(a) Interpretation: The titration of the aqueous solution of amount 50.0mL contains 2.500g of the unknown base which is represented as RNH. The equivalence point of the titration is reached by the 0.925M of HCl of amount 59.90mL. In the titration of the second experiment 50.00mL of the unknown base is used which is identical to the base which is used in the previous experiment. In the solution of the second experiment, 0.925M of HCl of amount 29.95 mL is used. The pH value of the solution is 10.77 after the addition of the HCl. The molar mass of the unknown base is to be determined. Concept introduction: The molarity of a solution is calculated as follows: Molarity= number of moles of solute volume The number of moles can be calculated as follows: number of moles= mass molar mass
(a) Interpretation: The titration of the aqueous solution of amount 50.0mL contains 2.500g of the unknown base which is represented as RNH. The equivalence point of the titration is reached by the 0.925M of HCl of amount 59.90mL. In the titration of the second experiment 50.00mL of the unknown base is used which is identical to the base which is used in the previous experiment. In the solution of the second experiment, 0.925M of HCl of amount 29.95 mL is used. The pH value of the solution is 10.77 after the addition of the HCl. The molar mass of the unknown base is to be determined. Concept introduction: The molarity of a solution is calculated as follows: Molarity= number of moles of solute volume The number of moles can be calculated as follows: number of moles= mass molar mass
The titration of the aqueous solution of amount 50.0mL contains 2.500g of the unknown base which is represented as RNH. The equivalence point of the titration is reached by the 0.925M of HCl of amount 59.90mL. In the titration of the second experiment 50.00mL of the unknown base is used which is identical to the base which is used in the previous experiment. In the solution of the second experiment, 0.925M of HCl of amount 29.95 mL is used. The pH value of the solution is 10.77 after the addition of the HCl.
The molar mass of the unknown base is to be determined.
Concept introduction:
The molarity of a solution is calculated as follows:
Molarity=number of moles of solutevolume
The number of moles can be calculated as follows:
number of moles=massmolar mass
Interpretation Introduction
(b)
Interpretation:
The constant of equilibrium of the base, Kb is to be determined.
Concept introduction:
The formulas which will be used are-
At the mid-way equivalence point
pKa=−logKa
The relation between equilibrium constant of acid and base -
Kb=KwKa
Interpretation Introduction
(c)
Interpretation:
The constant equilibrium of acid Ka for RNH2 + is to be determined.
(c) The following data have been obtained for the hydrolysis of sucrose, C12H22O11, to
glucose, C6H12O6, and fructose C6H12O6, in acidic solution:
C12H22O11 + H2O → C6H12O6 + C6H12O6
[sucrose]/mol dm³
t/min
0
0.316
14
0.300
39
0.274
60
0.256
80
0.238
110
0.211
(i) Graphically prove the order of the reaction and determine the rate constant of the
reaction.
(ii) Determine the half-life, t½ for the hydrolysis of sucrose.
(III) adsorbent
(b) Adsorption of the hexacyanoferrate (III) ion, [Fe(CN)6] ³, on y-Al2O3 from aqueous
solution was examined. The adsorption was modelled using a modified Langmuir
isotherm, yielding the following values of Kat pH = 6.5:
(ii)
T/K
10-10 K
280
2.505
295
1.819
310
1.364
325
1.050
Determine the enthalpy of adsorption, AadsHⓇ.
If the reported value of entropy of adsorption, Aads Se = 146 J K-1 mol-1 under the above
conditions, determine Aads Gº.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell