analytical chemistry chem 3811 chapter 10
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ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 10 ACID-BASE TITRATIONS. STRONG ACID – STRONG BASE. - Write balanced chemical equation between titrant and analyte - PowerPoint PPT PresentationTRANSCRIPT
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ANALYTICAL CHEMISTRY CHEM 3811
CHAPTER 10
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences
Clayton state university
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CHAPTER 10
ACID-BASE TITRATIONS
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STRONG ACID – STRONG BASE
- Write balanced chemical equation between titrant and analyte
- Calculate composition and pH after each addition of titrant
- Construct a graph of pH versus titrant added
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- Consider titration of base with acid
H+ + OH- → H2O
K = 1/Kw = 1/10-14 = 1014
- Equilibrium constant = 1014
- Reaction goes to completion
STRONG ACID – STRONG BASE
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H+ + OH- → H2O
At equivalence point
moles of titrant = moles of analyte
(V titrant)(M titrant) = (V analyte)(M analyte)
STRONG ACID – STRONG BASE
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Consider 50.00 mL of 0.100 M NaOH with 0.100 M HCl
- Three regions of titration curve exists
- Before the equivalence point where pH is determined by excess OH- in the solution
- At the equivalence point where pH is determined by dissociation of water (H+ ≈ OH-)
- After the equivalence point where pH is determined byexcess H+ in the solution
STRONG ACID – STRONG BASE
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First calculate the volume of HCl needed to reach the equivalence point
(V HCl)(0.100 M) = (50.00 mL)(0.100 M)
Volume HCl = 50.00 mL
STRONG ACID – STRONG BASE
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Before the equivalence point
Initial amount of analyte (NaOH) = 50.00 mL x 0.100 M = 5.00 mmol
After adding 1.00 mL of HClmmol H+ added = mmol OH- consumed
mmol H+ = (1.00 mL)(0.100 M) = 0.100 mmolmmol OH- remaining = 5.00 – 0.100 = 4.90 mmol
STRONG ACID – STRONG BASE
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Before the equivalence point
Total volume = 50.00 mL + 1.00 mL = 51.00 mL[OH-] = 4.90 mmol/51.00 mL = 0.0961 M
pOH = - log(0.0961) = 1.017
pH = 14.000 - 1.017 = 12.983
STRONG ACID – STRONG BASE
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- Repeat calculations for all volumes added
- Increments can be large initally but must be reduced just beforeand just after the equivalence point (around 50.00 mL in this case)
- Sudden change in pH occurs near the equivalence point
- Greatest slope at the equivalence point
STRONG ACID – STRONG BASE
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At the equivalence point
- pH is determined by the dissociation of water
H2O ↔ H+ + OH-
Kw = x2 = 1.0 x 10-14
x = 1.0 x 10-7
pH = 7.00 (at 25 oC)
STRONG ACID – STRONG BASE
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After the equivalence point
- Excess H+ is present
After adding 51.00 mL of HClExcess HCl present = 51.00 – 50.00 = 1.00 mL
Excess H+ = (1.00 mL)(0.100 M) = 0.100 mmol
Total volume of solution = 50.00 + 51.00 = 101.00 mL[H+] = 0.100 mmol/101.00 mL = 9.90 x 10-4 M
pH = -log(9.90 x 10-4) = 3.004
STRONG ACID – STRONG BASE
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STRONG ACID – STRONG BASEp
H
Volume of HCl added (mL)
7
50.00
Equivalence point(maximum slope or point of inflection)
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pH
Volume of NaOH added (mL)
7
50.00
Equivalence point(maximum slope or point
of inflection)
TITRATION CURVE
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Compare titration of HCl with NaOH and H2SO4 with NaOH(same volume and same concentration of acid)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Net ionic equation in both cases:H+(aq) + OH-(aq) → H2O(l)
orH3O+(aq) + OH-(aq) → 2H2O(l)
1 mol HCl conctributes 1 mol H3O+ 1 mol H2SO4 contributes 2 mol H3O+
STOICHIOMETRY AND TITRATION CURVE
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pH
Volume of NaOH added (mL)
STOICHIOMETRY AND TITRATION CURVE
HCl H2SO4
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WEAK ACID – STRONG BASE
Consider 50.00 mL 0f 0.0100 M acetic acid with 0.100 M NaOH
pKa of acetic acid = 4.76
HC2H3O2 + OH- → C2H3O2- + H2O
For the reverse reactionKb = Kw/Ka = 5.8 x 10-10
Equilibrium constant = 1/Kb = 1.7 x 109
So large that we can assume the reaction goes to completion
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WEAK ACID – STRONG BASE
