We now have an official Exam Procedure page. Check it out and know what to expect on Tuesday.
Coverage for Exam 3: 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. Remember to check out extra practice assignments (5 and 6) on canvas as well.
Length / Time for Exam 3: You should expect 25 multiple choice questions. The questions will have a range of point values from 2 points to 5 points. Remember that the point values are included with all questions. The exam is given during class time (9:30-10:45, 11-12:15) which is 75 minutes total which includes the handout time.
Concepts • Equations - 10 Acid/Base Equilibria
acid / base theory
acid = a proton donor
base = a proton acceptor
(Lowry-Bronsted definition)
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-][BH+]
[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...