Chem Chem 103 103 Lecture 2b Acids and Bases3 Last time: Last time: Today: Today: 1. Acids and Bases 2: a) self ionization of water b) Relative strengths of acids/conjugates c) K a and K b equilibria d) pH scale 1. Summary of previous lecture 2. Calculating pH: 5 scenarios 3. Acid base titration No office hours this Friday I will not be holding office hours this coming Friday. Ch 103 Seating arrangement In preparation for group work and midterm exams: Seating arrangement enforced for extra credit work. (if in wrong row, no extra credit) If your last name starts with please sit in row A, B or C A (front) D, E, F, G, H, I, J B K, L, M C N, O, P, Q, R, Sa D Si, T, U, V, W, X, Y, Z E Summary (so far…) [H 3 O + ] [ OH - ] = 1.0 x 10 -14 pH = -log[H 3 O + ] and pOH = -log[OH - ] pH + pOH = 14.00 K a = [H + ][A - ]/[HA] K b = [HA][OH - ]/[A - ] K a K b = 1.0 x 10 -14 [H + ] = 10 -pH [OH - ] = 10 -pOH pK a + pK b = 14.00 Sample problem A solution has a pH of 2.10. Determine the following: a) [H + ]=? [H + ] = 10 -2.10 = 7.9x10 -3 M b) pOH =? pOH = 14.00 - pH = 14.00 -2.10 = 11.90 c) [OH - ] = ? [OH - ] = 10 -pOH = 10 -11.90 = 1.3 x 10 -12 M
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Chem 103 - Cal State LA · 2013-09-27 · Chem 103 Lecture 2b Acids and Bases3 Last time: Today: 1.Acids and Bases 2: a)self ionization of water b)Relative strengths of acids/conjugates
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Chem Chem 103103Lecture 2b
Acids and Bases3
Last time:Last time:
Today:Today:
1. Acids and Bases 2:a) self ionization of waterb) Relative strengths of acids/conjugatesc) Ka and Kb equilibriad) pH scale
1. Summary of previous lecture2. Calculating pH: 5 scenarios3. Acid base titration
No office hours this FridayI will not be holding office hours this coming Friday.
Ch 103 Seating arrangementIn preparation for group work and midterm exams:
Seating arrangement enforced for extra credit work.(if in wrong row, no extra credit)
If your last name starts with please sit in row
A, B or C A (front)
D, E, F, G, H, I, J B
K, L, M C
N, O, P, Q, R, Sa D
Si, T, U, V, W, X, Y, Z E
Summary (so far…)[H3O+] [ OH-] = 1.0 x 10-14
pH = -log[H3O+] and pOH = -log[OH-]
pH + pOH = 14.00
Ka = [H+][A-]/[HA] Kb = [HA][OH-]/[A-]
KaKb = 1.0 x 10-14
[H+] = 10-pH [OH-] = 10-pOH
pKa + pKb = 14.00
Sample problem
A solution has a pH of 2.10. Determine the following:
a) [H+]=?
[H+] = 10-2.10 = 7.9x10-3 M
b) pOH =?
pOH = 14.00 - pH = 14.00 -2.10 = 11.90
c) [OH-] = ?
[OH-] = 10-pOH = 10-11.90 = 1.3 x 10-12 M
Calculating pH of a solution: 5Calculating pH of a solution: 5scenariosscenarios