To calculate the new pH, use the Henderson- Hasselbalch equation: 1141.
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Transcript of To calculate the new pH, use the Henderson- Hasselbalch equation: 1141.
![Page 1: To calculate the new pH, use the Henderson- Hasselbalch equation: 1141.](https://reader036.fdocuments.us/reader036/viewer/2022062500/56649f575503460f94c7c4ad/html5/thumbnails/1.jpg)
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To calculate the new pH, use the Henderson-Hasselbalch equation:
H]CO[CH
] CO[CHlogpKpH23
-23a
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To calculate the new pH, use the Henderson-Hasselbalch equation:
= 4.6
H]CO[CH
] CO[CHlogpKpH23
-23a
M]1.10M0.90log4.7
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To calculate the new pH, use the Henderson-Hasselbalch equation:
= 4.6
Note in this example that the [H+] changed from [H+] = 10-pH = 10-4.7 = 2.0 x 10-5 M to [H+] = 10-pH = 10-4.6 = 2.5 x 10-5 M
H]CO[CH
] CO[CHlogpKpH23
-23a
M]1.10M0.90log4.7
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To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.
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To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.
The initial [H+] = 1.0 x 10-7 M (from self-dissociation of water)
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To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.
The initial [H+] = 1.0 x 10-7 M (from self-dissociation of water)
After the HCl is added, [H+] = 1.0 x 10-1 M , so there is a million fold increase in [H+]!
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To appreciate the effectiveness of the buffer in this example, consider what happens to the pH when 0.10 mols of gaseous HCl is added to 1.0 liter of H2O.
The initial [H+] = 1.0 x 10-7 M (from self-dissociation of water)
After the HCl is added, [H+] = 1.0 x 10-1 M , so there is a million fold increase in [H+]!
Whereas for the buffer, [H+] changes from 2.0 x 10-5 M to 2.5 x 10-5 M.
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Distribution Curve
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Distribution Curve Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component present in solution as a function of the pH.
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Distribution Curve Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component present in solution as a function of the pH.
Example: Consider the acetic acid/acetate buffer.
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Distribution Curve Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component present in solution as a function of the pH.
Example: Consider the acetic acid/acetate buffer. At low pH: H+ + CH3CO2
- CH3CO2H
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Distribution Curve Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component present in solution as a function of the pH.
Example: Consider the acetic acid/acetate buffer. At low pH: H+ + CH3CO2
- CH3CO2H
At high pH: OH- + CH3CO2H CH3CO2-
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Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:
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Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:
1pKpH a
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Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:
For the acetic acid/acetate ion buffer (pKa =4.7) the
buffer range is pH = 3.7 – 5.7 .
1pKpH a
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Buffer Range: The pH range in which a buffer is effective. The range is sometimes defined as follows:
For the acetic acid/acetate ion buffer (pKa =4.7) the
buffer range is pH = 3.7 – 5.7 . Note that the buffer functions best at pH = 4.7, i.e. when
[CH3CO2H] = [CH3CO2-]
1pKpH a
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Exercise: Comment on the following as possible buffer systems.
(a) NaCl(aq)/HCl(aq)
(b) NH3(aq)/NH4Cl(aq)
(c) H2PO4-(aq)/ HPO4
2-(aq)
(d) NaHCO3(aq)
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(a) NaCl(aq)/HCl(aq)
To be a buffer, it needs to be able to react with both added H+ and added OH-. If H+ is added, the following reaction does not occur to any significant extent: H+
(aq) + Cl-
(aq) HCl(aq)
(because HCl is a strong acid).
That means that (a) cannot be a buffer system.
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(b) NH3(aq)/NH4Cl(aq)
Addition of H+: H+(aq) + NH3(aq) NH4
+(aq)
Addition of OH-: OH-
(aq) + NH4+
(aq) NH3(aq) + H2O
Hence (b) is a buffer system.
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(b) NH3(aq)/NH4Cl(aq)
Addition of H+: H+(aq) + NH3(aq) NH4
+(aq)
Addition of OH-: OH-
(aq) + NH4+
(aq) NH3(aq) + H2O
Hence (b) is a buffer system.
(c) H2PO4-(aq)/ HPO4
2-(aq)
Addition of H+: H+(aq) + HPO4
2-(aq) H2PO4
-(aq)
Addition of OH-: OH-
(aq) + H2PO4-(aq) HPO4
2-(aq) + H2O
Hence (c) is a buffer system.
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(d) NaHCO3(aq)
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(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)
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(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3
-(aq) H2CO3(aq)
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(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3
-(aq) H2CO3(aq)
Addition of OH-: OH-
(aq) + HCO3-(aq) CO3
2-(aq) + H2O
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(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3
-(aq) H2CO3(aq)
Addition of OH-: OH-
(aq) + HCO3-(aq) CO3
2-(aq) + H2O
Hence (d) is a buffer system.
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(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3
-(aq) H2CO3(aq)
Addition of OH-: OH-
(aq) + HCO3-(aq) CO3
2-(aq) + H2O
Hence (d) is a buffer system. Note that in this example, the HCO3
- ion functions as both the acid and the base.
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(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3
-(aq) H2CO3(aq)
Addition of OH-: OH-
(aq) + HCO3-(aq) CO3
2-(aq) + H2O
Hence (d) is a buffer system. Note that in this example, the HCO3
- ion functions as both the acid and the base.
