exam 3
11/7
exam 3
11/7
There was an "Exam 3 - Bonus" given on Canvas. Here are the Solutions to Exam 3 - Bonus.
The exam will be on Wednesday, 11/7, from 7:30 pm to 9:00 pm in our regular class room, BUR 106.
What we provide on Exams We will provide all students with:
Note that the periodic table handout is available on the gchem site in the appendix under "Exam Preparation". Here is a direct link to the Periodic Table Handout for Exam 3.
Coverage: Exam 3 covers all the material that was covered on LE's 18-23 and HW05 and HW06. The exam covers the rest of Chapter 10 (buffers, titration curves, and indicators) PLUS Chapter 8 on solubility equilibria.
Questions: The exam has 25 multiple choice questions. The questions will have an equal weight of 4 points each. The exam is fairly calculation heavy - so bring your calculator and come prepared. We will only grade you by what is bubbled in on the answer sheet. We will not look at your exam copy for answers, nor consider them in any way. Bubble carefully and correctly.
You now will want to use the entire page for Exam 3. All the information is relevant.
Concepts • Equations - 10 Acid/Base Equilibria
acid / base theory
(Lowry-Bronsted definition)
acid = a proton donor
base = a proton acceptor
buffer = a solution that resists pH change
water
Kw = [H+][OH-]
pH = -log[H+][H+] = 10-pH
pOH = -log[OH-][OH-] = 10-pOH
weak acids / weak bases
acid reaction:
HA(aq) ⇌ H+(aq) + A-(aq)
Ka =
[H+][A-]
[HA]
base reaction:
B(aq) (+ H2O) ⇌ OH-(aq) + BH+(aq)
Kb =
[OH-][B+]
[B]
conjugate pairs: Kw = KaKb
buffer composition
a buffer consists of a weak acid AND its conjugate base, or a weak base AND its conjugate acid. BOTH conjugates must be present. You cannot have a buffer with any strong acid and its conjugate or strong base and its conjugate. Buffers MUST come from weak acids and bases.
Henderson Hasselbalch
pH = pKa + log
[base]
[acid]
Concepts • Equations - 8 Solubility Equilibria
For a saturated solution of the salt MxAy
MxAy(s) = xM+y(aq) + yA-x(aq)
Ksp = [M+y]x · [A-x]y
None of this will be printed on the exam. You need to KNOW this on your own.
| salt ratio | salt formula | aqueous cation | aqueous anion | equilibrium formula | Ksp(x) | molar solubility | ||
|---|---|---|---|---|---|---|---|---|
| 1:1 | MX(s) | ⇌ | M+(aq) | + | X–(aq) | Ksp = [M+][X–] | \(K_{\rm sp} = x^2\) | \(x = \sqrt{K_{\rm sp}}\) |
| 1:2 | MX2(s) | ⇌ | M2+(aq) | + | 2 X–(aq) | Ksp = [M2+][X–]2 | \(K_{\rm sp} = 4x^3\) | \(x = \root 3 \of {K_{\rm sp}\over 4}\) |
| 1:3 | MX3(s) | ⇌ | M3+(aq) | + | 3 X–(aq) | Ksp = [M3+][X–]3 | \(K_{\rm sp} = 27x^4\) | \(x = \root 4 \of {K_{\rm sp}\over 27}\) |
| 2:3 | M2X3(s) | ⇌ | 2 M3+(aq) | + | 3 X2–(aq) | Ksp = [M3+]2[X2–]3 | \(K_{\rm sp} = 108x^5\) | \(x = \root 5 \of {K_{\rm sp}\over 108}\) |
forward\(\rightarrow\) \(Q < K\) which means more solid will dissolve if there is any or nothing happens because the solution is undersaturated.
equil\(\rightleftharpoons\)ibrium \(Q = K\) which means the solution concentrations will hold constant and there is no net change.
\(\leftarrow\)reverse \(Q > K\) means the solution is oversaturatured and a precipitation will occur until the concentrations drop to saturation level.
Note that mainly outcomes 10-16 will be on the exam for the Acid/Base chapter. However, you still need to know many of the concepts and definitions from outcomes 1-9.
Students will be able to...
Students will be able to...