Name (Last, First): _____________________________
ID Number: _____________________
High School Name: _____________________________________________________________
Washington University Chemistry Tournament
April 2, 2016
Individual Exam 2: Chemical and Physical Equilibria
Please write your full name, ID number, and high school at the top of every page.
45 minutes is allowed for this examination. This exam is 6 questions long, and has 14 numbered
pages in total. Please check to make sure that your exam is complete, and report to the exam
proctors if any pages are missing.
Necessary equations and constants, as well as a periodic table, can be found at the end of the
exam. Do not write on the scoring sheet on the last page of the exam.
Scratch paper will not be permitted during this exam. If you run out of room on the front of the
page, you can continue to work on the back of the page, provided that all answers for parts are
clearly labelled. Only write work on the backs of pages that correspond to the question on the
front of the page.
Make sure to circle or box your final answer if appropriate. Correct answers with appropriate
work will receive full credit. If work contains reasoning or justification that is partially correct,
partial credit may be awarded. Explanations should fully answer the question and provide
supporting evidence and logic. These should also be given in complete sentences. Correct
answers without reasonable supporting work will not receive credit.
If you cannot answer a question or the entire question, it is advised that you do not spend too
much time on that question and proceed onto other parts of the exam. No electronics of any kind
are allowed during the exam, with the exception of a non-programmable scientific calculator.
Cell phones must be turned off, and watches must be removed. A clock will be projected in
the exam room. The scoring policy can be found on the WUCT website. Any appeals must be
made in writing in the appeals room, Lab Sciences 400.
Cheating will not be tolerated on this exam. The cheating policy that has been listed on the
WUCT website will be followed. Violators of this policy will be referred to the directors of the
competition.
1
Name (Last, First):_________________________ ID Number: __________________________
1. In human blood, one major mechanism for pH regulation is the phosphate buffer system,
which consists of hydrogen phosphate (HPO2−
4 , aqueous) in equilibrium with dihydrogen
−
phosphate (H2 PO4 , aqueous). If the dihydrogen phosphate concentration is 0.015 M, and
the K a of dihydrogen phosphate at 37 ℃ is 6.23 × 10—8, find the pH of this buffer
system. Assume the concentration of hydrogen phosphate ions is 15 times higher than the
concentration of dihydrogen phosphate ions, as is typical under physiological conditions.
Show work to support your answer, and circle your final answer.
[HPO2−
4 ]
pH = pK a + log
[H2 PO−
4]
pH = − log(6.23 × 10−8 ) + log(15) = 8.382
2
Name (Last, First):_________________________ ID Number: __________________________
2. One of the many similarities between soda and blood is that both contain dissolved
carbonic acid.
a. Give the balanced decomposition reaction for carbonic acid (H2 CO3 ) in water that
contributes to the carbonation of sodas.
H2 CO3(aq) → H2 O(l) + CO2(g)
b. Give a balanced dissociation reaction for each of the following species in water:
i. carbonic acid
+
−
H2 CO3(aq) ⇄ HCO3(aq)
+ H(aq)
H3O+ is accepted in place of H+ as long as H2O appears on the left as well.
ii. bicarbonate (hydrogen carbonate)
+
2−
HCO−
3(aq) ⇄ CO3(aq) + H(aq)
H3O+ is accepted in place of H+ as long as H2O appears on the left as well.
c. Find the pH of a 0.05 M solution of bicarbonate ion in pure water at 25 ℃, given the
following information. The K a of carbonic acid is 4.5 × 10−7 , and the K a of
bicarbonate ion is 4.68 × 10−11 at 25 ℃.
1
For an amphiprotic salt, pH = (pK a1 + pK a2 )
2
K a1 is the dissociation of carbonic acid, and K a2 is the dissociation of bicarbonate.
pK a1 = − log(4.5 × 10−7 ) = 6.347
pK a2 = − log(4.68 × 10−11 ) = 10.330
1
pH = (6.347 + 10.330) = 8.34
2
3
Name (Last, First):_________________________ ID Number: __________________________
3. Under certain conditions, ions in solution can be separated on the basis of their solubility.
This is referred to as “selective precipitation.” In a selective precipitation experiment, an
aqueous solution of sodium chloride is slowly added to another aqueous solution that is
2.0 × 10−7 M in silver (I) ions and 0.0160 M in barium ions. Assume that there is no
volume change upon the addition of sodium chloride.
a. Identify the two precipitates that would form if sodium chloride were added in large
excess to the solution described above.
