analytical chemistry - clutch 1e ch.12 - advanced topics in...
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ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
CONCEPT: SYSTEMATIC APPROACH – ACID-BASE SYSTEMS
Strong Acids and Bases are considered _________________ Electrolytes so they ionize completely in water.
• In general, the ___________ the Ka the stronger the acid and the ___________ the concentration of H+.
When calculating the pH of a solution we must take into consideration the concentration of the strong acid or base.
Concentration ≥10−6M
The concentrations of either of H+ and OH– are significant enough to determine pH and pOH directly.
1.5 x 10-3 M 1.5 x 10-3 M 1.5 x 10-3 M 0.00075 M 0.00075 M 0.00075 MHNO3 (aq) H+ (aq) + NO3
– (aq) NaOH (aq) Na+ (aq) + OH– (aq)
H2O H2O
Concentration ≤10−8M
The concentrations either of H+ and OH– are too small to be significant and so pH equals _______.
8.4 x 10-11 M 8.4 x 10-11 M 8.4 x 10-11 M 7.0 x 10-9 M 7.0 x 10-9 M 7.0 x 10-9 MHNO3 (aq) H+ (aq) + NO3
– (aq) NaOH (aq) Na+ (aq) + OH– (aq)
H2O H2O
Between10−6M to10−8M
The concentrations of H+ and OH– must compete with the auto-ionization of water so a systematic approach is used.
0.200 M 0.200 M 0.200 M
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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PRACTICE: SYSTEMATIC APPROACH – ACID-BASE SYSTEMS CALCULATIONS 1 EXAMPLE: Determine the pH of a 3.5 x 10-8 M HBr. PRACTICE: Determine the pH of a 6.7 x 10-8 M NaOH.
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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CONCEPT: FRACTIONAL COMPOSITION Fractional compositions are used as an illustration for the amount of acid or base species at a specific pH.
Monoprotic Systems The dissociation of a weak monoprotic acid creates an equilibrium and expression that can be tied to its mass balance.
HA H+ +A– Ka =[H+ ][A− ][HA] Mass Balance : F = [HA]+[A– ]
The fraction of HA molecules is represented by αHA .
pH
(frac
tion
in e
ach
form
)α
0
0
.25
0.50
0.7
5
1.0
0
Diprotic Systems From the derived equilibrium expressions we can determine the mass balance.
H2A H+ +HA– HA– H+ +A2–
Ka2=[H+ ][A2− ][HA− ]
⇒ [A2− ]=[HA− ][Ka2
][H+ ]
=[H2A]Ka1
Ka2
[H+ ]2Ka1=[H+ ][HA][H2A]
⇒ [HA− ]=[H2A][Ka1
][H+ ]
F = [H2A] [H2A]⋅[H+ ]2 +[H+ ]Ka1
+Ka1Ka2
[H+ ]2⎤
⎦⎥
⎡
⎣⎢⎢
The fractions of the three major diprotic forms can be seen as:
[HA]= [H+ ]F[H+ ]+Ka
Dividing by F αHA =HAF
=[H+ ]
[H+ ]+Ka
The fraction in the conjugate base form, A–, can be represented as αA− .
αA− =
A−
F=
Ka
[H+ ]+Ka
pH
0
0
.25
0.50
0.7
5
1.0
0(fr
actio
n in
eac
h fo
rm)
α
αH2A=H2AF
=[H+ ]2
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
αHA− =
HA−
F=
Ka1[H+ ]
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
αA2−
=A2−
F=
Ka1Ka2
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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PRACTICE: FRACTIONAL COMPOSITION CALCULATIONS 1 EXAMPLE 1: A dibasic compound, B, has pKb1 = 5.00 and pKb2 = 8.00. Find the fraction of the acidic form when the pH = 9.00. EXAMPLE 2: What fraction of tyrosine (pKa1 = 2.37, pKa2 = 8.67) exists in all of its forms at pH = 10.00? PRACTICE: Calculate the fraction of the intermediate for sulfurous acid, H2SO3, at pH = 11.00? pKa1 = 1.80, pKa2 = 7.19.
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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CONCEPT: FRACTIONAL COMPOSITION AND CONCENTRATIONS Monoprotic Systems
Recall that the fraction of HA and A– molecules are represented by αHA and αA− .
pH
(frac
tion
in e
ach
form
)α
0
0
.25
0.50
0.7
5
1.0
0
When taking into account the formal concentrations of HA and A– molecules we now restructure the equations:
[HA]= αHAFHA =[H+ ]FHA[H+ ]+Ka
[A− ]= αA−FHA =
KaFHA[H+ ]+Ka
Diprotic Systems From the derived equilibrium expressions we can determine the mass balance.
