Download UILChemistryProblemsPart2

index
UIL Science
Chemistry Problems
index
Index
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Oxidation Rules
Oxidation Number/State Prob D/R
Neutralization Reaction Info
Neutralization Rxn Problems D/R/S
Equilibrium Constant Kc Info
Equilibrium Constant Problems R/S
Weak Acid/ Weak Base Info
Weak Acid/Base Problems D/R/S
Light & Energy – Electrons Info
Light & Energy – Electrons Problems
Solubility & Ksp Info D/R/S
Solubility & Ksp Problems D/R/S
•
•
Reaction Rates Info
Reaction Rates Problems
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Assigning Oxidation Numbers Rules
Rules
1. Uncombined elements =0
2. Sum of oxidation numbers
In a compound = 0
In an ion = charge of the ion
3.
Group 1 = +1
Group 2 = +2
Group 13 = +3
4.
F = -1
H = +1
index
Assigning Oxidation Numbers
Rules cont’d
5. O = - 2 ( except in peroxides O = -1, superoxides: O = 1/2 )
6. In binary compounds only (electrically neutral)
Group 17 = -1
Group 16 = -2
Group 15 = -3
(Follow the rules in order always restart at rule 1)
Shortcut: H = +1, sum = charge, O = -2 will get you the answer in
most problems.
index
Oxidation Number/State Problems 1
D1 2008
21. What is the oxidation number assigned to the phosphorous atom
in sodium phosphate?
a)
b)
c)
d)
e)
District 1 2008
0
-3
+7
+5
+1
index
Oxidation Number/State Problems 1
21. What is the oxidation number assigned to the phosphorous atom
in sodium phosphate?
1. Write the formula of the compound or just phosphate
Na+ PO43Na3PO4
2. Use the oxidation rules. (apply first rule that applies, always
restart at rule 1)
-8
-2
Na3 P O4
Rule 5: oxygen is -2 (4 oxygens = -8)
Rule 2: sum of poly. Ion = charge. Phosphorous
Na3 P O4 plus -8 must equal -3 therefore P = +5
(1X – 8 = -3 solve for X)
+5 -8
+5 -2
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Oxidation Number/State Problem 2
D2 2008
21. The oxidation state of chromium in potassium dichromate is
_____.
a)
b)
c)
d)
e)
3.5
2.5
12
6
5
index
Oxidation Number/State Problem 2
D2 2008
21. The oxidation state of chromium in potassium dichromate is
_____.
-2
K2Cr2O7
Step1: K 2Cr2 O7
-14
-2
Step2 : K 2Cr2 O7
+12 -14
-2
Step3 : K 2 Cr2 O7
+12 -14
+6 -2
Step4 : K 2 Cr2 O7
index
Neutralization Reactions Info
• Acid + base = ionic compound + water (HOH)
• H from acid combines with OH from base to make water. The
anion (-) from the acid combines with cation (+) from the base
to make the ionic compound.
• Molarity (M) = #moles/liters of solution
• pH = -log [H+]
• pOH = - log [OH]
[H+] = 10-pH
M=
n
Vsolution
n = MV
[OH-] = 10-pOH
• pH + pOH = 14
• If the concentration of the H+ or OH- is greater than 1x 10-6 M
we can ignore the contribution by water.
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Neutralization Info
• Strong acids = HCl (hydrochloric), HBr (hydrobromic), HI
(hydroiodic), HNO3 (nitric), H2SO4 (1st H only) (sulfuric),
HClO4 (perchloric).
• Strong bases = (group 1 hydroxides and Ca, Ba, & Sr) NaOH,
KOH, LiOH, Ca(OH)2, Ba(OH)2, Sr(OH)2
Dilution equation: C1V1=C2V2 where C = concentration in
molarity (M), molality (m), or % and V = volume
index
Naming Acids
Steps for Naming Acids:
1. Determine the ionic name.
2. Using the ending of the negative ion determine
which prefix/suffix to use.
• -ide = hydro- -ic acid
• -ate =
-ic acid
• -ite =
-ous acid
3. add the suffix and prefix to the root of the anion
name (chlor, fluor, brom, nitr, sulfur, phosphor
4. If ionic name had Per- or hypo- add it to the acid
name.
index
Acids ending in -ide
1. “-ide” ending is changed to “hydro- -ic”
Formula
HCl
HBr
HF
-ide
Hydrogen
Chloride
Hydrogen
bromide
Hydrogen
fluoride
Hydro- -ic
Hydrochloric acid
Hydrobromic acid
Hydrofluoric acid
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Acids ending in -ate
1. “-ate” ending is changed to “-ic” acid
Formula
HClO3
HNO3
H2CO3
H2SO4
-ate
Hydrogen
chlorate
Hydrogen
nitrate
Hydrogen
carbonate
Hydrogen
sulfate
-ic
Chloric acid
Nitric acid
Carbonic acid
Sulfuric acid
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Acids ending in -ate
1. “-ate” ending is changed to “-ic” acid
Formula
H3PO4
CH3COOH
-ate
Hydrogen
phosphate
Hydrogen
acetate
HC2H3O2
HClO4
-ic
Phosphoric acid
Acetic acid
Acetic acid
Hydrogen
perchlorate
Perchloric acid
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Acids ending in -ite
1. “-ite” ending is changed to “-ous” acid
Formula
HNO2
HClO2
H2SO3
HClO
-ite
Hydrogen
nitrite
Hydrogen
chlorite
Hydrogen
sulfite
Hydrogen
hypochlorite
-ous
Nitrous acid
Chlorous acid
Sulfurous acid
Hypochlorous acid
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Neutralization Reaction 1
D1 2008
29. 15.0 mL of a 0.250 M H2SO4 solution exactly neutralizes 20.0
mL of an NaOH solution. What is the molarity of the NaOH
solution?
a)
b)
c)
d)
e)
0.667 M
0.375 M
0.188 M
0.0938 M
0.250 M
District 1 2008 Q29
index
Neutralization Reaction 1
15.0 mL of a 0.250 M H2SO4 solution exactly neutralizes 20.0
mL of an NaOH solution. What is the molarity of the NaOH
solution?
1. Write the reaction
2. Balance it
3. Calculate the number of moles of acid and convert mL to L,
then use the coefficients of the reaction to convert to moles of
NaOH, divide by volume and covert to L
0.250molH 2 SO4 15.0mL 1l 2molNaOH
1000mL
=0.375MNaOH
l
1000ml 1molH 2 SO4 20.0mL
1l
Shortcut: calculate moles of acid, double it since it will need two
OH to neutralize the two H, calculate molarity of NaOH
index
Neutralization Reaction 2
D1 2008
35. Calculate the pH of a solution prepared by adding
80.0 mL of 0.100 M NaOH to 100 mL of 0,100 M
HNO3 solution.
a)
b)
c)
d)
e)
2.24
2.08
2.16
2.02
1.95
District 1 2008 Q35
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Neutralization Reaction 2
Calculate the pH of a solution prepared by adding 80.0 mL of
0.100 M NaOH to 100 mL of 0.100 M HNO3 solution.
1. This a limiting reagent problem where you need to calculate
2.
the amount of excess reactant. 0.100molNaOH 1L 80.0mL
n=
= 0.00800 mol NaOH
L
1000mL
Write the balanced equation
0.100molHNO
1L 100.mL
n=
= 0.0100 mol HNO
L
1000mL
Calculate the moles of each.