Precipitate #1: ______ 𝐀𝐠𝐂𝐥(𝐬) , 𝒐𝒓 𝑺𝒊𝒍𝒗𝒆𝒓 (𝑰) 𝒄𝒉𝒍𝒐𝒓𝒊𝒅𝒆_____________
Precipitate #2: ______ 𝐁𝐚𝐂𝐥𝟐(𝐬) , 𝒐𝒓 𝑩𝒂𝒓𝒊𝒖𝒎 𝒄𝒉𝒍𝒐𝒓𝒊𝒅𝒆 _____________
b. The Ksp of the silver (I) precipitate is 1.8 × 10−10 and the Ksp of the barium
precipitate is 2.6 × 10−9 at 25 ℃. When the sodium chloride solution is slowly added
to the solution of silver (I) and barium, what is the identity of the first precipitate to
be formed at 25 ℃? Show work to support your answer.
For AgCl, K sp = [Ag + ][Cl− ], so necessary [Cl− ] =
(K sp ) 1.8 × 10−10
=
[Ag + ] 2.0 × 10−7
= 9.0 × 10−4 M
For BaCl2 , K sp = [Ba2+ ][Cl− ]2 , so necessary [Cl− ]2 =
K sp
2.6 × 10−9
=
[Ba2+ ]
0.0160
= 1.625 × 10−7 M. So [Cl− ] = √1.625 × 10−7 = 4.03 × 10−4 M
The necessary chloride concentration to precipitate barium chloride is smaller than that of silver
(I) chloride, so it will be the first precipitate to be formed.
(This question continues on the next page.)
4
Name (Last, First):_________________________ ID Number: __________________________
c. Selective precipitation can be defined as being achieved when 100% of one ion
remains in solution and no more than 1% of the other ion remains in solution.
i. Using your answer in part (b) (ii), calculate the percentage of the ion that
precipitated first remains in solution at the point when the second precipitate starts
forming?
K sp(BaCl2 ) = [Ba2+ ][Cl− ]2
Using chloride concentration from part (b),
[Ba2+ ] =
K sp(BaCl2 )
2.6 × 10−9
=
= 3.21 × 10−3 M Ba2+
[Cl− ]2
(9.00 × 10−4 )2
[Ba2+ ] currently 3.21 × 10−3
=
∙ (100%)
original [Ba2+ ]
0.0160
= 20.06 % Ba remains
% Ba remaining =
ii. Based on the definition of selective precipitation given above, is it possible to use
sodium chloride to separate silver (I) from barium in solution? Explain your
answer in one complete sentence on the lines provided below.
No, there is much more than 1% of the original barium ion left in solution when the silver (I)
chloride begins to precipitate.______________________________________________________
______________________________________________________________________________
5
Name (Last, First):_________________________ ID Number: __________________________
4. Magnesium hydroxide, also called Milk of Magnesia, is an insoluble, white solid that can
be used as both a fire retardant and an antacid, though preferably not at the same time.
a. Give the balanced reaction for the dissolution of magnesium hydroxide in water.
2+
−
𝑀𝑔(𝑂𝐻)2(𝑠) ⇄ 𝑀𝑔(𝑎𝑞)
+ 2𝑂𝐻(𝑎𝑞)
b. 500 mL of a saturated solution of magnesium hydroxide is prepared in a flask at
25 ℃. When equilibrium is established, 2.00 × 10−5 moles of solid magnesium
hydroxide are left undissolved.
The saturated magnesium hydroxide solution was then titrated with a 0.002 M
solution of hydrochloric acid at 25 ℃. 102.78 mL of acid were needed to reach the
equivalence point. At the equivalence point, no solid magnesium hydroxide remained
in the flask.
What was the pH of the original solution?
𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻 +
1 𝑚𝑜𝑙 𝑂𝐻 −
(0.10278 𝐿 𝐻𝐶𝑙) ∗ (0.002
)∗(
)∗(
)
𝐿
1 𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻 +
= 2.056 × 10−4 𝑚𝑜𝑙 𝑂𝐻 − 𝑡𝑜𝑡𝑎𝑙
2 𝑚𝑜𝑙 𝑂𝐻 −
(2.00 × 10−5 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2 ) ∗ (
)
1 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2
= 4.00 × 10−5 𝑚𝑜𝑙 𝑂𝐻 − 𝑓𝑟𝑜𝑚 𝑠𝑜𝑙𝑖𝑑 𝑀𝑔(𝑂𝐻)2
(2.056 × 10−4 𝑚𝑜𝑙 𝑂𝐻 − 𝑡𝑜𝑡𝑎𝑙) − (4.00 × 10−5 𝑚𝑜𝑙 𝑂𝐻 − 𝑓𝑟𝑜𝑚 𝑠𝑜𝑙𝑖𝑑)
= 1.656 × 10−4 𝑚𝑜𝑙 𝑂𝐻 − 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1.656 × 10−4 𝑚𝑜𝑙 𝑂𝐻 −
= 3.311 × 10−4 𝑀
0.500 𝐿
𝑝𝑂𝐻 = − 𝑙𝑜𝑔(3.311 × 10−4 ) = 3.480
𝑝𝐻 = 14 − 𝑝𝑂𝐻 = 10.52
(This question continues on the next page.)
6
Name (Last, First):_________________________ ID Number: __________________________
c. Using your answer to part (b), calculate K sp for magnesium hydroxide at 25 ℃.
1
From the dissociation reaction and part (b), [𝑀𝑔2+ ]𝑠𝑜𝑙𝑛 = 2 [𝑂𝐻 − ]𝑠𝑜𝑙𝑛 =
3.311×10−4
2
=
1.656 × 10−4 𝑀
𝐾𝑠𝑝(𝑀𝑔(𝑂𝐻)2 ) = [𝑀𝑔2+ ][𝑂𝐻 − ]2 = (1.656 × 10−4 )(3.311 × 10−4 )2 = 1.815 × 10−11
d. Determine the solubility (in mg/L) of magnesium hydroxide in a buffered solution of
pH=13. If you were unable to answer part c, use the value K sp = 2.0 × 10−11 for
partial credit. Indicate clearly if you use the alternate Ksp value.
𝑝𝐻 𝑜𝑓 13 𝑚𝑒𝑎𝑛𝑠 𝑝𝑂𝐻 = 1. 𝑇ℎ𝑢𝑠, [𝑂𝐻 − ] = 10−𝑝𝑂𝐻 = 10−1 = 0.10 𝑀
[𝑀𝑔2+ ][0.102 ] = 1.815 × 10−11
[𝑀𝑔2+ ]
1.815 × 10−9 𝑚𝑜𝑙
=
𝐿
1.815 × 10−9 𝑚𝑜𝑙 𝑀𝑔2+ 1 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 58.32 𝑔 𝑀𝑔(𝑂𝐻)2 1000 𝑚𝑔
∗
∗
∗
𝐿
1 𝑚𝑜𝑙 𝑀𝑔2+
1 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2
1𝑔
−4
1.059 × 10 𝑚𝑔
=
𝐿
Alternate solution:
[𝑀𝑔2+ ][0.102 ] = 2.0 × 10−11
[𝑀𝑔2+ ] =
2.0 × 10−9 𝑚𝑜𝑙
𝐿
2.0 × 10−9 𝑚𝑜𝑙 𝑀𝑔2+ 1 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 58.32 𝑔 𝑀𝑔(𝑂𝐻)2 1000 𝑚𝑔
∗
∗
∗
𝐿
1 𝑚𝑜𝑙 𝑀𝑔2+
1 𝑚𝑜𝑙 𝑀𝑔(𝑂𝐻)2
1𝑔
−4
1.166 × 10 𝑚𝑔
=
𝐿
7
Name (Last, First):_________________________ ID Number: __________________________
5. In most introductory chemistry courses, the concentration of water is not included in
equilibrium constant calculations. This is because most problems deal with aqueous
solutions in which the concentration of water is significantly greater than that of its
solutes. However, the concentration of water can become important in some situations.
