PRACTICE EXAM 2
PART A ( Equivalence Point)
a) What volume of a 0.219 M hydrochloric acid solution is required to titrate to equivalence 26.0 mL of a 0.125
M potassium hydroxide solution ? And what is the resulting pH
MV=MV
(0.219)*x = 0.125*0.026
x= 0.0148L  14.8ml
Strong Acid/Strong Base ==> No excess and Ions are Neutral pH=7
b) What volume of a 0.219 M sulfuric acid solution is required to titrate to equivalence 26.0 mL of a 0.125 M
barium hydroxide solution ?
And what is the resulting pH
MV=MV
2*(0.219)*x = 2* 0.125*0.026
x= 0.0148L  14.8ml
Strong Acid/Strong Base ==> No excess and Ions are Neutral pH=7
c) What volume of a 0.219 M nitric acid solution is required to titrate to equivalence 26.0 mL of a 0.125 M
ammonium hydroxide (pKb= 4.75) solution ? And what is the resulting pH
MV=MV
(0.219)*x = 0.125*0.026
x= 0.0148L  14.8ml
Strong Acid/WEAK base ==> No excess Acid/Base and BUT contains Salt of Weak Base (Acidic)
Molarity of NH4+ = moles of NH4/(Total Volume) = (0.026*0.125)/(0.0148+0.026) = 0.0797 M
pH= 7- ½ *(4.75+log(0.0797)) = 5.17
d) What volume of a 0.219 M nitrous acid (pKa=3.35) solution is required to titrate to equivalence 26.0 mL of
a 0.125 M Calcium Hydroxide solution ? And what is the resulting approximate pH
0.219*V=0.026*0.250
V=
0.02968
29.68
Total V=
55.68
L
ml
ml
pH= 7+0.5*(3.35+LOG(0.219*0.0297/0.0557))
8.21
PART B: EQUILIBRIA
1) Consider the Following OVERALL Reaction (20 pts)
A + 2B < == > C + 2D
In which the Rate of the forward reaction is equal to
Rf = kf [A][B]
and the Rate of the reverse (backward) reaction is equal to
Rb = kb [C] [D]2 /[ B]
a)
b)
c)
d)
Under what conditions would the above reaction be defined as at Equilibrium? (2 pts)
When the Forward Rate = Backward Rate ; Rf = Rb
What defines the Equilibrium Constant Keq in terms of the Rate Laws(2 pts)
Keq= kf / kb
What defines the Equilibrium Constant Keq in terms of CONCENTRATIONS(2 pts)
Keq = [C] [D]2/ ( [B] 2[A])
If AT EQUILIBRIUM, one finds that the concentrations of B and D are 0.135M,
while the concentrations of A and C are 0.0214 M and 0.0107 M respectively;
What would be the NUMERIC VALUE of the Equilibrium Constant? (5 pts)
Keq = =0.0107*0.135^2/(0.0214*0.135^2 )= 0.5
e) Are these concentrations UNIQUE, meaning are there other conditions that could arrive at the same K eq?
Illustrate by example(2 pts)
No, the concentrations can vary and are NOT UNIQUE; as long as the overall ratio as defined by
the equilibrium constant is equal to the proper value
f)
Based on this value, does the reaction FAVOR the REACTANTS or the PRODUCTS and WHY? (2 pts)
Because the Ratio is less than one; it favors the REACTANTS; more reactants than products makes the
ratio smaller than one.
g)
Staring with the concentrations as defined above (d); if one removed 0.01 moles/L of D from the reaction; How
would that change the Reaction Quotient Q. NUMERICALLY and what would THEN happen to Q? (2 pts)
Q = 0.0107*0.125^2/(0.0214*0.135^2 = 0.429 Q will diminish thus reaction(Q) will move back toward Keq
h)
And how would that shift the Overall Reaction (Left or Right) (2 pts)
Reaction will move to the Right
i)
Now in a more QUANTITAVE manner, using the Keq, show how one would solve the above problem to obtain
the final equilibrium concentration of [A]. SETUP (4 pts) (SOLVE EXACTLY 5Extra Credit Pts)
= (0.0107+x)*(0.125+2x)2/{(0.0214-x)*(0.135-2x)2} =Keq = 0.5
thus x= 0.000772
12) Using the fact that pKa(HF) = 3.17; show at least TWO ways to construct a BUFFER solution at pH=3.17 using
the following:
0.5M solution of NaOH
0.5M solution of NaF
0.5M solution of HCl
0.5M solution of NaCl
0.5M solution of HF
1) Equal Volumes both HF and NaF
2) One Volume of HF and ½ that volume of NaOH making F3) One Volume of NaF and ½ that volume of HCl to make HF
\
3) Which of the following acids, if in solutions of equal concentration, is the least acidic? (2pt) most acidic
a) Phenol, Ka = 1.3 x 10-10
b) Lactic acid, Ka = 1.4 x 10-4
-12
c) Saccharin, Ka = 2.1 x 10
d) Boric acid, Ka = 5.8 x 10-10
e) All of these acids are equally acidic because they are all of equal concentration.