Determine which one is limiting by taking the number of
moles and dividing by the coefficient. (NaOH is Limiting
reagent)
Use ICE (Initial, change, equilibrium) method to solve for
excess acid.
Since nitric acid is strong [HNO3] = [H+]
Calculate molarity then pH
3
3.
4.
5.
6.
7.
3
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Neutralization Reaction 2
Calculate the pH of a solution prepared by adding 80.0 mL of
0.100 M NaOH to 100 mL of 0.100 M HNO3 solution.
I
0.00800 mol
C
-0.00800
E
0
M=
0.0100 mol
0
-0.00800
+ 0.00800
0.00200
0.00800
0.00200molH +
0
+0.00800
0.00800
1000mL
= 0.011111M [H + ]
180mL 1L
pH = -log 0.00111111 = 1.9542
index
Neutralization Reactions 3
D2 2008
29. One millimole of HCl would neutralize exactly _______
millimole(s) of barium hydroxide?
a)
b)
c)
d)
e)
½
2
3
1
3/2
index
Neutralization Reactions 3
D2 2008
29. One millimole of HCl would neutralize exactly _______
millimole(s) of barium hydroxide?
Write the formula for barium hydroxide. Ba2+ OH- = Ba(OH)2
2. Write the balanced chem. reaction
3. Covert moles of HCl to moles of Ba(OH)2 using coefficients of the
reaction.
1.
1 mmol HCl 1 mol Ba(OH)2
= 1/2 mol Ba(OH)2
2 mol HCl
Shortcut: Since only 1H and 2OH’s. 1 H will only neutralize half of it.
index
Neutralization Reaction 4
D2 2008
35. What is the pH at the point in a titration at which 20.0
mL of 1.00 M KOH has been added to 25.0 mL of 1.00 M
HBr?
a) 0.95
b) 2.71
1.22
d) 1.67
e) 3.84
c)
index
Neutralization Reaction 4
35. What is the pH at the point in a titration at which 20.0 mL of
1.00 M KOH has been added to 25.0 mL of 1.00 M HBr?
1. Write the balanced equation.
n=
1.00molKOH 1L 20.0mL
= 0.0200 mol KOH
L
1000mL
n=
1.00molHBr 1L 25.0mL
= 0.0250 mol HBr
L
1000mL
2. Calculate moles of each.
3. Determine limiting reagent. (KOH is LR)
4. Use ICE to determine amount of excess reactant.
5. Calculate molarity and pH.
index
Neutralization Reaction 4
35. What is the pH at the point in a titration at which 20.0 mL of
1.00 M KOH has been added to 25.0 mL of 1.00 M HBr?
n=
1.00molKOH 1L 20.0mL
= 0.0200 mol KOH
L
1000mL
n=
1.00molHBr 1L 25.0mL
= 0.0250 mol HBr
L
1000mL
I
0.0250 mol
0.0200 mol
C
- 0.0200 mol
- 0.0200 mol
E
0.0050 mol
M=
0.0050molHBr
0
0
+ 0.0200mol
+0.0200 mol
0.0200
0.0200
1000mL
= 0.1111111M
45.0mL 1L
Since HBr is strong then [HBr ] = [ H + ]
pH = -log[0.111111] = 0.954242
0
index
Neutralization Reaction 5
D1 2009
28. What is [Cl-] in the solution formed by mixing 50 mL of 0.20
M hydrochloric acid (aq) with 50 mL of 0.30 M sodium
hydroxide (aq)?
a)
b)
c)
d)
e)
0.25 M
0.30 M
0.10 M
0.20 M
0.50 M
index
Neutralization Reaction 5
28. What is [Cl-] in the solution formed by mixing 50 mL of 0.20
M hydrochloric acid (aq) with 50 mL of 0.30 M sodium
hydroxide (aq)?
Not really a neutralization, but looks like one.
2. Write the formula of hydrochloric acid.
3. Calculate moles of HCl, Write dissociation reaction for HCl and
covert mol HCl to moles of chloride
4. Since no additional Cl- from NaOH just calculate new molarity
based on new volume of solution.
1.
Shortcut: Since HCl has 1 Cl then
[HCl] = [Cl-] than just use dilution equation
C1V1=C2V2
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Neutralization 6
R 2009
28. How many milliliters of 0.250 M HCl are needed to
neutralize 5.00 g of Na2CO3?
2 HCl + Na2CO3  2 NaCl + CO2 + H2O
a)
b)
c)
d)
e)
377 mL
189 mL
Impossible to calculate
94.3 mL
754 mL
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Neutralization 6
R 2009
28. How many milliliters of 0.250 M HCl are needed to neutralize
5.00 g of Na2CO3?
2 HCl + Na2CO3  2 NaCl + CO2 + H2O
Determine moles of sodium carbonate by dividing by molar mass.
2. Convert moles of sodium carbonate to moles HCl using coefficients
of the reaction
3. Calculate volume of HCl using concentration.
1.
5.00gNa2CO3 1molNa2CO3
2molHCl
= 0.0943301..molHCl
106.01gNa2CO3 1molNa2CO3
M=
n
V
V=
n 0.0943307molHCl
L HCl
1000mL
=
= 377.322mL
M
0.250molHCl 1L
index
Neutralization 7
S 2009
28. It required 25.0 mL of 0.333 M sodium hydroxide solution to
completely neutralize 15.0 mL of sulfuric acid solution. What
was the molarity of the sulfuric acid solution?
a)
b)
c)
d)
e)
0.200 M
0.278 M
0.555 M
1.11 M
0.430 M
index
Neutralization 7
S 2009
28. It required 25.0 mL of 0.333 M sodium hydroxide solution to
completely neutralize 15.0 mL of sulfuric acid solution. What
was the molarity of the sulfuric acid solution?
Calculate the number moles of NaOH used.
2. Write the balanced reaction.
3. Convert moles NaOH to moles H2SO4 using the coefficients of
the reaction.
4. Calculate the molarity of the solution
1.
index
Equilibrium Constants Info
mass action =
1.
2.
3.
4.
5.
6.
[HI]
2
[ H 2 ][ I2 ]
No matter what the starting concentrations, at equilibrium the mass action
expression is always equal to a certain number, the equilibrium constant (Kc) for
that reaction at a specified temp. (known as equilibrium law for that reaction.)
The reaction quotient (Q) is also equal to the value of the mass action expression.
At equilibrium Q = Kc. If Q > Kc then reaction is on the right side of equation
moving toward the left. If Q< Kc reaction is on the left moving toward the right.
