28-3-2011
DESCRIPTION
28-3-2011. Diprotic and Polyprotic Acids. Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation. Species concentrations of diprotic acids. - PowerPoint PPT PresentationTRANSCRIPT
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28-3-2011
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Diprotic and Polyprotic Acids
Have more than one acidic proton.
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.
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H2CO3H+HCO3-
Initial0.1000
Change-xX+X+
Equil.0.1-xxx
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H3PO4H+H2PO4-
Initial0.1000
Change-xX+X+
Equil.0.1-xxx
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Species concentrations of diprotic acids
Evaluate concentrations of species in a 0.10 M H2SO4 solution.
Solution:H2SO4 H+ + HSO4
– completely ionized
(0.1–0.1) 0.10+y 0.10-y
HSO4–
H+ + SO42– Ka2 = 0.012
0.10–y 0.10+y y Assume y = [SO42–]
(0.10+y) y————— = 0.012(0.10-y)
[SO42–] = y = 0.01M
[H+] = 0.10 + 0.01 = 0.11 M; [HSO4
–] = 0.10-0.01 = 0.09 M
Y2 + 0.112 y – 0.0012 = 0
- 0.112+0.1122 + + 4*0.0012y = —————————————— = 0.0098
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Another approach: Two steps method
Solution:H2SO4 H+ + HSO4
– completely ionized
(0.1–0.1) 0.10 0.10
In the 2nd equilibrium, substitute in [H+] = 0.1 M and [HSO4-] = 0.1 M
HSO4–
H+ + SO42– Ka2 = 0.012
0.10–x 0.10+x x
(0.10+x) x————— = 0.012, assume 0.1>>x, RE = {0.012/0.1}*100 = 12%
(0.10-x) Assumption is not OK
[SO42–] = y = 0.01M
[H+] = 0.10 + 0.01 = 0.11 M; [HSO4
–] = 0.10-0.01 = 0.09 M
Y2 + 0.112 y – 0.0012 = 0
- 0.112+0.1122 + + 4*0.0012y = —————————————— = 0.0098
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Species concentrations of weak diprotic acids
Evaluate concentrations of species in a 0.10 M H2S solution.
Solution:H2S H+ + HS– Ka1 = 1.02*10-7
(0.10–x) x+y x-y Assume x = [HS–]
HS– H+ + S2– Ka2 = 1.0*10-13
x–y x+y y Assume y = [S2–]
(x+y) (x-y) (x+y) y————— = 1.02*10-7 ———— = 1.0*10-13
(0.10-x) (x-y)
[H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = x y = 1.0*10–4 M; [S2–] = y = 1.0*10-13 M
0.1>> x >> y: x+ y = x-y = x
x = 0.1*1.02*10-7 = 1.00*10-4
y = 1*10-13
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Another approach: Two steps methodSolution:
H2S H+ + HS– Ka1 = 1.02*10-7 HS–
H+ + S2– Ka2 = 1.0*10-13
Ka1/Ka2 >102 , therefore the 2nd equilibrium can be neglected for the moment. From the 1st equilibrium:
H2S H+ + HS– Ka1 = 1.02*10-7
0.1 – x x x 1.02*10-7 = x2/(0.1-x), x = 10-4 and RE = {10-4/0.1}*100 = 0.1% (OK)
Therefore, [H+] = [HS-] = 10-4M
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From the second equilibrium, we have:
HS– H+ + S2– Ka2 = 1.0*10-13
10-4 -x 10-4 +x x
1.0*10-13 = (10-4 +x)x/(10-4 -x)
Assume 10-4 >>x
X = 1.0*10-13 which is very small compared to 10-4
Therefore, the concentrations of the different species are:
[H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = 10-4 - x = 10–4 M; [S2–] = x = 1.0*10-13 M
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NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Weak Bases and Base Ionization Constants
Kb =[NH4
+][OH-][NH3]
Kb is the base ionization constant
Kb
weak basestrength
Solve weak base problems like weak acids except solve for [OH-] instead of [H+].
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Ionization Constants of Conjugate Acid-Base Pairs
HA (aq) H+ (aq) + A- (aq)
A- (aq) + H2O (l) OH- (aq) + HA (aq)
Ka
Kb
H2O (l) H+ (aq) + OH- (aq) Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Ka = Kw
Kb
Kb = Kw
Ka
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Ka Kb and Kw
H+ + Base Conjugate_acid of Base+
Acid H+ + Conjugate_base of Acid-
For example:
NH3 + H2O NH4+ + OH-
Ka for NH4+ = Kw / Kb for NH3
HA H+ + A- Kb for A- = Kw / Ka for HA
Thus, Ka Kb = Kw for conjugate pairs.
A- + H2O HA + OH-
[HA] [OH-] [H+]Kb = ————— ———
[A-] [H+]
[HA] = ———— [OH-] [H+] [A-] [H+]
1 = —— Kw Ka
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Ka Kb and Kw – another way
HA + H2O (l) H3O+ (aq) + A– (aq) Ka of HA +) A– + H2O (l) OH–(aq) + HA (aq) Kb of A–
2 H2O (l) H3O+ (aq) + OH– (aq) Kw = Ka Kb
When you add two equations to get a third, what are the relationship between the K’s?
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Weak Bases
Bases react with water to produce hydroxide ion.
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pH of Basic Solutions
What is the pH of a 0.15 M solution of NH3?
[NH4+] [OH−]
[NH3]Kb = = 1.8 10−5
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
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NH3(aq) + H2O(l) NH4+ (aq) + OH−(aq)
Initial 0.15 0.00 0.00Change -x +x +x
Equilibrium 0.15 – x x x
1.8*10-5 = x2/(0.15 – x)Kb seems very small, therefore assume that 0.15>>x
X = 1.6*10-3
Relative error = {1.6*10-3/0.15}*100% = 1.1%, which is acceptable.[OH-] = 1.6*10-3 M, pOH = 2.79, therefore pH = 14-2.79 = 11.2
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Aniline is an organic chemical used in dye synthesis and is a weak base, with a Kb of 3.8 x 10-10. What is the pH of a 1.5 M ArNH2 solution?
ArNH2 (aq) + H2O(l) ArNH3+
(aq) + OH-(aq)
Kb = [ArNH2]
Concentration (M) ArNH2 H2O ArNH3+ OH-
Initial 1.5 ---- 0 0Change -x ---- +x +xEquilibrium 1.5 - x ---- x x
making the assumption: since Kb is small: 1.5 M - x = 1.5 M
[ArNH3+] [OH-]
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Substituting into the Kb expression and solving for x:
Kb = = = 3.8 x 10-10[ArNH3
+] [OH-]
[ArNH2]
(x)(x)
1.5 - x
x = 2.4 x 10-5 = [OH-] = [ArNH3+]
Calculating pH:
[H3O+] = = = 4.2 x 10-10Kw
[OH-]
1.0 x 10-14
2.4 x 10-5
pH = -log[H3O+] = - log (4.2 x 10-10) = 9.4Or calculate pOH = 4.6 pH = 14 – 4.6 = 9.4
x = 2.4 x 10-5 , RE = {2.4*10-5/1.5}*100% = 1.6*10-3%