ph = -log[hvandenbout.cm.utexas.edu/courses/ch302s08/files/dvb020708a.pdfph log scale. useful when...
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![Page 1: pH = -log[Hvandenbout.cm.utexas.edu/courses/ch302s08/files/dvb020708a.pdfpH Log scale. Useful when dealing with very small or very large number (big ranges of numbers) every "pH" unit](https://reader033.vdocuments.us/reader033/viewer/2022052010/6020024d54d7d7437a0f365b/html5/thumbnails/1.jpg)
Principles of Chemistry II © Vanden Bout
pH
Log scale. Useful when dealing with very small
or very large number (big ranges of numbers)every "pH" unit is 10x larger or smaller [H+]
pH = -log[H+]
pH= 7[H+] =10-7
pH= 2[H+] =10-2
pH= 13[H+] =10-13
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Principles of Chemistry II © Vanden Bout
linear [H+]1M0 M
10-1M pH =1
10-2 M pH =2
pH 3-14
log [H+]
1M10-110-210-310-410-510-610-610-610-710-8
pH=8 pH=0
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Principles of Chemistry II © Vanden Bout
Strong Acids and Bases
"Strong" means one thing
The substance dissociates 100% in water
Strong Acid
HCl(aq) H+(aq) + Cl-(aq)
Ka =[H+][Cl-]
[HCl]≈∞
Strong Electrolyte
NaCl(s) Na+(aq) + Cl-(aq)
Ksp = [Na+][Cl-]≈∞
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Principles of Chemistry II © Vanden Bout
Strong Acids
HCl HydrochloricHBr HydrobromicHI Hydroiodic
HClO4 PerchloricHClO3 Chloric
H2SO4 SulfuricHNO3 Nitric
All Dissociate 100%
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Principles of Chemistry II © Vanden Bout
What is the pH of a 0.1 M solution of Nitric Acid
HNO3(aq) H+(aq) + NO3-(aq)
0.1 M acid makes a solution with [H+] = 0.1M
pH =-log(0.1) = 1
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Principles of Chemistry II © Vanden Bout
What is the pH of a 0.5M solution of HBr?
A. 0.5
B. 1
C. 0.3 D. 0
E. 12
[H+] =0.5 0<pH<1
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Principles of Chemistry II © Vanden Bout
We can ignore the conjugate base of a strong acid
HCl(aq) H+(aq) + Cl-(aq)
Ka =[H+][Cl-]
[HCl]≈∞
equilibrium constant is so large,even if we add Cl- the shift back to
HCl will be negligible
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Principles of Chemistry II © Vanden Bout
For this reaction which has a higher entropy?
H2O(l) H+(aq) + OH-(aq)
A. the products
B. the reactants
C. they are the same
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Principles of Chemistry II © Vanden Bout
For this reaction which has a lower enthalpy?
H2O(l) H+(aq) + OH-(aq)
A. the products
B. the reactants
C. they are the same
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Principles of Chemistry II © Vanden Bout
For this reaction which has a lower free energy?
H2O(l) H+(aq) + OH-(aq)
A. the products
B. the reactants
C. they are the same
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Principles of Chemistry II © Vanden Bout
Liquid Water will spontaneously dissociate to a small extent
H2O(l) H+(aq) + OH-(aq)
K = [H+][OH-]
1
Kw =[H+][OH-] = 10-14
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Principles of Chemistry II © Vanden Bout
Pure Water
H+ OH-
I
C
E
O
+x
+x
O
+x
+x
Kw = 10-14 = [H+][OH-] = (x)(x)
x = 10-7 [H+]=[OH-]=10-7
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Principles of Chemistry II © Vanden Bout
pH of pure water at 25°C
x = 10-7 [H+]=[OH-]=10-7
pH = -log[H+]=-log(10-7) = 7
Neutral
[H+]=[OH-]
at 25°C
pH = 7pOH = 7
Acidic
[H+]>[OH-]
at 25°C
pH < 7pOH > 7
[H+]<[OH-]
at 25°C
pH > 7pOH > 7
Basic
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Principles of Chemistry II © Vanden Bout
H2O(l) H+(aq) + OH-(aq)
This reaction is endothermic.Given that information what do you think
the pH is for pure water at 60°C?
A. 6.5
B. 7
C. 7.5
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Principles of Chemistry II © Vanden Bout
If pure water has a pH = 6.5 at 60°C is it Acidic?
A. Yes
B. No[H+]=[OH-] its neutral Kw = 9x10-14
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Principles of Chemistry II © Vanden Bout
Let's look at three possible solutions
Weak AcidConjugate Base of the Weak Acid
Weak Acid + Conjugate Base
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Principles of Chemistry II © Vanden Bout
Weak Acid
HA(aq) H+(aq) + A-(aq)
HA H+ A-
I
C
E
C
-x
C-x
O
+x
+x
Ka =[H+][A-]
[HA]
O
+x
+x
(x)(x)
C-x= x ~ √KaC
really 10-7
assuming x << C
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Principles of Chemistry II © Vanden Bout
A-(aq) +H2O(l) HA(aq) + OH-(aq)
Weak Base
Kb =[HA][OH-]
[A-]
identical result as before (same assumptions)
[OH-] = √KbC
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Principles of Chemistry II © Vanden Bout
Buffer Both HA and A-
HA(aq) H+(aq) + A-(aq)
HA H+ A-
I
C
E
CHA
-x
CHA-x
O
+x
+x
Ka =[H+][A-]
[HA]
CA-
+x
CA- + x
(x)(CA-+x)
CHA-x=
really 10-7
assuming x << C=(x)(CA)
CHA