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Lecture

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Department
Chemistry
Course Code
CHM135H1
Professor
Kris Quinlan

Page:
of 3
LECTURE 9 – Acids and Bases Equilibria
Slide No. Notes
Last slide
of last
lecture
-k small – not many ions
-all types of equilibrium equations, all have similar behaviour
starting at a point not at equilibrium, then tending to equilibrium
1-H2S is the acid
-NH3 is the base
-Conjugate base is the HS, conjugate acid is the NH4
-Conjugate acid-base pairs: the H2S and HS, and the NH3 and NH4
-Each acid-base pair related by one proton
-Water can both give proton, and receive proton
2-from left to right: acid base
-H3O+ is also the acid
-Acid weaker than H3O+ therefore equilibrium shifts to products side
-Strong acid wants to give protons, strong bases want to receive
protons
-Strong acids are more reactive
-The stronger the acid, the weaker the conjugate base
3-Ans: A
-HNO2 is the stronger acid
-Direction of reaction is towards the products’ side
-Cannot have stronger A or B and weaker A or B on one side
-Must be stronger acid and stronger base reacting to form weaker acid
and weaker base
4-the ion-product constant for water holds for any aqueous solution
-if either of [H3O+] or [OH-] increases, the other decreases
5-ACIDS section
-Water acts as base to accept proton
-The stronger the acid, the larger the Ka value
-MEMORIZE all six acids
-BASES section
-Water acts as acid
-Ka and Kb exclusive to solutions of water
7- Ans: B
9-Ans: C
-Ans2: C
-***weak acids don’t necessarily have strong bases
10 -Ans1: B because strong acids dissociate more than weak acids
-Ans2: A
-Ans3: B
11 -can have negative pH
-large [H3O+] – small pH – large pOH – small [OH-]
12 -know HCl is a strong acid
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-dissociates completely so that concentrations of Cl- and H3O+ are
both 0.050M
-pH = -log[H3O+] = 1.30
13 -HNO2 + H2O -> NO2- + H3O+
-Note that water is not included in calculations because of liquid state
-I: 0.50 M -> O M + O M
-C: -x -> + x + +x
-E: 0.50-X -> x + x
-Ka = [NO2-][H3O+]/[HNO2] = 7.1x10^-4 = x^2/ (0.50-x)
- Solve for x using quadratic equation – x = 0.018M
-double check
-pH = -log[0.018M] = 1.73
14 -to continue from previous example
-percent dissociation = 0.018M/0.50M times 100% = 3.6% of acid
reacts, the rest remain intact
15 -pyridine can accept protons because of lone pair on nitrogen
-C5H5N + water -> OH- + C5H6N+
-I: 0.10M 0M 0M
-C: -x +x +x
-E: 0.10M-x x x
-Kb = [OH-][C5H6N+]/[C5H6N] = 1.7x10^-9 = x^2/ (0.10-x)
-The smaller the Kb, the more confident you can be about making the
assumption that x is much, much smaller than 0.10M
-Solve for x
-X = 1.3 x 10^-5 M
-Assumption correct
- [OH-] = 1.3x10^-5 M
-pOH = -log[1.3x10^-5) = 4.88
-pH = 14-pOH = 9.12
16 - more difficult to remove proton from increasingly negative molecules
17 -need to consider what is really in solution when salt dissolves in
water
-Cl- and Na+ both do not effect pH of solution because neither can
accept or donate protons
19 - conjugate base of weak acid – can accept proton from water to form some
OH- therefore BASIC solution
20 -similarly, when H3O+ is produced, solution becomes ACIDIC
-Al3+ to Al2+ - proton lost
21 -NH4F(s) -> NH4+ + F-
-So take the conjugate acid and base and react them with water
-NH4 reaction with water produces H3O+
-F- reaction with water products OH-
-What decides acidity or basicity is how the concentrations of each
compare to each other
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-Given Ka of HF – use to find Kb of F- = Kw/Ka of HF = 1.5x10^-11
-Given Kb of NH3 – use to find Ka of NH4+ = Kw/ Kb of NH3 =
5.7x10^-10
-Therefore Ka > Kb therefore more H3O+ therefore ACIDIC
21 -NaHCOO -> Na+ + HCOO-
-Na+ has no effect on pH
-HCOO- reaction with water produces OH- therefore B
-To actually quantitatively calculate the Kb:
-HCOO- plus water -> HCOOH + OH-
-I: 0.24M 0 0
-C: -x x x
-E: 0.24-x x x
-Kb = [HCOOH][OH-]/[HCOO-] = Kw/Ka
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