When a reaction starts with only reactants (no products), Q will increase as it
approaches equilibrium. Q is a number that changes as the concentrations of the
reactants and products change (as the reaction proceeds toward equilibrium)
Q shows where the reaction is in reference to equilibrium, it usually uses the
starting concentrations or whatever concentrations are available
Kc is a constant the equilibrium expression uses only the concentrations at
equilibrium
For equilibrium reactions only gases and aqueous species count. Solids and
liquids ignored.
Equilibrium constants change with temperature.
index
Equilibrium Constants Info
mass action =
Predicting the Equilibrium Law:
[ F ] f [G ] g
mass action =
[ D] d [ E ] e
At equilibrium
[F ] f [ G ] g
[F ] f [ G ] g
Kc =
and Q =
[ D] d [ E ] e
[ D] d [ E ] e
Not at equilibrium
[F ] f [ G ] g
Q =
[ D] d [ E ] e
Ignore liquids and solids.
Use only gases and aqueous (aq).
[HI]
2
[ H 2 ][ I2 ]
index
Equilibrium Constants Info
Manipulating Equations for Chemical Equilibria
So Kc’ = 1/Kc
If Reaction 1 is combined with Reaction 2 to make reaction 3 then
K3 = K1xK2
Significance of the Magnitude of K:
1.
2.
3.
K is very large: reaction proceeds far toward completion, The position of
equilibrium lies far toward the products
K ≈ 1: the concentrations of reactants and products are nearly the same.
K is very small: Extremely small amounts of products are formed, position
of equilibrium lies far toward the reactants
index
Equilibrium Constants Info
Kp = equilibrium constant using partial pressure of gases
T in Kelvin (+273K)
R = Gas Constant
index
Equilibrium Constant Problem 1
D1 2008
34. In which of the following cases will the least time be
required to arrive at equilibrium
a)
b)
c)
d)
e)
Kc is a very small number
Cannot tell since the time to arrive at equilibrium does not
depend on Kc
Kc is about 1
Cannot tell without knowing the value of Kc
Kc is a very large number
index
Equilibrium Constant Problem 1
D1 2008
34. In which of the following cases will the least time be
required to arrive at equilibrium
Cannot tell since the time to arrive at equilibrium does not
depend on Kc
index
Equilibrium Constant Problem 2
D1 2010
39. Given:
H3O+(aq) + NO2-(aq)  HNO2(aq) + H2O
Kc = 3.78 x 103 at T = 298 K
Calculate Kc for the reaction:
NO2- (aq) + H2O (l) 
a)
b)
c)
d)
e)
0.30 x 10-11
0.83 x 10-9
0.57 x 10-9
0.15 x 10-10
0.38 x 10-10
HNO2(aq) + OH-(aq)
index
Equilibrium Constant Problem 2
D1 2010
39. Given:
H3O+(aq) + NO2-(aq)  HNO2(aq) + H2O
Kc = 3.78 x 103 at T = 298 K
Calculate Kc for the reaction:
NO2- (aq) + H2O (l) 
HNO2(aq) + OH-(aq)
1. Need to add a reaction to it that we know the Kc for that changes H3O+
to OH- like 2H2O = H3O+ + OH- .
2. When you add reactions the Kc of the new reaction will equal the K’s
multiplied together.
index
Equilibrium Constant Problem 2
D1 2010
39. Given:
H3O+(aq) + NO2-(aq) 
HNO2(aq) + H2O
Kc = 3.78 x 103 at T = 298 K
Calculate Kc for the reaction:
NO2- (aq) + H2O (l) 
HNO2(aq) + OH-(aq)
When adding reactions, like
substances on opposite sides of
the arrow cancel.
K c3 = 3.78x103 (1x10-14 ) = 3.78x10-11 = 0.378x10-10
index
Equilibrium Constant Problem 3
D2 2010
39. Kc= 1.506 x 10-5 at T = 100 C for the reaction A (g)  2B (g).
What is Kc for the reaction: 2B(g)  A (g) at T = 100 C?
a)
b)
c)
d)
e)
119500
93000
66400
26600
46500
index
Equilibrium Constant Problem 3
D2 2010
39. Kc= 1.506 x 10-5 at T = 100 C for the reaction A (g)  2B (g).
What is Kc for the reaction: 2B(g)  A (g) at T = 100 C?
The Kc of the reverse of a reaction is 1/Kc of the forward reaction.
1
1
K¢c =
=
-5 = 66401.06242 = 66400
K c 1.506x10
index
Equilibrium Constant Problem 4
R 2009
34. What is the form of the equilibrium constant for the reaction
2 HgO (s)  2 Hg (l) + O2 (g) ?
a)
b)
c)
d)
e)
Kc = [Hg]2[O2]
Kc = [O2]
None of the other answers is correct
Kc = [O2]/[HgO]2
Kc = [Hg]2[O2]/[HgO]2
index
Equilibrium Constant Problem 4
R 2009
34. What is the form of the equilibrium constant for the reaction
2 HgO (s)  2 Hg (l) + O2 (g) ?
At equilibrium
[F ] [ G ]
Kc =
[ D] d [ E ] e
f
g
Ignore liquids and solids.
Only use gases and aq
Since we ignore liquids and solids use
only the oxygen gas when writing the
equilibrium constant expression.
Kc = [O2]
index
Equilibrium Constant Problem 5
S 2009
34. Given the reaction 2 CO (g) + O2 (g)  2 CO2 (g) which is
the relationship between Kc and Kp ?
a)
b)
c)
d)
e)
Kp = Kc(RT)-2
Kp = Kc
Kp = KcRT
Kp = Kc(RT)-1
Kp = Kc(RT)2
index
Equilibrium Constant Problem 5
S 2009
34. Given the reaction 2 CO (g) + O2 (g)  2 CO2 (g) which is
the relationship between Kc and Kp ?
index
Equilibrium Constant Problem 6
R 2010
34. If Kc = 0.040, at 450 C, what is the Kp for
PCl5 (g)  PCl3 (g) + Cl2 (g) ?
a)
b)
c)
d)
e)
6.7 x 10-4
5.2 x 10-2
0.40
2.4
0.64
index
Equilibrium Constant Problem 6
R 2010
34. If Kc = 0.040, at 450 C, what is the Kp for
PCl5 (g)  PCl3 (g) + Cl2 (g) ?
index
Equilibrium Constant Problem 7
S 2010
34. The equilibrium constant, Kc, for the gaseous reaction
HCHO (g)
 H2 (g) + CO (g) has the numerical value 0.50 at 600C. A mixture of
HCHO, H2, and CO is introduced into a flask at 600C. After a short
time, analysis of a small sample of the reaction mixture shows the
concentration to be [HCHO] = 1.5, [H2] = 0.5 and [CO] = 1.0 moles/liter.
Which of the following statements is true?
a)
b)
c)
d)
e)
The reaction mixture is not at equilibrium , but will move toward equilibrium
by using up more HCHO
The reaction mixture is not at equilibrium, but no further reaction will occur.