H2A H+ +HA– HA– H+ +A2–
Ka2=[H+ ][A2− ][HA− ]
⇒ [A2− ]=[HA− ][Ka2
][H+ ]
=[H2A]Ka1
Ka2
[H+ ]2Ka1=[H+ ][HA][H2A]
⇒ [HA− ]=[H2A][Ka1
][H+ ]
F = [H2A] [H2A]⋅[H+ ]2 +[H+ ]Ka1
+Ka1Ka2
[H+ ]2⎤
⎦⎥
⎡
⎣⎢⎢
The formal concentrations of the three major diprotic forms can be restructured as:
[HA]= [H+ ]F[H+ ]+Ka
Dividing by F αHA =HAF
=[H+ ]
[H+ ]+Ka
αA− =
A−
F=
Ka
[H+ ]+Ka
pH
0
0
.25
0.50
0.7
5
1.0
0(fr
actio
n in
eac
h fo
rm)
α
αH2A=H2AF
=[H+ ]2
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
αHA− =
HA−
F=
Ka1[H+ ]
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
αA2−
=A2−
F=
Ka1Ka2
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
[H2A]= αH2AFH2A =
[H+ ]2 FH2A[H+ ]2 +[H+ ]Ka1
+Ka1Ka2
[HA− ]= αHA−FH2A =
Ka1[H+ ]FH2A
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
[A2− ]= αA2−FH2A =
Ka1Ka2FH2A
[H+ ]2 +[H+ ]Ka1+Ka1
Ka2
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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PRACTICE: FRACTIONAL COMPOSITION AND CONCENTRATIONS CALCULATIONS 1 EXAMPLE 1: A dibasic compound, B, has pKb1 = 4.00 and pKb2 = 6.00. Find the concentration of the intermediate form when FH2A = 0.150 M and the pH = 8.00.
EXAMPLE 2: Calculate the concentration of the acidic form for 0.230 M tartaric acid, H2C4H4O6, at pH = 6.00? pKa1 = 3.00, pKa2 = 4.34.
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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CONCEPT: DAVIES EQUATION We learned that the activity coefficient and ionic strength of a solution could be closely and accurately related by using the extended Debye-Huckel equation:
log γ = − 0.51z2 µ
1+ α µ305
⎞
⎠⎟⎟
⎛
⎝⎜⎜
When the size parameter of the ion is unknown we can instead use the Davies Equation. • Because of the lack of a size parameter this formula is most useful for monovalent ions.
From the Davies Equation, all ions with the same magnitude in charge will have the same activity coefficient. EXAMPLE: Calculate the activity coefficient of Ca2+ in 0.025 M Ca3(PO4)2.
log γ = − 0.51z2 µ
1+ µ− 0.3µ
⎞
⎠⎟
⎛
⎝⎜⎜ Ionic Strength ±1 ± 2 ±3
0.001 0.97 0.87 0.730.005 0.93 0.74 0.510.010 0.90 0.66 0.400.050 0.82 0.45 0.160.100 0.78 0.36 0.100.200 0.73 0.28 0.060.500 0.69 0.23 0.040.700 0.69 0.23 0.04
Ionic Charge (z)
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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CONCEPT: DEPENDENCE OF SOLUBILITY ON PH
Recall that ionic compounds are composed of an anion and cation, either of which can create an acidic, basic or neutral solution.
Cations +
Transition Metals___ or higher charge will be acidic, less than ___ will be neutral
MnI5
Main Group Metals___ or higher charge will be acidic, less than ___ will be neutral
AlF3
Positive Amines
CH3NH3Cl
Positively charged amines are acidic
Cations can create solutions that are either acidic or neutral.
• ______________ the pH increases the solubility of sparingly acidic salts.
Anions –Add an H+ to the anion and if you create a weak acid then your negative ion is basic.
NaNO2H+
Add an H+ to the anion and if you create a strong acid then your negative ion is neutral.
KClH+
Anions can create solutions that are either basic or neutral.
• ______________ the pH increases the solubility of sparingly basic salts.
AmphotericAcidic BasicHSO4
– HSO3– H2PO4
– HCO3– HS – HPO4
2-
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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PRACTICE: DEPENDENCE OF SOLUBILITY ON PH CALCULATIONS 1
EXAMPLE 1: BaCO3 is the slightly soluble ionic salt that results from the reaction between Ba(OH)2 and H2CO3. Identify the effect of increasing acidity on the solubility of the given compound.
EXAMPLE 2: Which salts will be more soluble in an acidic solution than in pure water?
a. CuBr
b. Ag2SO4
c. BaSO3
d. Sn(OH)2
e. KClO4
ANALYTICAL CHEMISTRY - CLUTCH 1E
CH.12 - ADVANCED TOPICS IN EQUILIBRIUM
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