4) Write a net ionic equation to show that Pyridine, C5H5N, behaves as a Bronsted-Lowry base in WATER. (2pt)
BL base
C5H5N
BL acid
+ H2O
< == >
BL acid
BL base
C5H5NH+
OH-
5) Why is it impossible to make a NaOH solution with a pOH= 7.82? (3pts)
Solution has a pH= 6.18 which makes it slightly acidic and it is impossible to make a Acid solution from a Strong Base
PART C: Le Chatelier's principle
Consider the Reaction:
3Cl2 (gas) + N2(gas) < == > 2 NCl3(gas)
is found to be EXOTHERMIC and have a Kpeq = 2.18 x 10-2 atmospheres
a) Discuss three ways that one can use Le Chatelier's principle (to include Reactants/Products, Temperature and Pressure)
to produce the maximum quantity of NCl3. (3pts)
1) Increase Reactants and/or Remove Product
2) Increase Pressure of system which will push reaction to right in order to reduce pressure(fewer moles on right
(NOT COVERED IN CLASS BUT VALUABLE TO KNOW!)
3) Because EXOTHERMIC; reduce the temperature which will pull reaction to the right
Extra Credit: Looking at the direction that temperature would push this Exothermic Reaction. Since Keq is temperature
dependent, how would this shift alter Q(Reaction Quotient) and thus the new Keq(T). (5pts)
Adding Heat to this reaction would actually push the reaction to the LEFT thus REDUCING Q and actually Keq
b) If one began with only Cl2 and N2 gas both at 0.3 atm partial pressures; what would be the final pressure of NCl3.
SETUP only (5pts)
3Cl2 (gas) + N2(gas) < == > 2 NCl3(gas)
Kpeq=2.18 x 10-2 =(2x)2 /{ (0.3-3x)3*(0.3-x) } = > xexact= 0.006
c) Assume now that the one guesses the change x to be 0.006 in Cl2 and N2 in achieving equilibria; utilizing this value, is
the resultant too big or too small in it’s effect and how should x be changed?(2pts)
Exactly correct gives the Keq value
PART D ( Titration)
Draw a reasonable titration curve for the reaction of 100 ml 0.2M Isophthalic acid (pKa= 3.46) with 0.1M
Barium hydroxide (assume totally soluble). Contrast this with the titration curve for a 0.2M Hydrochloric acid
Note the THREE(3) important pH’s that help define the shape of this graph
GRAPH TO INCLUDE S shaped surve for weak acid with 3 key pH points at 0, 50 100mls of Ba(OH)2
pH= ½ (pKa-log(HA)) =
ISPHTH=2.08
HCl= 0.70
pH= pKa –log( [HA]/[A])=pKa
3.46 (1/2 EQUIVALENCE)
pH(salt)= 7+ ½ (pKa+log[A-]
8.23 (REMEMBER DILUTION for A)
1) Calculate the pH of a 0.0120 M aqueous solution of nitrous acid (HNO2, Ka = 4.5e-4). HINT: Check relationship of
Ka and C)
[H3O+]approx=
pH
0.00232
2.63
pH exact = 2.68
2) What is the pH of an aqueous solution of 0.547 M pyridine (a weak base with the formula C5H5N) pKb= 8.79
BASE: pOH= ½(Kb-log(0.547) = 4.54 
pH= 9.46
3) What concentration of Calcium hydroxide (assume STRONG and it all dissolves) is needed to give an aqueous
solution with a pH of 9.730
pOH=
14-9.73
[OH] =
=10^-4.27
but [Ca(OH)2 ] = 0.5[OH]=
4.27
5.37E-05
2.69E-05
4) What is the pH of a 0.207 M aqueous solution of ammonium nitrate, NH4NO3 ? pKb(NH3)=4.75
SALT of WEAK BASE: pH= 7-1/2(pKb+log(0.207)) = 4.97
5) If the pH of a solution of NaF is adjusted for pH. Using the fact that pKa(HF) = 3.17; when is half of the
total fluoride in the form of F- and when is 75% of the total fluoride in the form of F-?
pH= pKa + log (0.5/0.5) = pKa = 3.17
pH = pKa + log(0.75/0.25) = 3.65
7) Which of the following aqueous solutions are good buffer systems ?