This question examines the concentration of water and its chemical significance.
g
a. Given that the density of water is 1.000 mL at 25 ℃, determine the concentration of
water (in
mol
L
1.000
) at 25 ℃.
𝑔
𝑔
𝑚𝑜𝑙
𝑚𝐿
÷ 18.02
= 0.05549
∗ 1000
= 55.49𝑀
𝑚𝐿
𝑚𝑜𝑙
𝑚𝐿
𝐿
b. If K w of water is 10−14 at 25 ℃, what is the pK a of water at 25 ℃?
𝐻2 𝑂 ⇌ 𝐻 + + 𝑂𝐻 −
[𝐻 + ][𝑂𝐻 − ]
𝐾𝑤
10−14
𝐾𝑎 =
=
=
= 1.802 ∗ 10−16 → 𝑝𝐾𝑎 = − 𝑙𝑜𝑔(𝐾𝑎 ) = 15.7
[𝐻2 𝑂]
[𝐻2 𝑂] 55.49
𝐻2 𝑂 ⇌ 𝐻 + + 𝑂𝐻 −
𝐾𝑎 = [H + ][OH − ] = K w = 10−14 → 𝑝𝐾𝑎 = − 𝑙𝑜𝑔(𝐾𝑎 ) = 14
Both are accepted as correct answer if accompanying work is correct
c. A very important reaction in organic synthesis is the Fischer Esterification, through
which an alcohol and a carboxylic acid are joined to form an ester, releasing water.
An example of a Fischer esterification is the reaction between ethanol (CH3 CH2 OH)
and acetic acid (CH3 COOH) as shown below:
CH3 CH2 OH + CH3 COOH ⇄ CH3 COOCH2 CH3 + H2 O
K was determined experimentally to be 4.0 at 25 ℃.
i. Assume the reaction takes place in a non-aqueous, inert solvent. If the
initial concentrations of acetic acid and ethanol are both 1.0 M, what are
the concentrations of ethyl acetate and water in the final product mixture?
Using an ICE table and substituting into the expression for K:
𝐾=
𝑥2
,
(1−𝑥)2
where x is the final concentration, in M, of water and ethyl acetate.
𝑥
2
= √𝐾 = 2 → 𝑥 =
1−𝑥
3
(This question continues on the next page.)
8
Name (Last, First):_________________________ ID Number: __________________________
ii. After the reaction achieves equilibrium, anhydrous magnesium sulfate
(which absorbs water) is added to the solution. In which direction would
the equilibrium shift upon this addition? Explain on the lines provided
below.
Magnesium sulfate would remove water, decreasing concentration of the products. The
equilibrium would shift to the right (direction of the products) in accordance with Le Chatelier’s
Principle. _____________________________________________________________________
d. Suppose you react 1.0 mole of ethanol with 1.0 mole of acetic acid in 1.0 L of an
inert, non-aqueous solvent. What mass (in g) of magnesium sulfate you should add if
you want to isolate 75.0 g of ethyl acetate, given that one magnesium sulfate reacts
with 7 molecules of water.