The reaction mixture is at equilibrium
The forward rate of this reaction is the same as the reverse rate at those
concentrations
The reaction mixture is not at equilibrium, but will move toward equilibrium by
forming more HCHO
index
Equilibrium Constant Problem 7
S 2010
34. The equilibrium constant, Kc, for the gaseous reaction
HCHO (g)  H2 (g) +
CO (g) has the numerical value 0.50 at 600C. A mixture of HCHO, H2, and CO is
introduced into a flask at 600C. After a short time, analysis of a small sample of the
reaction mixture shows the concentration to be [HCHO] = 1.5, [H2] = 0.5 and [CO]
= 1.0 moles/liter. Which of the following statements is true?
[ F ] f [G ] g
mass action =
[ D] d [ E ] e
At equilibrium
[F ] f [ G ] g
[F ] f [ G ] g
Kc =
and Q =
[ D] d [ E ] e
[ D] d [ E ] e
Not at equilibrium
[F ] f [ G ] g
Q =
[ D] d [ E ] e
Ignore liquids and solids.
Use only gases and aqueous (aq).
1. Calculate Q (reaction quotient) and compare it
to the value of Kc
H 2 ][CO] 0.5M× 1.0M
[
Q=
=
= 0.33333M
1.5M
[ HCHO]
2. Since Q < Kc, (too much HCHO) reaction is
not at equilibrium and will move towards
equilibrium by using up HCHO
index
Weak Acid – Weak Base Info
1. Kw = [H+][OH-] = 1 x 10-14 (at 25° C)
2. Acid ionization constant, (Ka ) shows the strength of acids (the
amount of dissociation) (usually only for weak acids)
1. Base ionization constant (Kb)
Equilibrium constants (Kc, Ka, Ksp,
Kb) only involve aqueous (aq) and
gases (g). No solids, no liquids.
index
Weak Acid – Weak Base Info
1. Weak Acids
H O ][ F ]
[
=
+
Ka
-
3
[HF]
2. Conjugate Base of a weak acid will produce OHKb =
3. Ka x Kb = Kw
like HF and F-
[ HF ][OH - ]
[F - ]
( Kw= 1 x 10-14) for a conjugate acid –base pair
index
Weak Acid – Weak Base Info
1. Weak Bases
2. Conjugate acid of a weak base will produce H+
index
Weak Acid – Weak Base Info
1. Ionic compounds with conjugate acids or conjugate base of strong
acids or bases will produce neutral solutions. Ex: NaCl, KNO3,
2. Ionic compounds with only conjugate bases of weak acids will
produce basic solutions. Examples: KF, NaCH3COO,
3. Ionic compounds with only conjugate acids of a weak base will
produce acidic solutions. Examples: NH4Cl
4. Ionic compounds with both – pH is determined by relative
strengths
5. The higher Ka the stronger the acid the higher the Kb the stronger
the base.
index
Weak Acid – Weak Base Problem 1
D2 2008
34. Consider: H3O+ (aq) + OH- (aq) ↔ 2 H2O
If the initial concentrations are
[H3O+ (aq)] = 0.1 M
[OH- (aq)] = 0.7 M
at equilibrium, [OH- (aq)] will be _________.
a)
b)
c)
d)
e)
0.5 M
0.6 M
0.8 M
0.7 M
0.0 M
index
Weak Acid – Weak Base Problem 1
D2 2008
34. Consider: H3O+ (aq) + OH- (aq) ↔ 2 H2O
If the initial concentrations are [H3O+ (aq)] = 0.1 M [OH- (aq)] = 0.7 M at
equilibrium, [OH- (aq)] will be _________.
1. Write the reaction, use ICE (initial, change, & equilibrium) to
determine equilibrium values. Since volume doesn’t change we
can use molarity.
2. Use the equilibrium expression, Kw = 1 x 10-14 = [H3O+][OH-]
(equilibrium concentrations) to solve for X.
K w = [0.1 - X ][0.7 - X ] = 1x10 -14
0.07 - 0.8X + X 2 = 1x10 -14
X 2 + 0.8X + 0.07 = 0
X = 0.7 X = 0.1
index
Weak Acid – Weak Base Problem 1
D2 2008
34. Consider: H3O+ (aq) + OH- (aq) ↔ 2 H2O
If the initial concentrations are [H3O+ (aq)] = 0.1 M [OH- (aq)] = 0.7 M at
equilibrium, [OH- (aq)] will be _________.
K w = [0.1 - X ][0.7 - X ] = 1x10 -14
0.07 - 0.8X + X 2 = 1x10 -14
X 2 - 0.8X + 0.07 = 0
X = 0.7 X = 0.1
Since X can’t be .7 it must be .1 therefore 0.7-X = 0.6
0.6 M
index
Weak Acid – Weak base 2
D1 2009
35. For a given weak acid, HA, the value of Ka ________.
a)
b)
c)
d)
e)
None of the other answers are correct
Cannot be less than 10-7
Will change with pH
Does not change with temperature
Cannot be greater than 10-7
index
Weak Acid – Weak base 2
D1 2009
35. For a given weak acid, HA, the value of Ka ________.
a)
b)
c)
d)
e)
None of the other answers are correct
Cannot be less than 10-7
Will change with pH
Does not change with temperature
Cannot be greater than 10-7
A. None of the other answers are correct
index
Weak Acid – Weak Base 4
35. Given the following acid ionization constants, Ka
HF
7.2 x 10-4
HNO2
4.5 x 10-4
CH3COOH
1.8 x 10-5
HClO
3.5 x 10-8
HCN
4.0 x 10-10
Which is the weakest acid?
a)
b)
c)
d)
e)
HClO
HNO2
HF
CH3COOH
HCN
index
Weak Acid – Weak Base 4
35. Given the following acid ionization constants, Ka
HF
7.2 x 10-4
HNO2
4.5 x 10-4
CH3COOH
1.8 x 10-5
HClO
3.5 x 10-8
HCN
4.0 x 10-10
Which is the weakest acid?
HCN has the smallest Ka
index
Weak Acid – Weak Base 5
R 2009
35. 0.50 moles of HCN are added to a liter of water. What is the
pH if Ka for HCN = 4.0 x 10-10?
a)
b)
c)
d)
e)
5.35
4.85
9.40
4.35
4.69
index
Weak Acid – Weak Base 5
R 2009
35. 0.50 moles of HCN are added to a liter of water. What is the
pH if Ka for HCN = 4.0 x 10-10?
1. Write the balanced dissociation reaction for HCN
2. Calculate the molarity of HCN, since volume doesn’t change
you can use molarity.
3. Use ICE to determine equilibrium values (use X for the
change)
4. Write the equilibrium expression and use the value of Ka to
solve for X
4. Use this to determine molarity of H3O+, then calculate pH
index
Weak Acid – Weak Base 5
R 2009
35. 0.50 moles of HCN are added to a liter of water. What is the
pH if Ka for HCN = 4.0 x 10-10?
*Simplification can be
made can be made if
[HA]initial ≥ 400 Ka
[H 3O+ ][CN - ]
Ka =
[HCN]
[X][X]
4.0x10 -10 =
0.50M
X2
-10
4.0x10 =
0.50 M
2.0x10 -10 = X 2
*Since K a is so small (0.50 - X) » 0.50
X = 1.414213 x 10 -5 M
= [H 3O+ ]
pH = -log(1.414213x10 -5 M) = 4.849485
index
Weak Acid/Weak Base 6
R 2009
40. A solution is prepared by adding 2.0 moles of acetic acid
and 1.0 mole of sodium acetate to enough water to make
1.0 L of solution. Finally, that solution was diluted to a
final volume of 5.0L. What is the pH of the final solution?