Determine volume of base at equivalence point
mmol HC2H3O2 ≈ mmol OH-
(V NaOH)(0.100 M) = (50.00 mL)(0.0100 M)
Volume NaOH = 5.00 mL
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WEAK ACID – STRONG BASE
Four types of calculations to be considered
Before OH- is added
- pH is determined by equilibrium of weak acid
HA ↔ H+ + A-
x]-[F
][xK
2
a
x = 4.1 x 10-4
pH = 3.39
x]-[0.0100
][x2
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WEAK ACID – STRONG BASE
Before equivalence point
By adding OH- a buffer solution of HA and A- is formed
After adding 0.100 mL OH-
HA + OH- → A- + H2OInitial mmol 0.500 0.0100 0Final mmol 0.490 0 0.0100
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WEAK ACID – STRONG BASE
Before equivalence point
[HA]
][AlogpKpH
-
a
07.3[0.490]
[0.0100]log4.76pH
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WEAK ACID – STRONG BASE
Before equivalence point
If volume of OH- added is half the volume at equivalence pointHA = A- = 0.250 mmol
[0.250]
[0.250]log4.76pH
pH = pKa = 4.76
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WEAK ACID – STRONG BASE
At equivalence point
Volume of OH- = 5.00 mL
mmol OH- = (5.00 mL)(0.100 M) = 0.500 mmol
HA is used up and [HA] = 0
mmol A- = 0.500 mmol
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WEAK ACID – STRONG BASE
At equivalence point
Only A- is present in solution
A- + H2O ↔ HA + OH-
[A-] = (0.500 mmol)/(50.00 mL + 5.00 mL) = 0.00909 M
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WEAK ACID – STRONG BASE
At equivalence point
x]-[F
][xK
2
b
Kb = Kw/Ka = 5.8 x 10-10
x = [OH-] = 2.3 x 10-6
pOH = 5.64pH = 14.00 – 5.64 = 8.36
pH is not 7.00 but greater than 7.00(pH at equivalence point increases with decreasing strength of acid)
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WEAK ACID – STRONG BASE
After equivalence point
- Strong base (OH-) being added to weak base (A-)
- pH is determined by the excess [OH-] (approximation)
- After adding 5.10 mL OH-
[OH-] = (0.10 mL)(0.100 M)/(50.00 mL + 5.10 mL)= 1.81 x 10-4
pH = 14.00 – pOH = 14.00 – 3.74 = 10.26
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WEAK ACID – STRONG BASEp
H
Volume of NaOH added (mL)
8.36
5.00
Equivalence point(maximum slope or point of inflection)
pH = pKa
Minimum slope
pKa
2.50
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STRONG ACID – WEAK BASE
- The reverse of weak acid and strong base
B + H+ → BH+
- Similarly assume reaction goes to completion
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Consider 50.00 mL of 0.0100 M pyridine with 0.100 M HClKb of pyridine = 1.6 x 10-9
Determine volume of acid at equivalence point
mmol pyridine ≈ mmol H+
(V HCl)(0.100 M) = (50.00 mL)(0.010 M)
Volume HCl = 5.00 mL
STRONG ACID – WEAK BASE
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Four types of calculations to be considered
Before H+ is added
- pH is determined by equilibrium of weak base(determined using Kb)
B + H2O ↔ BH+ + OH-
STRONG ACID – WEAK BASE
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Before equivalence point
- By adding H+ a buffer solution of B and BH+ is formed
STRONG ACID – WEAK BASE
][BH
[B]logpKpH a
- When volume of H+ added = half the volume at equivalent point
pH = pKa (for BH+)
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At equivalence point
- B has been converted into BH+
- B is used up and [B] = 0
pH is calculated by considering BH+
BH+ ↔ B + H+
pH is not 7.00 but less than 7.00
STRONG ACID – WEAK BASE
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After equivalence point
- Strong acid (H+) is being added to weak acid (BH+)
pH is determined by the excess [H+] (approximation)
STRONG ACID – WEAK BASE
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STRONG ACID – WEAK BASEp
H
Volume of HCl added (mL)
5.00
Equivalence point(maximum slope or point of inflection)
pH = pKa
Minimum slopepKa
2.50
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END POINT
Use of Indicators
- Indicators are acids or bases so a few drops of dilute solutions are used to minimize indicator errors
- Acidic color if pH ≤ pKHIn - 1- Basic color if pH ≥ pKHIn + 1
- A mixture of both colors if pKHIn - 1 ≤ pH ≤ pKHIn + 1
- Use an indicator whose transition range overlaps the steepestpart of the titration curve
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END POINT
Use of pH Electrodes
- The end point is where the slope of the curve is greatest
- The end point is the volume at which the first derivative of a titration curve is maximum
- The end point is the volume at which the second derivative of a titration curve is zero