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CHEM160 General Chemistry II Lecture Presentation Aqueous Equilibria: Acids and Bases. Chapter 16. Dr. Daniel Autrey. Why study acids and bases? . Acids and bases are common in the everyday world as well as in the lab. Some common acidic products
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Chapter 16
Dr. Daniel Autrey
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
acid conj. base + H+
base + H+ conj. acid
Chapter 16
HX + H2O H3O+ + X
HX + H2O H3O+ + X
Chapter 16
NaOH Na+ + OH
B + H2O BH+ + OH
Chapter 16
Strongest HClO4 ClO4 Weakest
Acids H2SO4 HSO4 Bases
HCl Cl
HNO3 NO3
H3O+ H2O
HSO4 SO42
HNO2 NO2
HClO ClO
Weakest NH4+ NH3 Strongest
Acids H2O OH Bases
OH O2
H2 H
Strongest acid that can exist in H2O
Strongest base that can exist in H2O
Chapter 16
or
Chapter 16
Kw = [H3O+][OH]
so
Kw = 1 x 1014 (25 °C)
Chapter 16
Chapter 16
(ans.: [H3O+] = 0.0035 M, [OH] = 2.9 x 1012 M)
(b) 0.22 M NaOH
(ans.: [OH] = 0.22 M, [H3O+] = 4.5 x 1014 M)
Chapter 16
Chapter 16
Chapter 16
logKw = log([H3O+][OH])
logKw = log[H3O+]  log[OH]
pKw = pH + pOH
14.00 = pH + pOH ( at 25 °C)
Chapter 16
(ans.: pH = 3.55, pOH = 10.45)
Chapter 16
log
inv
ln
yx
cos
sin
tan
Exp
8
7
9
x
5
4
6
2
1
3
+
+/
0
·

pH Calculations with CalculatorsChapter 16
Chapter 16
log
inv
ln
yx
cos
sin
tan
Exp
8
7
9
x
5
4
6
2
1
3
+
+/
0
·

Calculating [H3O+] from pHChapter 16
[H3O+] [OH] pH pOH
neutral = 107 M = 107 M = 7.0 = 7.0
acidic > 107 M < 107 M < 7.0 > 7.0
basic < 107 M > 107 M > 7.0 < 7.0
Chapter 16
(ans.: [H3O+] = 0.29 M, [OH] = 3.5 x 1014 M, pH = 0.54, pOH = 13.46)
Chapter 16
Chapter 16
Chapter 16
(ans.: [OH] = 0.11 M, [H3O+] = 9.0 x 1014 M, pOH = 0.96, pH = 13.04)
Chapter 16
Chapter 16
HA + H2O <=> H3O+ + A
Ka = [H3O+][A]/[HA]
(Note that these are all equilibrium concentrations.)
Chapter 16
(Any equilibrium constant can be used to calculate equilibrium concentrations for solution species.)
Chapter 16
Chapter 16
(ans.: [HOAc] = 0.497 M, [OAc] = [H3O+]
= 0.0030 M, [OH] = 3.3 x 1012 M, 0.60%)
Chapter 16
H2A + H2O <=> H3O+ + HA Ka1 = [H3O+][HA]/[H2A]
HA + H2O <=> H3O+ + A2 Ka2 = [H3O+][A2]/[HA]
Chapter 16
(ans.: [H2CO3] = 0.250 M, [HCO3] = [H3O+] = 0.00033 M,
[CO32] = 5.6 x 1011 M], [OH] = 3.0 x 1011 M, pH = 3.48)
Chapter 16
B + H2O <=> BH+ + OH
Kb = [BH+][OH]/[B]
(Note that these are all equilibrium concentrations.)
Chapter 16
Chapter 16
NH4+ + H2O <=> NH3 + H3O+
weak acid
F + H2O <=> HF + OH
weak base
Chapter 16
Kw = KaKb
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Chapter 16
Al3+ + 6H2O Al(H2O)63+
Lewis acid Lewis base
Chapter 16
e shift
Chapter 16
Al(H2O)63+ + H2O <=> Al(H2O)5(OH)2+ + H3O+
Zn2+, Fe3+, Co2+, Cr3+, Ni2+, Cu2+
Chapter 16
Chapter 16