For acetic acid Ka = 1.8 x 10-5
a)
b)
c)
d)
e)
2.6
4.0
4.4
5.0
7.5
index
Weak Acid/Weak Base 6
R 2009
40. A solution is prepared by adding 2.0 moles of acetic acid
and 1.0 mole of sodium acetate to enough water to make
1.0 L of solution. Finally, that solution was diluted to a
final volume of 5.0L. What is the pH of the final solution?
For acetic acid Ka = 1.8 x 10-5
1. Calculate the molarity of acetic acid and sodium acetate.
2. Write the balanced dissociation equation for acetic acid. Use
ICE to determine the equilibrium values.
3. Write the equilibrium expression and use the value of Ka to
solve for X
4. Use this to solve for molarity then pH
index
Weak Acid/Weak Base 6
R 2009
40. A solution is prepared by adding 2.0 moles of acetic acid and 1.0 mole of sodium
acetate to enough water to make 1.0 L of solution. Finally, that solution was diluted
to a final volume of 5.0L. What is the pH of the final solution? For acetic acid Ka
= 1.8 x 10-5
*Simplification can be
made can be made if
[HA]initial ≥ 400 Ka
[H 3O+ ][CH 3COO- ]
Ka =
[CH 3COOH]
[X][0.2M]
0.40M
0.2MX
1.8x10 -5 =
0.40M
7.2x10 -6 = 0.2MX
1.8x10 -5 =
X = 3.6 x 10 -5 M
*Since K a is so small (0.40 - X) » 0.40 and (0.20 + X) » 0.20
= [H 3O+ ]
pH = -log(3.6x10 -5 M) = 4.44369
index
Weak Acid/Weak Base 7
S 2009
35. Ka for HX is 1.31 x 10-3 at 25oC. What are [H3O+] and
[HX] for a 0.0200 M solution of HX?
[H3O+]
a)
b)
c)
d)
e)
4.5 x 10-3
5.1 x 10-3
5.3 x 10-3
5.7 x 10-3
6.0 x 10-3
[HX]
1.6 x 10-2
1.5 x 10-2
1.5 x 10-2
1.4 x 10-2
1.4 x 10-2
index
Weak Acid/Weak Base 7
S 2009
35. Ka for HX is 1.31 x 10-3 at 25oC. What are [H3O+] and
[HX] for a 0.0200 M solution of HX?
2. Write the balanced dissociation equation for HX. Use ICE to
determine the equilibrium values.
3. Write the equilibrium expression and use the value of Ka to
solve for X. [HX]/Ka = 15.2 simplification doesn’t work.
4. Use this to solve for molarity.
index
Weak Acid/Weak Base 7
S 2009
35. Ka for HX is 1.31 x 10-3 at 25oC. What are [H3O+] and
[HX] for a 0.0200 M solution of HX?
-b ± b 2 - 4ac
x=
2a
[H 3O+ ][X - ]
Ka =
[HX]
[X][X]
1.31x10 -3 =
0.020M - X
*Since
X2
1.31x10 =
0.020 M - X
2.62x10 -5 -1.31x10 -3 X = X 2
[HX]
= 15.267 is less than 400 can't cancel X must use quadratic formula
Ka
-3
X = 0.0045097
X 2 +1.31x10 -3 X - 2.62x10 -5 = 0
X = -0.0058097 X can't be negative therefore
X = .0045097 M = [H 3O+ ] = 4.5x10 -3 M
[HX] = (0.020 M - X) = 0.0154903 M = 1.5 x 10 -2 M
index
Weak Acid/Weak Base 8
R 2010
40. Given that Kw = 3.995 x 10-13 at some temperature, what is
the pH of pure water at the temperature?
a)
b)
c)
d)
e)
6.2
6.6
7.0
7.4
7.8
index
Weak Acid/Weak Base 8
R 2010
40. Given that Kw = 3.995 x 10-13 at some temperature, what is
the pH of pure water at the temperature?
Since Kw = [H3O+][OH-]
K w = [H 3O+ ][OH - ]
3.995x10 -13 = [X][X]
3.995x10 -13 = X 2
X = 6.320601x10 -7 M = [H 3O+ ]
pH = -log(6.320601x10 -7 ) = 6.1992416 = 6.1992
index
Light & Energy – Electrons Info
1. Speed of light (c) = 3.00 x 108 m/s
2. Wavelength = λ, frequency = ν (hz, 1/s)
1m = 109nm
1m = 1010A (angstrom)
3. Energy: E = hν = hc/λ
c = λ×ν
h = Planck’s constant
4. spectrum: (low freq/energy to high freq/energy)
radio (short-TV-micro)-Ir-Vis-UV-Xrays-Gamma Rays
ROY G BIV
5. Line Spectrum of Hydrogen (Finds wavelength of light emission
from hydrogen)
æ1
1ö
= R H ç 2 - 2 ÷ where R H = 109678cm -1 (constant)
l
è n1 n2 ø
n1 & n 2 are variables, whole numbers, quantum numbers, orbits in hydrogen, 1- infinity, n 2 > n1
1
index
Light & Energy – Electrons Info
6. Bohr’s equation – calculates energy of orbits of hydrogen
7. De Broglie’s equation – calculates the wavelength of electrons
h
l=
mn
m = mass of electron v = velocity
1J = 1Nm 1N = 1Kg m/s2
index
Principle Quantum number (n)
Main energy level (shell)
1. Allowable value: positive integer
(1,2,3,4…)
2. As n  energy 
3. n2 = # orbitals in that shell
(since 1 orbital will hold 2 e- )
4. 2n2 = # of electrons in that shell
5. n = # of sublevels or subshells
index
Angular momentum quantum number (l)
Subsidiary quantum number
Subshell or sublevel, Shape of the orbital
1. Allowable value: l = 0….(n-1) all positive integers
between 0 and n-1
2. The value of l = shape
l=0 “s” shape
l=1
“p” shape
l=2
“d” shape
l = 3 “f” shape
3. As l  energy 
index
Magnetic Quantum number (m)
Which orbital – orientation around the nucleus
1. Allowable values: m = ( -l …0…+l) all integers
between –l and +l including zero.
2. “s” subshell – 1 orbital (2 e-)
3. “p” subshell – 3 orbitals (6 e-)
4. “d” subshell – 5 orbitals (10 e-)
5. “f” subshell – 7 orbitals
(14 e-)
6. “g” subshell – 9 orbitals (18 e)
ground state)
(not occupied in
index
Spin Quantum number
1. Allowable value: + ½ or – ½
2. Electrons exist in two states of spin. One spins one
way the other spins the other way.
