  (point D)
• The plot of the pH of the HCOOH solution,
  against the volume of the added NaOH

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• pH at the stoichiometric (equivalent) point pH
  after the hydrolysis of formate anion
• number of moles of HCOOH 25.00 / 1000 .
  0.100 2.50 . 10-3
• so the number of moles of HCOO- at the
  equivalent point 2.50 . 10-3
• volume of the NaOH solution to reach the
  stoichiometric point
• V/1000 . 0.150 M 2.5 . 10-3 so V16.7 ml
• HCOO- in 41.7 ml (16.7 25.0) 2.5.
  10-3. 1000 /41.7 ml 0.060 M
• HCOO- H2O HCOOH OH-
• 0.060 - x x
  x
• Kb x . x / 0.060 5.6 . 10-11
• x OH- 1.8 . 10-6 M (approxim.)
• H 5.6 . 10-9 M
• pH 8.25 (point s)

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• In the middle of the way to equivalent point
• HCOOH HCOO-
• Ka H3O HCOO- / HCOOH
• H3O
• -log Ka -log H3O
• pKa pH
• In our example, the midway pH is 3.75, so pKa of
  formic acid is 3.75 (point C).
• This is one of the practical ways to determine
  the pKa of acids or bases.
• The same is true for a weak base.

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Indicators

• A weak organic acid with different colours in
  acid (HIn) and its conjugate base forms (In-).
  The colour change results from the effect of the
  acidic proton in the chemical structure of the
  acid form (HIn).
• The colour change can be explained in the context
  of Br?nsted Lowry concept.
• HIn H2O
  In- H3O

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• There are many naturally occuring indicators. A
  single compound is responsible for the colours of
  red poppies (a) and blue cornflowers (b). The pH
  of the sap is different in the two plants.

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• HIn H2O
  In- H3O
• KIn H3O
  In- / HIn
• Each indicator has its own KIn.
• In the case of Litmus KIn 10-7
• HIn
  H2O In- H3O
• RED
  BLUE
• 10-7
  H3O In- / HIn
• 10-7
  / H3O In- / HIn
• When pH 5 Litmus is RED, so 10-7 / 10-5
  1 / 100 In- / HIn
• When pH 8 Litmus is BLUE ,so 10-7 / 10-8
  10 / 1 In- / HIn
• When pH is between 5 and 8 Litmus is purple.

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Selcting An Indicator

• The end point of a titration is defined as the
  point at which the concentrations of HIn and
  In- are equal, HIn In- so
• KIn H3O
• That is, the colour change occures when
• pKIn pH
• It is important to select an indicator with
  an end point close to the stoichiometric point of
  a titration. In practice
• pKIn pH
  (stoichiometric) 1

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• For a strong acid strong base titration,
  litmus (pKIn 6.5) is the best indicator,
  however, the change in pH is so abrupt that
  phenolphthalein (pKIn 9.4) can also be used.

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• Phenolphthalein (pKIn 9.4) can be used for
  titrations with stoichiometric point near pH 9,
  such as a titration of a weak acid with a strong
  base.

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• Methyl orange (pKIn 3.4) can be used for a weak
  base strong acid titration.

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Imagine a solution with equal concentrations of
acetic acid and acetate ion

• CH3COOH
  H CH3COO-
• 1.00 M
  ? 1.00 M
• 
  H 1.82 . 10-5
• pH
  pKa 4.742
• If we add 0.01 mol of HCl to 1 liter of this
  solution the change in pH
• Before 1.00 M
  1.82 . 10-5 M 1.00 M
• After 1.01 M
  ? 0.99 M
• H CH3COO- /
  CH3COOH 1.82 . 10-5
• 
  H(0.99) / (1.01) 1.82 . 10-5
• 
  H 1.86 . 10-5
• 
  pH 4.731 ( 0.011 unit change)

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• If we add 0.01 mol of NaOH to 1 liter of this
  solution the change in pH
• CH3COOH
  H CH3COO-
• Before 1.00 M
  1.82 . 10-5 M 1.00 M
• After 0.99 M
  ? 1.01 M
• H CH3COO- /
  CH3COOH 1.82 . 10-5
• 
  H(1.01) / (0.99) 1.82 . 10-5
• 
  H 1.78 . 10-5
• 
  pH 4.749 ( 0.009 unit change)
• The same would happen with a solution of equal
  amounts of a weak base and its conjugate acid
• NH3 H2O
  NH4 OH-
• NH4 OH-
  / NH3 1.82 . 10-5
• 
  OH- 1.82 . 10-5
• 
  pOH 4.74
• 
  pOH 9.26

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• So a solution of equal amounts of a weak
  acid and its conjugate base or a weak base and
  its conjugate acid can keep the pH unchanged at
• pH pKa or pH
  14 pKb
• Imagine again a solution with equal
  concentrations of acetic acid and acetate ion
• CH3COOH
  H CH3COO-
• 1.00 M
  1.82 . 10-5 M 1.00 M
• If we add 0.30 mol of HCl to 1 liter of this
  solution the change in pH
• After 1.30 M
  ? 0.70 M
• H CH3COO- /
  CH3COOH 1.82 . 10-5
• 
  H(0.70) / (1.30) 1.82 . 10-5
• 
  H 3.38 . 10-5
• 
  pH 4.471 ( 0.271 unit change)
• Therefore, such a solution can play this role
  only against little amounts of strong acids or
  bases.

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• If the amounts of the weak acid and its conjugate
  base are not equal this system can protect the pH
  against added strong acids or bases but in an
  amount different from the pKa of the acid
• HA H2O
  H3O A-
• Ka H3O
  A- / HA
• H3O Ka
  HA / A-
• -log H3O -log Ka
  log HA / A-
• pH pKa - log
  HA / A-
• pH pKa log A-
  / HA
• This equation determines the pH of a
  solution with different concentrations of HA and
  A- with the ability to control the pH of a
  solution.
• The same equation is true for a weak base
  and its conjugate acid pH
  pKa log B / BH
• This equation is called Henderson
  Hasselbalch equation.
• The solution of a weak acid and its
  conjugate base or a weak base and its conjugate
  acid is called a Buffer solution.

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• What is the pH of a buffer solution of CH3COONa
  0.040 and CH3COOH 0.080 at 25C ? Ka
  1.82 . 10-5
• pH pKa log A- / HA
• pH 4.74 log
  0.040/0.080 pH 4.44

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Buffer Capacity

• The buffer capacity is and indication of the
  amount of the acid or base that can be added
  before the buffer loses its ability to resist the
  change in pH.
• The buffer becomes exhausted when most of the
  weak base has been converted to acid or when most
  of the weak acid has been converted to base.
• A buffer has a high capacity when the amount of
  the base present is about 10 or more of the
  amount of acid, for otherwise the base gets used
  up quickly. Similarly, the amount of the acid
  present should be about 10 or more of the amount
  of base, for otherwise the acid gets used
  quickly.
• So the pH range would be calculated by
  Henderson Hasselbach equation
• pH
  pKa log A- / HA
• pH
  pKa log (A- / 10A- )
• pH
  pKa log (1 / 10) pKa 1
• pH
  pKa log (10HA / HA)
• pH
  pKa log (10/ 1) pKa 1

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• Some Buffer Systems
• 
  pKa
• CH3COOH / CH3COO-. 4.47
• HNO2 / NO2- 3.37
• HClO2 / ClO2- 2.00
• NH4 / NH3 9.25
• (CH3)3NH / (CH3)3NH 9.81
• H2 PO4- / HPO42- 7.21