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Review for Midterm 3 Chem 2
Chapter 15 (Ksp)
To begin:
 Solubility (S)
 Ksp (Solubility product constant)
In a reaction: (Cation)n (Anion)m(s)  n Cation(aq) + m Anion(aq):

n

Ksp or Q = [Cation] [Anion]

m

At a certain temperature: All salts have a given constant Ksp
If:
Q= Ksp : Solution is saturated
Q> Ksp : A solid precipitate is formed in the reaction
Q< Ksp : Solution is considered unsaturated
Solubility – Measure of how to make saturated solution
Either gram solubility (how many grams in a given volume) or MOLAR solubility (S, how many
moles per liter of saturated solution)
Relationships between S and Ksp
1
...
Write an expression to help solve for Ksp
3
...
Substitute the concentrations into the Ksp expression
Example:
Say you are given AgCl (Ksp value = 1
...
8*10^(-10) = S2 so, S=1
...

 Effect of pH (acidic solutions): as you add acid the acidic solution gets neutralized (X- is
removed from the solution)
...
If delta H is positive heat is taken away from the surroundings and it is
considered exothermic
2
...
If S > 0, then there is an increase in randomness
...
If S < 0, then there is a decrease in randomness
...

1
...
If G < 0, then it is spontaneous and does not need an external driving force

Standard State: Is a state that is defined when a pure solid, liquid, or gas is at 1 atmosphere
pressure and 1M concentration
...
For review purposes, the
law of conservation is that: energy cannot be created or destroyed, but only changed from
one form into another or transferred from one object to another
...
Gases have higher entropy than liquids, which have a higher entropy than solids, so g>l>s
2
...
Between substances in the same phase: If there are more atoms and/or the substance is
heavier, the greater the entropy is
4
...
Increase in temperature = increase in entropy, decrease in pressure = increase in entropy

Relationship between Entropy, Enthalpy, and Free Energy
H

S

G

+

+

positive at low temp, negative at high temperature

+

-

always positive

-

+

always negative

-

-

negative at low temp, positive at high temperature

K
>>1

G
very negative

>1

negative

1
<1

0
positive

<<1

very positive

Equations:
 Hreaction

[Hf(products)] – [Hf(reactants)]

=

Hf of elements in natural state = 0
 Sreaction

[Sf(products)] – [Sf(reactants)]

=

 Greaction
=
[Gf(products)] – [Gf(reactants)]
Gf of elements in natural state = 0
For Entropy
 H = -TS(surroundings)
 S(universe) = S(reaction) + S(surroundings)
 G = -TS(universe)
For Delta G:
 G = H - TS

(T in K; get ALL quantities in either Joules or Kilojoules)

 G = -RTlnK

(R = 8
...
Oxidation and Reduction
1
...
Reduction
 Substance gains electrons: Cl2 + 2e- → 2Cl Substance loses Oxygen: HgO → Hg
 Substance gains Hydrogen: CH2CH2 → CH3CH3
B
...
Ionic Redox Reactions
a
...
Balancing:
Na + Cl2 → Na+ + 2 Cl0

0

1

-2

2Na+ Cl2 → Na+ + 2 Cl0
Total charge:

0

2

0

-2
0

For non-ionic redox reactions:


There has to be an equal number of electrons lost and gained


1
...

3
...
Non-ionic Redox Reaction
 Identification: CH3CH2OH → CH3CHO
 For easier identification, we can write it as C2H6O → C2H4O
 For finding oxidation numbers for C, H, and O in C2H6O:
1
...
First one we can assign oxidation number to is Hydrogen: H = +1
3
...
Since we cannot assign the rest of the oxidation numbers from here,
we can assign Oxygen because it is closest to Fluorine (O = -2)
5
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Now we can get oxidation number for Carbon
7
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2C = -4
9
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Let’s assign oxidation numbers
a) KClO4 : K = +1 O = -2
(-2 x 4) + (1) + (Cl) = 0; Cl = +7
b) Mg(s) = 0
c) H = +1 Br = -1 HBr = -1; HBr = 0
Products
a) Cl2 (g) = 0
b) K = +1 Br = -1; KBr = 0
c) Mg = +2 Br = -1; MgBr2 = 0
d) H2O = 0
2
...
Balance electron transfer
 Balance redoxed atoms using appropriate coefficients on whole species
 Chlorine needs to gain 7 electrons
 Mg needs to lose 2 electrons
*In order to have the same charges on each side, Chlorine must gain the same amount of
electrons that Magnesium needs to lose*




Find the least common multiple (LCM) between 2 and 7, which is 14
So we add the coefficient 2 to KClO4, 7 to Mg(s), and 7 to MgBr2 to balance out the
charges
This gives us
a) 2KClO4 (aq) + 7 Mg (s) + HBr (aq) → Cl2 (g) + KBr (aq) + MgBr2 (aq) + H2O
b) Now we need to balance the rest, but we cannot mess with KClO4, Mg, or Cl2
because if we do, we need to change the charge accordingly

c) 2 KClO4 (aq) + 7 Mg (s) + 16 HBr (aq) → Cl2 (g) + 2 KBr (aq) + 7 MgBr2 (aq) + 8
H2O
Simple redox reactions (half equations)
1
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0 V)
Or

2 H2O + 2e-  H2 + 2 OH-

(hydrogen gas evolved, basic solution)

(will happen if metal is group 1 or group 2 or Al)

ANODE

either Anion is oxidized to element (if E less negative than about –1

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