3. An orbital will hold 2 e- - to be in the same orbital they
must have opposite spins (Pauli Exclusion Principle)
4. Spin is usually represented by an up or down arrow.
 - spin in one direction
 - spin in the other direction
index
Rules of Electrons
1. Each row on the periodic table represents a
shell.
2. Outer shell (ring) can have a max of 8 e3. Shell #1 (n=1): 1 subshell (s), 1orbital, & can
have a max of 2 e4. Shell #2 (n=2): 2 subshells (s,p), 4 orbitals, &
can have a max of 8 e5. Shell #3 (n=3): 3 subshells (s,p,d), 9 orbitals,
& can have a max of 18 e6. Shell #4 (n=4): 4 subshells (s,p,d,f), 16
orbitals, & can have a max of 32 e-
index
Some Patterns in the Rules:
1.
2.
3.
4.
5.
6.
# of orbitals in a subshell: 1, 3, 5, 7
# of electrons in a subshell: 2, 6, 10, 14
n = shell number = row in the periodic table
# of subshells = n
# of orbitals in a shell = n2
(1, 4, 9, 16…)
# of electrons in a shell = 2n2 (2, 8, 18, 32)
index
s
p
n = row#
n=row
d
n = (row# -1)
f
n = (row# - 2)
index
Light & Energy – Electrons Info
index
Light & Energy – Electrons Problem 1
D1 2008
23. The number of valence electrons in one tin atom is ______
a) 0
b) 4
2
d) 14
e) 8
c)
index
Light & Energy – Electrons Problem 1
D1 2008
23. The number of valence electrons in one tin atom is ______
a) 0
b) 4
2
d) 14
e) 8
c)
index
Light & Energy – Electrons Problem 2
D1 2008
25. ________equals Planck’s constant times the frequency of the
electromagnetic radiation
a) Velocity
b) Wavelength
Intensity
d) Quantum
e) Energy
c)
index
Light & Energy – Electrons Problem 2
D1 2008
25. ________equals Planck’s constant times the frequency of the
electromagnetic radiation
a) Velocity
b) Wavelength
Intensity
d) Quantum
e) Energy
c)
index
Light & Energy – Electrons Problem 3
D1 2008
26. When n=4, l = 2, and ml= -1, what atomic orbital type is this?
a) 4d
b) 2p
2d
d) 4p
e) 5d
c)
index
Light & Energy – Electrons Problem 3
D1 2008
26. When n=4, l = 2, and ml= -1, what atomic orbital type is this?
a) 4d
b) 2p
2d
d) 4p
e) 5d
c)
index
Light & Energy – Electrons Problem 4
D2 2008
23. The number of valence electrons in a gallium atom is _____
a) 13
b) 1
2
d) 31
e) 3
c)
index
Light & Energy – Electrons Problem 4
D2 2008
23. The number of valence electrons in a gallium atom is _____
a) 13
b) 1
2
d) 31
e) 3
c)
index
Light & Energy – Electrons Problem 5
D2 2008
25. The higher the energy of electromagnetic radiation, the ____.
a) Higher the velocity of light
b) Greater its mass
Longer its wavelength
d) Lower its frequency
e) Shorter its wavelength
c)
index
Light & Energy – Electrons Problem 5
D2 2008
25. The higher the energy of electromagnetic radiation, the ____.
a) Higher the velocity of light
b) Greater its mass
Longer its wavelength
d) Lower its frequency
e) Shorter its wavelength
c)
index
Light & Energy – Electrons Problem 6
D2 2008
26. Is n=5, l =-3, ml=-2, ms= + ½ a valid set of quantum numbers
for a hydrogen atom?
a) Yes
b) No, because l cannot be negative
No, because m cannot be negative
d) No, because n cannot be as large as 5
e) No, because l must be equal to n-1
c)
index
Light & Energy – Electrons Problem 6
D2 2008
26. Is n=5, l =-3, ml=-2, ms= + ½ a valid set of quantum numbers
for a hydrogen atom?
No, because l cannot be negative
index
Light & Energy – Electrons Problem 7
D1 2010
24. What is the maximum number of spatial orbitals there can be
with n=2 and l=2?
a) 1
b) 5
2
d) None
e) 3
c)
index
Light & Energy – Electrons Problem 7
D1 2010
24. What is the maximum number of spatial orbitals there can be
with n=2 and l=2?
None. Acceptable values of l = 0…(n-1)
index
Light & Energy – Electrons Problem 8
D1 2010
23. _____light has the highest energy per photon.
a) Infrared
b) Red
Ultraviolet
d) Green
e) Blue
c)
index
Light & Energy – Electrons Problem 8
D1 2010
23. _____light has the highest energy per photon.
a) Infrared
b) Red
Ultraviolet
d) Green
e) Blue
c)
index
Light & Energy – Electrons Problem 9
D2 2010
23. What is the frequency of photons with wavelength = 6580 A?
a) 4.6 x1015 /s
b) 8.0 x 1014 /s
3.0 x 1014 /s
d) 4.6 x 1014 /s
e) 8.7 x 1014 /s
c)
index
Light & Energy – Electrons Problem 9
D2 2010
23. What is the frequency of photons with wavelength = 6580 A?
index
Light & Energy – Electrons Problem 10
D2 2010
24. The number of spatial orbitals with the quantum numbers
n=6, l=2, and ml = -1
a) 3
b) 1
9
d) 5
e) 7
c)
index
Light & Energy – Electrons Problem 10
D2 2010
24. The number of spatial orbitals with the quantum numbers
n=6, l=2, and ml = -1
a) 3
b) 1
9
d) 5
e) 7
c)
index
Light & Energy – Electrons Problem 11
D1/D2 2010
25. The Lanthanide and actinide series (inner transition metals)
consists of elements filling what orbitals?
a) g
b) d
p
d) s
e) f
c)
index
Light & Energy – Electrons Problem 11
D1/D2 2010
25. The Lanthanide and actinide series (inner transition metals)
consists of elements filling what orbitals?
a) g
b) d
p
d) s
e) f
c)
index
Light & Energy – Electrons Problem 12
R 2009
23. What is the wavelength of an electron traveling one tenth the
speed of light?
a) 7.3 x 10-7 m
b) 1.6 x 10-19m
3.0 x 108 m
d) 2.4 x 10-15 m
e) 2.4 x 10-11 m
c)
index
Light & Energy – Electrons Problem 12
R 2009
23. What is the wavelength of an electron traveling one tenth the
speed of light?
1. Use De Broglie equation.
l=
h
mn
m = mass of electron v = velocity h = Planck's constant
Kg× m
6.626x10 -34 J× s 1N× m s2
l=
-31
8 m =
1J
1N 9.11x10 kg 0.1× 3.00x10 s
= 2.42444...x10 -11 m
index
Light & Energy – Electrons Problem 13
S 2009/R 2010
23/22. The number of unpaired electrons in the lowest energy
electron configuration of an isolated sulfur atom?
a) 1
b) 3
4
d) 0
e) 2
c)
index
Light & Energy – Electrons Problem 13
S 2009/R 2010
23. The number of unpaired electrons in the lowest energy electron
configuration of an isolated sulfur atom?