Various anions of multi-proton acids can function as buffer systems with a single species present.
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IONIC EQUILIBRIUMSolubility Products
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IONIC EQUILIBRIUMSolubility Products
A saturated solution of an insoluble salt is a heterogeneous equilibrium.
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IONIC EQUILIBRIUMSolubility Products
A saturated solution of an insoluble salt is a heterogeneous equilibrium.
Example: In a saturated solution of AgCl solution, the following equilibrium is present:
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IONIC EQUILIBRIUMSolubility Products
A saturated solution of an insoluble salt is a heterogeneous equilibrium.
Example: In a saturated solution of AgCl solution, the following equilibrium is present:
AgCl(s) Ag+(aq) + Cl-
(aq)
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IONIC EQUILIBRIUMSolubility Products
A saturated solution of an insoluble salt is a heterogeneous equilibrium.
Example: In a saturated solution of AgCl solution, the following equilibrium is present:
AgCl(s) Ag+(aq) + Cl-
(aq)
The equilibrium constant is:
] [AgCl]-][Cl[AgK
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Now
]][Cl[Ag] K[AgCl -
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Now
but [AgCl] is a constant, recall
]][Cl[Ag] K[AgCl -
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Now
but [AgCl] is a constant, recall
]][Cl[Ag] K[AgCl -
AgClmass molarAgCldensity[AgCl]
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Now
but [AgCl] is a constant, recall
We set where the subscript sp stands for solubility product.
]][Cl[Ag] K[AgCl -
AgClmass molarAgCldensity[AgCl]
] K[AgClKsp
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Now
but [AgCl] is a constant, recall
We set where the subscript sp stands for solubility product. Hence,
]][Cl[Ag] K[AgCl -
AgClmass molarAgCldensity[AgCl]
] K[AgClKsp
]][Cl[AgK -sp
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Examples: MgF2(s) Mg2+(aq) + 2 F-
(aq)
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Examples: MgF2(s) Mg2+(aq) + 2 F-
(aq)
2-2sp ]][F[MgK
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Examples: MgF2(s) Mg2+(aq) + 2 F-
(aq)
Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4
3-(aq)
2-2sp ]][F[MgK
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Examples: MgF2(s) Mg2+(aq) + 2 F-
(aq)
Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4
3-(aq)
2-2sp ]][F[MgK
2-34
32sp ][PO][CaK
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Examples: MgF2(s) Mg2+(aq) + 2 F-
(aq)
Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO4
3-(aq)
Note that the pure solids do not occur in the expression for Ksp.
2-2sp ]][F[MgK
2-34
32sp ][PO][CaK
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Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
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Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-
(aq) at 25 oC, we have one of the following situations:
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Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-
(aq) at 25 oC, we have one of the following situations:
unsaturated solution
10- 10x1.6]][Cl[Ag
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Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-
(aq) at 25 oC, we have one of the following situations:
unsaturated solution saturated solution
10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag
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Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-
(aq) at 25 oC, we have one of the following situations:
unsaturated solution saturated solution supersaturated solution
10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag
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Very small values for Ksp indicate very insoluble salts. For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-
(aq) at 25 oC, we have one of the following situations:
unsaturated solution saturated solution supersaturated solution Some AgCl precipitate will form until the product of
the ionic concentrations is equal to 1.6 x 10-10.
10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag 10- 10x1.6]][Cl[Ag
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Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3
2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.
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Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3
2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.
Ag2CO3(s) 2 Ag+(aq) + CO3
2-(aq)
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Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3
2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.
Ag2CO3(s) 2 Ag+(aq) + CO3
2-(aq)
= (2.54 x 10-4)2 (1.27 x 10-4) = 8.19 x 10-12
][CO][AgK -23
2sp
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Sample problem 1: In a saturated solution of Ag2CO3, the concentrations of the ions are [Ag+] = 2.54 x 10-4 M and [CO3
2-] = 1.27 x 10-4 M. Calculate the Ksp and the solubility of Ag2CO3 in g/liter.
Ag2CO3(s) 2 Ag+(aq) + CO3
2-(aq)
= (2.54 x 10-4)2 (1.27 x 10-4) = 8.19 x 10-12 The concentration of CO3
2-(aq) is equal to the number
of moles of Ag2CO3(s) that have dissolved.
][CO][AgK -23
2sp
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molar mass of Ag2CO3 = 275.8 g/mol
The solubility of Ag2CO3
= 1.27 x 10-4 mol l-1 275.8 g mol-1
= 0.0350 g/l
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Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.
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Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.
PbF2(s) Pb2+(aq) + 2 F-
(aq)
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Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.
PbF2(s) Pb2+(aq) + 2 F-
(aq)
The abbreviated ICE table for this problem looks like
Pb2+ F-
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Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.
PbF2(s) Pb2+(aq) + 2 F-
(aq)
The abbreviated ICE table for this problem looks like
Pb2+ F-
y 2y
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Sample problem 2: Calculate the solubility of PbF2 in g/l given Ksp = 4.1 x 10-8.
PbF2(s) Pb2+(aq) + 2 F-
(aq)
The abbreviated ICE table for this problem looks like
Pb2+ F-
y 2y
228sp ]-][F[Pb10x4.1K