Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question: Do you expect ∆S° for the dissolution of borax to be a positive or negative number?
Explain your reasoning.
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A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated
cylinder after the borax solution had cooled to a certain temperature T. The student rinses the sample into a small beaker using
distilled water, and then titrates the solution with a 0.500 M HCl solution. 12.00 mL of the HCI solution is needed to reach the
endpoint of the titration.
Calculate the value of Ksp for borax at temperature T.
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B405(OH)42- -
+ 2 HCI + 3 H2O => 4 B(OH)3
Moles of HCl = volume x concentration of HCI
= 12.00/1000 x 0.500 = 0.006 mol
Moles of B405(OH)4 = 1/2 x moles of HCl
= 1/2 x 0.006 = 0.003 mol
= moles/volume of B405(OH)42-
= 0.003/0.0085 = 0.35294 M
Na2B407.10H20 <=> 2 Na+ + B405(OH)4
+ 8 H20
[Na+] = 2 x [B405(OH)4²]
= 2 x 0.35294 = 0.70588 M
Ksp = [Na+]°[B405(OH)4²]
%3D
= 0.705882 x 0.35294
= 0.176](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F793c7cf3-f790-48f6-9e1d-49d8157e667b%2F5f2091c5-b4d4-403b-b050-031acbcf679f%2Ft9hh8xc_processed.png&w=3840&q=75)
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A student who is performing this experiment pours an 8.50 mL sample of the saturated borax solution into a 10 mL graduated
cylinder after the borax solution had cooled to a certain temperature T. The student rinses the sample into a small beaker using
distilled water, and then titrates the solution with a 0.500 M HCl solution. 12.00 mL of the HCI solution is needed to reach the
endpoint of the titration.
Calculate the value of Ksp for borax at temperature T.
Expert Answer
Chemtutor answered this
Was this answer helpful?
O 20
23,336 answers
B405(OH)42- -
+ 2 HCI + 3 H2O => 4 B(OH)3
Moles of HCl = volume x concentration of HCI
= 12.00/1000 x 0.500 = 0.006 mol
Moles of B405(OH)4 = 1/2 x moles of HCl
= 1/2 x 0.006 = 0.003 mol
= moles/volume of B405(OH)42-
= 0.003/0.0085 = 0.35294 M
Na2B407.10H20 <=> 2 Na+ + B405(OH)4
+ 8 H20
[Na+] = 2 x [B405(OH)4²]
= 2 x 0.35294 = 0.70588 M
Ksp = [Na+]°[B405(OH)4²]
%3D
= 0.705882 x 0.35294
= 0.176
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