1s2 2s2 2p6 3s2 3p4
3p
2
3p
3p
index
Light & Energy – Electrons Problem 14
S 2009
25. The n and l quantum numbers of the highest energy electrons
of an element are n=4 and l=2.
a) Nonmetal
b) Noble gas
d-transition metal
d) Representative element
e) f –transition metal
c)
index
Light & Energy – Electrons Problem 14
S 2009
25. The n and l quantum numbers of the highest energy electrons
of an element are n=4 and l=2.
n = 4, l=2 is 4d therefore d-transition metal
index
Light & Energy – Electrons Problem 15
R 2010
24. For a subsidiary quantum number value of 3, what are the
possible values for the angular momentum quantum number,
m l?
a) +2, +1, 0, -1, -1
b) None of the other choices are correct
+3, +2, +1, 0, -1, -2, -3
d) +3, +2, +1, 0
e) +2, +1, 0
c)
index
Light & Energy – Electrons Problem 15
R 2010
24. For a subsidiary quantum number value of 3, what are the
possible values for the angular momentum quantum number,
m l?
a) +2, +1, 0, -1, -1
b) None of the other choices are correct
+3, +2, +1, 0, -1, -2, -3
d) +3, +2, +1, 0
e) +2, +1, 0
c)
Subsidiary quantum number = l
l = 0….(n-1) = 0, +1, +2
index
Light & Energy – Electrons Problem 15
R 2010
29. A maximum of _____ electrons having a given spin can occupy
any given molecular orbital.
a) 0
b) ½
4
d) 2
e) 1
c)
index
Light & Energy – Electrons Problem 15
R 2010
29. A maximum of _____ electrons having a given spin can occupy
any given molecular orbital.
a) 0
b) ½
4
d) 2
e) 1
c)
Pauli exclusion principle says no two electrons can have the same 4
quantum numbers (can’t have two electrons in the same orbital with
the same spin.
index
Light & Energy – Electrons Problem 16
S 2010
23. In the Bohr atom, what is the wavelength of light emitted when
the electron goes from the n=4 level to the n=3 level? Rydberg’s
constant is 1.097x107 m-1
a) 9100 nm
b) 1900 nm
3000 nm
d) 4200 nm
e) 6200 nm
c)
index
Light & Energy – Electrons Problem 16
S 2010
23. In the Bohr atom, what is the wavelength of light emitted when
the electron goes from the n=4 level to the n=3 level? Rydberg’s
constant is 1.097x107 m-1
æ1
1ö
= RH ç 2 - 2 ÷
l
è n1 n 2 ø
1 1.097x10 7 æ 1 1 ö
=
ç - ÷ = 533263.888m -1
è 9 16 ø
l
m
1
1.8752 x10 -6 m 10 9 nm
l=
= 1875.244 nm
1m
index
Light & Energy – Electrons Problem 17
S 2010
24. How many spatial orbitals exist for a g sub-energy level?
a) 5
b) 1
9
d) 7
e) 3
c)
index
Light & Energy – Electrons Problem 17
S 2010
24. How many spatial orbitals exist for a g sub-energy level?
a) 5
b) 1
9
d) 7
e) 3
c)
1, 3, 5, 7, 9, 11… s = 1, p = 3, d = 5, f = 7, g = 9
index
Light & Energy – Electrons Problem 18
S 2010
25. If the n and l quantum numbers of the last electron added in
the Aufbau sequence for an element are n=4 and l=3, the
element is a(n) ________.
a) Noble gas
b) Representative element
Non-metal
d) f-transition metal
e) d-transition metal
c)
index
Light & Energy – Electrons Problem 18
S 2010
25. If the n and l quantum numbers of the last electron added in
the Aufbau sequence for an element are n=4 and l=3, the
element is a(n) ________.
a) Noble gas
b) Representative element
Non-metal
d) f-transition metal
e) d-transition metal
c)
n = 4, l = 3 is 4f therefore f-transition metal
index
Solubility & Ksp Info
1. Dissolved ionic compounds 100% dissociate.
2. When two solutions are mixed you must check the new
possible compounds (double replacement reaction) to see if
they are soluble or insoluble using the solubility rules. If
they are both soluble then no reaction (NR) occurs.
3. Only one of the products can be insoluble. If one is, then
the solution turns cloudy from the formation of the
precipitate.
4. Precipitate – tiny solid particles formed when formation of
an insoluble compound occurs
index
Solubility & Ksp Info
Solubility Product Constant, Ksp (another equilibrium constant)
1. All ionic compounds will dissolve at least a little bit (even the
ones we consider insoluble)
2. The Ksp tells us how soluble an ionic compound is. Higher Ksp
= more soluble.
3. Saturated solution = no more will dissolve.
4. When an ionic compound is dissolved and forms a saturated
solution an equilibrium is set up.
Precipitate will form if: ion product (Qsp) > Ksp (supersaturated)
No Precipitate will form if: ion product = Ksp (saturated)
ion product < Ksp (unsaturated)
index
Solubility Rules
Solubility Rules of Ionic compounds
1. All compounds w/ group 1metals or NH4+ (ammonium) are soluble.
2. All compounds with nitrate, acetate, perchlorate are soluble.
3. All compounds with chloride are soluble except AgCl, PbCl2,
Hg2Cl2 (mercury I) (Usually Iodides and bromides as well)
4. All sulfates are soluble except Ba2+, Ca2+, Pb2+, Hg2+
5. All compounds that contain PO43-, CO32-, SO32-, and S2- are insoluble
except those of group 1 and NH4+
6. All hydroxides and all metal oxides are insoluble except Group 1,
NH4+, Ca2+, Sr2+, and Ba2+
index
Dissociation Reaction
Ionic compounds  ions
index
Net Ionic Equations
NaCl( aq ) + AgNO3( aq ) ® AgCl( s ) + NaNO3( aq )
Net ionic equation
Na+ & NO3- : spectator ions
Total ionic
index
Solubility & Ksp Prob 1
R 2009
21 Which reactants will generate a precipitated product?
a) AgNO3 + CH3COONa
b) FeCl2 + Na2SO4
(NH4)2SO4 + KOH
d) BaCl2 + Na2SO4
e) NaH + H2O
c)
index
Solubility & Ksp Prob 1
R 2009
21 Which reactants will generate a precipitated product?
a) AgNO3 + CH3COONa = Sodium nitrate (sol) + silver acetate (sol)
b) FeCl2 + Na2SO4 = sodium chloride (sol) + iron (II) sulfate (sol)
(NH4)2SO4 + KOH all group 1 and NH4 are sol
d) BaCl2 + Na2SO4 = sodium chloride (sol) + barium sulfate (insol)
e) NaH + H2O (hydrides are very reactive with water)
c)
1. Switch the ions (double replacement reaction)
2. Use solubility rules to see which one is insoluble
D because all sulfates are soluble except Ba, Ca, Pb (II) , & Hg (II).
index
Solubility & Ksp Prob 2
R 2009
37. What is the concentration of Pb2+ and F- in a saturated
solution of PbF2? Ksp= 3.7 x 10-8
[Pb2+]
a) 2.1 x 10-3
b) 1.9 x 10-4
2.1 x 10-3
d) 3.3 x 10-3
e) 9.6 x 10-5
c)
[F-]
4.2 x 10-3
3.8 x 10-4
2.1 x 10-3
6.6 x 10-3
1.9 x 10-4
index
Solubility & Ksp Prob 2
R 2009
37. What is the concentration of Pb2+ and F- in a saturated
solution of PbF2? Ksp= 3.7 x 10-8
1. Write dissociation reaction.
2. Write equilibrium expression.
3. Use ICE to determine equilibrium values in terms of X.
Assume that you start with zero product.