At initial equilibrium, we can set up another ice table for addition of magnesium sulfate:
𝐶𝐻3 𝐶𝐻2 𝑂𝐻 + 𝐶𝐻3 𝐶𝑂𝑂𝐻 ⇄ 𝐶𝐻3 𝐶𝑂𝑂𝐶𝐻2 𝐶𝐻3 + 𝐻2 𝑂
[i]
0.33
0.33
0.67
0.67
[c1]
-y
[c2]
-x
-x
+x
+x
[e]
0.33 – x
0.33 – x
0.67 + x
0.67 – y + x
Thus,
(0.67 + 𝑥) ∗ (0.67 + 𝑥 − 𝑦)
=𝐾=4
(0.33 − 𝑥)2
This equation has two variables – to solve it, we need another equation. This can be
obtained by knowing that we need
𝑔
75.0 𝑔 ÷ 88.11
= 0.851 𝑚𝑜𝑙 𝑒𝑡ℎ𝑦𝑙 𝑎𝑐𝑒𝑡𝑎𝑡𝑒
𝑚𝑜𝑙
Therefore, 0.67 + 𝑥 = 0.851 → 𝑥 = 0.184 𝑚𝑜𝑙 and the equation becomes
0.854 ∗ (0.854 − 𝑦)
= 4 → 𝑦 = 0.754 𝑚𝑜𝑙
0.1462
If we want to remove 0.754 moles of water, we need
1 𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
𝑔 𝑀𝑔𝑆𝑂4
0.754 𝑚𝑜𝑙 𝐻2 𝑂 ∗
∗ 120.37
= 12.97 𝑔
7 𝑚𝑜𝑙 𝐻2 𝑂
𝑚𝑜𝑙 𝑀𝑔𝑆𝑂4
9
Name (Last, First):_________________________ ID Number: __________________________
6. Pyrophosphate has the chemical formula P2 O4−
7 (abbreviated PPi) and is very important in
biochemical processes. The hydrolysis of PPi to two inorganic hydrogen phosphate
molecules is an extremely exergonic reaction. This means that the hydrolysis of PPi can
be coupled to other reactions in order to drive them.
a) In PPi, the two phosphate atoms are linked by a single oxygen atom. The other
oxygen atoms are bonded to the phosphate atoms. Draw a valid Lewis structure that
obeys the octet rule for all atoms. Draw a resonance structure that does not obey the
octet rule for either phosphorus atom. Indicate all lone pairs and non-zero formal
charges on each structure. Clearly label one structure as “obeys octet rule,” and one
structure as “does not obey octet rule.”.
The Lewis structure on the left obeys the octet rule. The Lewis structure on the right
does not obey the octet rule.
(This question continues on the next page.)
10
Name (Last, First):_________________________ ID Number: __________________________
b) Give a balanced chemical reaction for the reaction of one molecular ion of PPi
(P2 O4−
7 ) with one water molecule to form two molecules of hydrogen phosphate.
𝑃2 𝑂74− (𝑎𝑞) + 𝐻2 𝑂(𝑙) → 2𝐻𝑃𝑂42− (𝑎𝑞)
c) The activation of fatty acids by Coenzyme A (CoA − SH) and ATP is given by the
following reaction:
Fatty Acid + ATP + CoA − SH ⇄ Acyl − CoA + AMP + PPi
kJ
∆G° = −15.0 mol for this reaction at 37 ℃. Calculate the equilibrium constant for
this reaction at 37 ℃.
∆𝐺° = −𝑅𝑇𝑙𝑛𝐾𝑒𝑞
𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
𝑒 (0.008314 𝑚𝑜𝑙∙𝐾)(310 𝐾)
15.0
𝐾𝑒𝑞 = 𝑒
−∆𝐺°
𝑅𝑇
=
= 𝑒 5.820 = 336.96
d) Write the overall reaction when activation of fatty acids is coupled with the
hydrolysis of PPi.
𝐹𝑎𝑡𝑡𝑦 𝐴𝑐𝑖𝑑 + 𝐴𝑇𝑃 + 𝐶𝑜𝐴 − 𝑆𝐻 + 𝐻2 𝑂 ⇄ 𝐴𝑐𝑦𝑙 − 𝐶𝑜𝐴 + 𝐴𝑀𝑃 + 2𝐻𝑃𝑂42−
kJ
e) The hydrolysis of PPi has a ∆G° = −19.2 mol at 37 ℃. Calculate the equilibrium
constant of the overall reaction you gave in part (d).
∆𝐺°𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = −15.0
𝑘𝐽
𝑘𝐽
𝑘𝐽
− 19.2
= −34.2
𝑚𝑜𝑙
𝑚𝑜𝑙
𝑚𝑜𝑙
∆𝐺° = −𝑅𝑇𝑙nK eq
K eq = e
−∆G°
RT
=
kJ
34.2
mol
kJ
(0.008314
)(310 K)
mol∙K
e
= e13.27 = 5.79 × 105
(End of Exam)
11