h.) The solution at the end of the synthesis of the coordination complex was ess
ID: 997776 • Letter: H
Question
h.) The solution at the end of the synthesis of the coordination complex was essentially an aqueous solution of potassium oxalate. Ammonium oxalate is an alternative salt that could have been used in this synthesis. Suppose a technician prepared a 0.200 M solution of ammonium oxalate in this synthesis. What would be the pH of this solution?
j.) Carbon dioxide can be converted to carbon monoxide according to the following equilibrium.
Suppose we introduce 0.0350 moles of the carbon dioxide formed from the coordination compound synthesis reaction into a 1.50 L Erlenmeyer flask.
i.) What would be the PCO at equilibrium?
ii.) What would be the corresponding value of Kc?
k.) The carbon monoxide from the previous reaction can be used to synthesize phosgene (a toxic gas used in some organic syntheses).
The following data were obtained in a kinetic study of its formation:
i.) Write the rate law for the formation of phosgene.
ii.) Calculate the average value of the rate constant.
l.) Oxalate complexes like the one synthesized in this reaction can be oxidized using Mn3+ ions in solution. The kinetics of this reaction have been studied, and the proposed mechanism is
i.) A student studies the course of the oxidation reaction by measuring the absorbance of the oxalate over time at 580 nm. In the study, Mn3+ ions are
in large excess so that the concentration of Mn3+ remains relatively constant over time. The student generates the following graphs.
Absorbance vs. Time
ln(absorbance) vs. Time
1/(absorbance)
It is known that the reaction is first order in terms of Mn3+. Using the graphs above, complete the rate law for this oxidation reaction.
ii.) Other than temperature, describe two factors that could influence the rate of this reaction.
iii.) Use the kinetics data below to determine the energy of activation for the first step in the reaction mechanism.
ii.) What would be the minimum change in entropy necessary for this reaction to be spontaneous?
n.) During the filtration process of this reaction, cold water and cold ethanol are used to wash the product. Use the information provided in the table below in regard to water, ethanol, and organic molecules similar in size to ethanol to answer the following questions.
H H
/
H - C - C - H
/
H H
H H
/
H - C - C - OH
/
H H
H
H - C - C = O
/
H H
H
H - C - C = O
/
H OH
i.) Indicate which of the molecules in the chart above would be soluble in water. Justify your answer.
ii.) Explain the observed trends in boiling points in terms of intermolecular forces.
i.) Construct a functioning galvanic cell made of copper, chromium, 1.00 M solutions of copper(II) nitrate and chromium(II) nitrate, and label the
diagram below to represent this galvanic cell.
ii.) Indicate the proper voltage in the cell diagram.
iii.) Indicate the flow of ions in the salt bridge.
v.) Indicate which electrode will lose mass and which will gain mass as the galvanic cell functions. Justify your answer.
vi.) Suppose you had 0.50 M solutions, rather than 1.00 M solutions. Would the voltage of the nonstandard cell be the same, less than, or greater than
the voltage of the standard cell? Justify your anwer.
vii.) How much time would be required to deposit 1.0 g of chromium metal on a lamp base from an aqueous Cr3+ solution?
Trial Initial [CO] (mole/L) Initial [Cl2] (mole/L) Initial Rate (mole/Lsec) 1 1.00 0.100 1.29 x 10-29 2 0.100 0.100 1.33 x 10-30 3 0.100 1.00 1.30 x 10-29 4 0.100 0.0100 1.32 x 10-31Explanation / Answer
Answer-h)
Ammonium oxalate is a salt of weak organic acid Oxalic acid and weak base ammonia NH3.
Ammonium oxalate when dissolved in water undergo hydrolysis reaction to form NH4+ and oxalate ions.
C2O4(NH4)2 ----------> C2O42– + 2 NH4+
0.2M 0.2 0.4 M
Each ion, C2O42– and NH4+ hydrolyzes water as,
A) C2O42– + 2H2O <--------------> H2C2O4 + 2OH-
Initial conc. 0.2 M 0 M 0M
Change -X +X +2X
Eqm conc. (0.2 – X) ‘X’ M ‘2X’ M
pKb = 10.19 implies Kb = 10–pKb = 10–10.19 = 6.46 x 10-11
To determine [OH-], write the equilibrium expression
kb = [OH]2[H2C2O4]/ [C2O42–]
6.46 x 10-11 = (2X)2 x (X) / (0.2 – X)
X <<< 0.2 hence small concentration assumption holds hence 0.2 – X = 0.2
6.46 x 10-11 = 4X3 / 0.2
X3 = 0.2 x 6.46 x 10-11 /4
X3 = 3.23 x 10-12
X = 1.48 x 10-4 M ……………… (on taking cube roots)
From ICE table,
[OH–] = 2 x X = 2 x 1.48 x 10-4 = 2.96 x 10-4 M
[OH–] = 2.96 x 10-4 M …………………………….(1)
B) NH4+ + H2O <--------------> NH4OH + H+
Initial conc. 0.4 M 0 0
Changes -X +X +X
Eqm. Conc. (0.4 – X) M ‘X’ M ‘X’ M
pKa = 9.25 implies Ka = 10–pKa = 10–9.25 = 5.62 x 10-10
Ka = [NH4OH][H+]/[NH4+]
5.62 x 10-10 = (X)(X) / (0.4 – X)
Small concentration assumpition holds,
5.62 x 10-10 = X2 / 0.4
X2 = 0.4 x 5.62 x 10-10
X2 = 2.25 x 10-10
X = 1.5 x 10-5 M
From ICE table,
[H+] = X = 1.5 x 10-5 M …………………….(1)
From equation (1) and (2) we have,
[OH–] > [H+]
It means [OH–] will neutralize [H+] and excess of [OH–] will make solution basic,
Let us calculate excess [OH–]
Excess [OH–] = [OH–] – [H+]
Excess [OH–] = (2.96 x 10-4) – (1.5 x 10-5)
Excess [OH–] = 2.81 x 10–4
pOH = – log [OH-]
pOH = – log (2.81 x 10–4)
pOH = 3.55
Hence pH = 14 – pOH
pH = 14 – 3.55
pH = 10.45
Hence pH of 0.2 M Ammonium oxalate is 10.45.
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