4. Use equilibrium expression and value of Ksp to solve for X.
5. Solve for concentrations.
index
Solubility & Ksp Prob 2
R 2009
37. What is the concentration of Pb2+ and F- in a saturated
solution of PbF2? Ksp= 3.7 x 10-8
index
Solubility & Ksp Prob 3
S 2009
22. Write the balanced, total ionic equation for the reaction of
calcium hydroxide with nitric acid using integers. What is
the sum of the stoichiometric coefficients of the ions on the
left side of the balanced equation?
a) Three
b) Four
Seven
d) Six
e) five
c)
index
Solubility & Ksp Prob 3
S 2009
22. Write the balanced, total ionic equation for the reaction of
calcium hydroxide with nitric acid using integers. What is
the sum of the stoichiometric coefficients of the ions on the
left side of the balanced equation?
7
index
Solubility & Ksp Prob 4
S 2009
37. If a solution is 1.00 x 10-5 M in manganese nitrate and 1.50
x 10-3 M in aqueous ammonia, will manganese hydroxide
precipitate? Kb (ammonia) = 1.8 x 10-5 Ksp (manganese
hyrdoxide)= 2.5 x 10-13
a) No, because Qsp>Ksp
b) No, because Ksp>Qsp
Yes, because Ksp>Qsp
d) Yes, because Qsp>Ksp
e) Yes, because Qsp=Ksp
c)
index
Solubility & Ksp Prob 4
S 2009
37. If a solution is 1.00 x 10-5 M in manganese nitrate and 1.50
x 10-3 M in aqueous ammonia, will manganese hydroxide
precipitate? Kb (ammonia) = 1.8 x 10-5 Ksp (manganese
hyrdoxide)= 2.5 x 10-13
1. Determine the concentration of OH formed by ammonia by
writing the dissociation reaction for ammmonia, using ICE
to determine the equilibrium values in terms of X, using
the value of Kb to solve for X, use X to determine the M
2. Write the dissociation reaction for Manganese hydroxide.
Write the Ksp expression for reaction. Use the starting
concentrations to calculate Qsp. Compare Qsp with Ksp to
see if it precipitates
index
Solubility & Ksp Prob 4
S 2009
*Simplification can be
made can be made if
[HA]initial ≥ 400 Ka
37. If a solution is 1.00 x 10-5 M in manganese (IV) nitrate and
1.50 x 10-3 M in aqueous ammonia, will manganese (IV)
hydroxide precipitate? Kb (ammonia) = 1.8 x 10-5 Ksp
(manganese hyrdoxide)= 2.5 x 10-13
No, because Ksp>Qsp
index
Solubility & Ksp Prob 5
R 2010
36. How many mL of a 0.0010 M chloride solution must be
added to a 100 mL solution of 7.2 x 10-5 M Ag+ solution for
AgCl to begin to precipitate? Ksp of AgCl = 1.8x10-10
a) Need to know the solution pH to solve this problem
b) 2.6 mL
0.25 mL
d) 0.030 mL
e) 4.1 mL
c)
index
Solubility & Ksp Prob 5
R 2010
36. How many mL of a 0.0010 M chloride solution must be
added to a 100 mL solution of 7.2 x 10-5 M Ag+ solution for
AgCl to begin to precipitate? Ksp of AgCl = 1.8x10-10
1. Write the dissociation reaction for AgCl.
2. Write Ksp expression for AgCl
3. Use Ksp expression and value of Ksp to calculate the M of Clto make a saturated solution.
4. Use this molarity to calculate how much Cl to add
index
Solubility & Ksp Prob 5
R 2010
36. How many mL of a 0.0010 M chloride solution must be
added to a 100 mL solution of 7.2 x 10-5 M Ag+ solution for
AgCl to begin to precipitate? Ksp of AgCl = 1.8x10-10
index
Solubility & Ksp Prob 6
S 2010
36. Magnesium hydroxide, which has a solubility product
constant of 1.5 x 10-11, dissolves in water when NH4Cl is
added to the water because __________
a) Mg2+ forms a very stable complex ion with ammonia
b) Hydroxide ion reacts with chloride ion to form the weak acid
HClO
c) Hydroxide ion is converted to NH4OH by reaction with NH4+
d) Part of the magnesium hydroxide is oxidized to form a
different species.
e) Magnesium chloride is a salt and completely ionized in
water solution.
index
Solubility & Ksp Prob 6
S 2010
36. Magnesium hydroxide, which has a solubility product
constant of 1.5 x 10-11, dissolves in water when NH4Cl is
added to the water because __________
a)
Mg2+ forms a very stable complex ion with ammonia
b)
Hydroxide ion reacts with chloride ion to form the weak acid HClO
c)
Hydroxide ion is converted to NH4OH by reaction with NH4+
d)
Part of the magnesium hydroxide is oxidized to form a different species.
e)
Magnesium chloride is a salt and completely ionized in water solution.
Since NH4+ is the conjugate acid of a weak base it will react
with some OH- to establish an equilibrium
index
Solubility & Ksp Prob 7
D1 2008
36. The value of Ksp for copper (I) iodide is 1.0 x 10-12. What
is the solubility for copper (I) iodide in grams per liter?
a) 1.0 x 10-12
b) 9.0 x 10-6
1.9 x 10-4
d) 1.0 x 10-6
e) 1.0 x 10-4
c)
index
Solubility & Ksp Prob 7
D1 2008
36. The value of Ksp for copper (I) iodide is 1.0 x 10-12. What
is the solubility for copper (I) iodide in grams per liter?
1.
index
Rate Laws Info
index
Solution Concentration Info
index
Types of Colloids
Sol: solid in liquid
(Paints, mud)
Gel: solid network in liquid (gelatin, Jello)
Liquid emulsion: liquid in liquid (milk, mayo)
Foam: gas in a liquid (Shaving/whipped cream)
Solid aerosol: solid in a gas (smoke, dust)
Liquid aerosol: liquid in a gas (fog, cloud, aerosols)
Solid emulsion: liquid in a solid (cheese, butter)
index
Concentration
The amount of solute dissolved in a specific
quantity of solvent or solution.
Different methods
1.
2.
3.
4.
moles of solute
Molarity (mol/L) liter of solution
moles of solute
Molality (mol/kg) kilogram of solvent
mass of solute
Mass percent (%) mass of solution x100%
milligrams of solute
Parts per million (ppm)
kilogram of solution
5. Parts per billion (ppb)
microgram of solute
kilogram of solution