0.24 M hydroiodic acid + 0.22 M potassium iodide
0.31 M ammonia + 0.34 M amonium hydroxide
0.04 M potassium fluoride + 0.47 M hydrofluoric acid
0.31 M ammonia + 0.34 M amonium chloride
0.36 M potassium bromide + 0.26 M barium bromide
0.18 M hydrocyanic acid + 0.19 M sodium cyanide
NO(strong acid)
NO(both ammonia)
NO(could be BUT KF too small
YES
NO(both salts)
YES
8) What is the pH of a solution contains 0.282 M ammonium iodide and 0.489 M ammonia. (pKb(NH3)= 4.75)
pH of BASE BUFFER: pOH= pKb - log ([B]/[BH+]) = 4.75 - log(0.489/0.282) = 4.51  pH= 9.49
9) An aqueous solution contains 0.31 M ammonium sulfate.
One Liter of this solution could be converted into a buffer by the addition of:…
(Assume that the volume remains constant as each substance is added.)
0.31 mol H2SO4
No Reactions(Both Acids)
0.075 mol Ba(OH)2 YES ( SB will convert 0.15M
NH4+ to NH3
0.30 mol Ba SO4
No Buffer/salt
0.30 mol NH3
YES (weak base/conjugate acid)
0.15 mol HNO3
No Reactions(Both Acids)
10) A buffer solution contains 0.481 M nitrous acid and 0.314 M sodium nitrite. Ka = 4.50 x 10-4
If 0.0309 moles of sodium hydroxide are added to 150 ml of this buffer, what is the pH of the resulting
solution? (Assume that the volume change does not change upon adding sodium hydroxide)
Initial
HA(Nitrous Acid)=
A-( Nitrite)
NaOH=0.0309/0.15
pH= pKa - log (0.275/0.520)
0.481
0.314
M
M
0.206
Final
=0.481-0.206
=0.314+0.206
M
3.35-log(0.275/0.520) =
3.62
E. Consider the following Titration Conditions
:
0.275
0.520
0.25 M
ACID
0.25 M
BASE
50 ml HNO3
100 ml KOH
50 ml
Chloroacetic
Acid
(pKa=2.85)
50 ml HCl
25 ml
Ca(OH)2
50 ml NH3
(pKb = 4.75)
80 ml
Nitrous
Acid
(pKa)=3.35
40 ml NaOH
75 ml of
H2SO4
75 ml of
Ca(OH)2
15ml of
Sulfuric
Acid
40 ml of
NaOH
30 ml
Formic Acid
(pKa)=3.17
moles OH
Molar
pOH
pH
=0.25*0.05-2*0.25*0.025=
No Acid
25 ml of
Methylamine
pKb=3.34
8.04
0.125
5.08
At Half Equivalence Point
pH= pKa-log([HA]/[A-]) but log(1)=0
thus pH= 3.35
pH=7
Winner Strong Base
=0.01*0.25/0.055
=0.5*(3.46-LOG(0.25))
25 ml of
Methylamine
pKb=3.34
Molar
0.166667 Anion
0 moles of Acid/Base
Salt of weak Acid
=7+0.5*(2.85+LOG(0.166)
)
pH=
[NH4+] =
Salt of weak Base
pH=7-0.5*(4.75+LOG(0.125))=
10 ml of
Ca(OH)2
No Base
25 ml of
HCl
0.0125
0.083
1.08
12.92
Winner Weak Acid
=0.01*0.25
20 ml of
Cyanic Acid
(HCNO)
pKa = 3.46
pH
Equation/work
pOH= 0.5*(3.34log(0.25))
pH=
Salt of Weak Base
pH=70.5*(3.34+LOG(0.125))
0.0455
1.34
12.66
M OH
pOH
pH
0.0025
moles
0.0625
Molar
0.125
Molar
pH=2.0
3
1.97103
12.0289
7
5.78
Acid
Format
e
pH= pka log(Acid/Conjugate Base)
pH=
3.47