Pre-lab Questions 1. A student who is performing this experiment pours an 8.50 m
ID: 706541 • Letter: P
Question
Pre-lab Questions 1. 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 HCI solution. 12.00 mL of the HCl solution is needed to reach the endpoint of the titration. Calculate the value of Kap for borax at temperature T. (Answer: 0.176 Show your work-) Here is a suggested procedure for doing this calculation: (a) Calculate the number of moles of HCI that were added during the titration (b) Use reaction (2rim the lab manual to relate the number of moles of HCI to the number of moles of tetraborate ion in the 8.50 mL sample Calculate the concentration of tetraborate ions in the 8.50 mL sample. Use reaction (1) in the lab manual to relate the concentration of tetraborate ions to the concentratiou of sodium ions. (c) (d) (e) Use the concentrations of tetraborate ions and sodium ions to calculate the equilibrium constant (Ksp) at temperature T for reaction (1) in the lab manual. 2. Do you expect ASo for the dissolution of borax to be a positive or negative number? Explain your reasoning. Additional Questions (for the finished laboratory report) 1. Calculate K for Reaction (I) at 65 °C and at 25 °C, using your values of AHo and AS 2. Calculate the solubility of borax in grams per liter at 65 °C and at 25 C, using your values of K at that temperature.Explanation / Answer
(a) 0.500 M HCl means 0.5 moles are present in one litre solution, hence 12.0 ml HCl contains
0.500/1 multiply by 12.0 ml=6 mmol
(b) using M1V1=M2V2
M1* 8.50= 0.006*12.0 =0.00847M
(c) moles of borax in 8.50 ml solution = 0.00847/1*8.50 =7.2 mmol
(d) Na2B4O7 . 10 H2O(s) = 2Na+ (aq) + B4O5(OH)42- (aq) + 8H2O(l)
therefore the concentration of sodium is twice of borate ion
(e) Ksp = 4 [B4O5(OH)42- ]3
because concentration of sodium also include in above equation.
Ksp= 0.500 mol H+ x 0.012.0 L x 1 mol B4O5(OH)42- L/ 2 mol H+ = 0.35 M B4O5(OH)42- /0.00850L
ln Ksp = ln (4(0.35)3 ) = 0.177
(2) Entropy of the solution after mixing is positive because number of ions increase in the solution after mixing hence creating more randomness. Therefore increase the entopy.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.