Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A 7.55 mg sample of a weak base (FW - 144.44), pKa = 8.200, is dissolve in a 45.

ID: 952242 • Letter: A

Question

A 7.55 mg sample of a weak base (FW - 144.44), pKa = 8.200, is dissolve in a 45.0 mL of a with a pH of 8.00. The resulting solution is extracted four (4) times with 10.00 mL of an organic solvent. The partition coefficient (K) for the base between the organic solved and the aqueous solution is 11.2. What is the distribution coefficient, D of the base in this extraction system? 11.2 5.60 28.95 4.33 How many milligrams of the weak base (total) will remain in the aqueous solution after the first time of extraction? 3.85 mg 2.16 mg 0.51 mg 0.051 mg How many milligrams of the weak bease (total) will remain in the aqueous solution after the four time of extraction? 3.85 mg 2.16 mg 0.51 mg 0.051 mg When a standard solution containing 234 mg of butanol and 312 mg of hexanol in 10.0 mL was separated by gas chromatography, the relative peak areas were butanol: hexanol - 1.00.1.45. A hexanol unknown sample containing 112 mg butanol spiked (as internal standard) was also separated with the same GC method, and the relative peak areas were butanol: hexanol= 97:161. What mass of hexanol was in the unknown sample solution? 34.9 mg 93.5 mg 202 mg 171 mg

Explanation / Answer

37) the distribtion coeffeceint of the base will be

D = [Base] in organic phase / [Base] water + [Dissociated base] water ......(1)

Let base is BOH , it will dissociate as

BOH --> b+ + OH-

Kb = [B+] [ OH-] / [BOH]undissociated

[B-] = Kb X [BOH]w / [OH-] .........(2)

substitute (2) in (1)

D = [B]org / [BOH]w + [B-]

D = [B]org / [BOH]w + Kb [BOH]w / [OH-]

D = [B]org [OH-]/ [BOH]w X [OH-] + Kb [BOH]w

D = [B]org [ OH-] / [BOH]w ( [OH-] + Kb)

D = Parition coeffeceint [OH-] / ([OH-] + Kb )

D = 11.2 X 10^-6 / 10^-6 + 1.58 X 10^-6

D = 11.2 / 2.58 = 4.34

38) Milligrams in the aqeuous phase after first time extraction

Qaq = (Vaq / D Vorg + Vaq)

Qaq = 45 / 10 X 4.33 + 45

Qaq = 0.509 mg Or 0.51 mg

39 ) Qaq = (Vaq / D Vorg + Vaq) ^4

Qaq = 0.509^4 = 0.0676

40)

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote