Transcript Test 4 Review - Ralph C. Mahar

Test 4 Review
Advanced Chemistry
Equilibrium
mA + nB g sP + rQ
Keq = [P]s[Q]r
[A]m[B]n
Ksp = [A+][B-] for dissolving of a solid
AB g A+ + BKa = [H+][A-] for the dissociation of an acid
[HA]
Ka is small for a weak acid, large for a strong acid.
Never include solids or pure liquids in a Keq.
Solubility Product Constant
AB(s) D A+(aq) + B-(aq)
Ksp = [A+][B-]
At 25°C the solubility product constant for
strontium sulfate, SrSO4, is 7.6 x 10-7.
What is the concentration of Sr2+ at 25°C?
[Sr2+][SO42-] = 7.6 x 10-7
[Sr2+]=[SO42-]
[Sr2+] = 8.7 x 10-4M
The solubility product constant for strontium
fluoride is 7.9 x 10-10. What is the molar
solubility of SrF2 at 25°C?
SrF2 D Sr+2 + 2F[Sr2+][F-]2=7.9 x 10-10
[Sr2+]= x
[F-]= 2x
(x)(2x)2=7.9 x 10-10
x = 5.8 x 10-4M


Strong bases:
hydroxides of groups 1 & 2 (except Be)
Strong acids:
HCl, HBr, HI
HClO4, H2SO4, HNO3
Ionization constant, Ka,
for a weak acid
HA D H+ + AKa = [H+][A-]
[HA]

[CH 3 NH 3 ][ OH-]
Kb =
[CH 3 NH 2 ]
Ka ∙Kb= Kw = 1.00 x 10-14
What is the [H+] in 0.100M formic acid?
Ka for formic acid is 1.77 x 10-4
HCOOH D H+ + COOHSince this is a weak acid, [HCOOH] ͌ 0.100M
Ka = [H+][COOH-] = 1.77 x 10-4
[HCOOH]
Let x = [H+] = [COOH-]
x2 = 1.77 x 10-4
0.100
X = 4.21 x 10-3M
Percent ionization
[amount ionized]
[original acid]
What is the percent ionization of [H+] from
the previous problem?
[H+] = 4.21 x 10-3M, [HCOOH] = 0.100M
4.21 x 10-3 = 4.21%
0.100
hydrolysis
The reaction of a salt
with water to form an
acidic or basic solution.
 Example:
FeCl3 D Fe3+(aq) + Cl-(aq)
Fe3+ + 3OH- D Fe(OH)3
H2O D H+ + OH
shifts right, creating more H+
cation
anion
solution
Strong
base
Strong
base
Weak
base
Weak
base
Strong
acid
Weak
acid
Strong
acid
Weak
acid
Neutral
Basic
Acidic
Neutral
[H+] ∙ [OH-] = 10-14
pH + pOH = 14.0
pH = -log[H3O+]
[H3O+] = antilog(-pH)
Find the pH of a solution with [H3O+] of
9.85 x 10-8M.
pH = -log (9.85 x 10-8)
= 7.01
What is the [H3O+] in a solution with pH 7.01?
[H3O+] = antilog(-7.01)= 9.85 x 10-8M
What is the pH of the following solutions?
[H ] = 10 , pH = 1
0.1M HCl
[OH ]=10 , pOH=1, pH=13
0.1M NaOH
K = 4.4 x 10 =
x
0.1M H2CO3
0.1
+
-1
-
a
X = 2.1x10-4
-1
-7
2
pH = 3.68
Titration


Standard solution- one whose
concentration is known
Endpoint- the point at which equivalent
amounts of reactants are present.


M∙V = moles
MaVa=MbVb
Titration curves
Calculating pH of a buffer


What is the pH of a buffer that is 0.12M lactic
acid (HC3H5O3) and 0.10M sodium lactate?
Ka = 1.4 x 10-4
HC3H5O3
Initial
Change
equilibrium
0.12M
-x
.12-x

H+
C3H5O3-
0
0.10M
+x
+x
x
.10+x
Henderson-Hasselbalch

pH = pKa + log [base]
[acid]
= -log(1.4x10-4) + log .10
.12
= 3.85 + (-.079)
= 3.77
Conjugate base
of the acid
Common Ion (Buffers)
Calculate the pH of a 0.100M solution of
formic acid and 0.020M sodium formate.
Ka= 1.77 x 10-4
HCOOH g H+ + COOH0.100
x
0.020
Ka = [H+][COOH-] = .020x
X = 8.85 x 10 M
pH = 3.05
[HCOOH]
0.100
-4


Redox reactions involve a change in
oxidation number
Oxidation


Loss of electrons
Reduction

Gain electrons
3Cu2+ + 2Fe g 3Cu + 2Fe3+
copper gains 2 electrons
(reduced)
(oxidized)
iron loses 3 electrons
Reducing agent- is oxidized (iron)
Oxidizing agent- is reduced (copper)
Steps for balancing redox reactions
using the half-reaction method
Write ionic equation for half reactions
Balance chemically
1.
2.
a)
b)
c)
3.
4.
5.
6.
Balance non- O and H atoms
Add H2O to balance O’s
Add H+ to balance H’s (in a basic solution add OH-)
Balance electrically- add e-’s to the more + side
Check for balanced charges on both sides
Combine half reactions and cancel common items
Add spectator ions and balance
Activity Series
More active
Less active
lithium
potassium
magnesium
aluminum
zinc
iron
nickel
lead
HYDROGREN
copper
silver
platinum
gold
Oxidizes
easily
The metal must be above (more
active than) the ion for it to be a
spontaneous reaction.
Reduces
easily
Voltaic Cell
Anode


attracts anions
where oxidation occurs
Cathode


attracts cations
where reduction occurs
Salt bridge


connects the two half cells
contains a strong electrolyte
Two half cells connected by a salt bridge
Zn Zn+ Cu2+ Cu in shorthand
Reduction half
reactions
F2 is the
strongest
oxidizing agent
Li is the strongest
reducing agent
Reduction Potentials (E)



If E is positive, the reaction is
spontaneous.
If E is negative, the reverse reaction is
spontaneous.
Eo is the standard electrode potential




all ions are 1M and gases are 1 atm
The net Eo is the sum of the Eo of the half
reactions
The stronger oxidizing agent reduces.
Reverse the sign of the substance
oxidized.
What is the voltage produced from the
reaction of Zn metal with Cu2+ ions?
Zn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s)
Zn2+ + 2e- g Zn
-0.7628
Cu2+ + 2e- g Cu
0.3402
-(-0.7628) + 0.3402 = 1.103 volts
Will happen spontaneously
Zn2+ + Ni(s) g Zn(s) + Ni2+
-(-0.23) + (-0.7628) = -.53
Will not occur spontaneously
Faraday’s Law

Coulombs = amperes x seconds
1 C = 1amp·1sec

96,485 coulombs = 1 mole e-

What mass of copper will be
deposited by a current of 7.89
amps flowing for 1200 seconds?
Cu2+ + 2e- g Cu at the cathode
7.89A x 1200s x 1C x 1 mol e- = .0981 mol eA·s 96,485C

.0981 mol e- x 1 mol Cu x 63.5g Cu = 3.1g Cu
2 mol e- 1 mole Cu
Nernst Equation

Eo
Voltage under standard conditions
(1M solutions at 25°C and 101.3kPa)
At non-standard conditions, use Nernst equation
E = E° - 0.05916 log [products]
n
[reactants]
n = no. of electrons transferred
Coefficients in front of reactants or products are used as powers of their concentrations.