Cryolite In Cryolite (sodium hexafluoroaluminate) the six fluoride ions are bound with decreasing strength. The consecut

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

Cryolite In Cryolite (sodium hexafluoroaluminate) the six fluoride ions are bound with decreasing strength. The consecut

Post by answerhappygod »

Cryolite In Cryolite Sodium Hexafluoroaluminate The Six Fluoride Ions Are Bound With Decreasing Strength The Consecut 1
Cryolite In Cryolite Sodium Hexafluoroaluminate The Six Fluoride Ions Are Bound With Decreasing Strength The Consecut 1 (25.92 KiB) Viewed 54 times
need 3 answerss asap pleasee, if you can answer only 1 , answer
13
Cryolite In Cryolite (sodium hexafluoroaluminate) the six fluoride ions are bound with decreasing strength. The consecutive constants, given as log K, are respectively 6.13, 5.02, 3.85, 2.74, X. 0.47, for n=1,2,3,4,5,6, respectively) 11) (6) Give the chemical formula and structure of sodium hexafluoroaluminate. 12) (9) Give estimates of lower and upper limits for log Ks = X using other given log K values (log Ke and log Ka) The K-constants were determined from a pH-measurement Ks was determined from the following single experiment and by knowledge of the other constants A solition including aluminum nitrate (c(AP)=1.00-10-³M), nitric acid (c(H+) = 9:10 10-M), potassium fluoride (c(F-) = 28.62 10-³M) was used. The pH of this solution was measured to 321 pka for hydrofluoric acid was measured in a corresponding solution to 294 13) (12) Calculate log Ks (Hint i) Account for protons and find the protonated- and the free fluoride actual concentrations ([HF]and [F] in unit M), i) determine the average ligand number and in) use the most simple approximation to determine log Ks, and iv) make a correction to log K's accounting for the first determined log K's being at least too small or too big
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply