Question: Do you expect ∆S° for the dissolution of borax to be a positive or negative number? Explain your reasoning.

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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
Transcribed Image Text:Home Study tools My courses Мy books Career Life 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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