HPLC = High-performance liquid chromatography Theophyline ( C7H8N4O2 / molar mas
ID: 508440 • Letter: H
Question
HPLC = High-performance liquid chromatographyTheophyline ( C7H8N4O2 / molar mass=180.164 g/mom) I had email saying "missing information" previously. If it is still lacking information, please let me know what element is missing. Thank you. A buffer solvent mixture for an HPLC procedure for determination of theophylline is required which has a final concentration of 20 mmol/L sodium acetate and 15% (v/v) methanol. a. Show calculations for preparation of 500 mL of the buffer solvent. The reagents available to make the solvent are 200 mmol/L sodium acetate, 100% methanol, and deionized water. (Round off answer to nearest mL.) b. Show calculations for preparation of one liter of 200 mmol/L sodium acetate from NaC2H302. (Round off answer to nearest 0.01 g.) c. Show calculations for preparation of two liters of 100 mmol/l sodium acetate from Nac2H302.3H20. (Round off answer to nearest 0.01 g.)
Explanation / Answer
a) We have reagents 200 mmol/L sodium acetate, methanol (100%) and water.
we need to prepare 500 mL of buffer solution with 20 mmol/L sodium acetate and 15% v/v methanol.
V1 = 500 mL ; M1 = 20 mmol/L
V2 = ? and M2 = 200 mmol/L
M1V1= M2V2
V2 = M1V1/M2 = 500 (mL)* 20 (mmol/L)/200 (mmol/L) = 50 mL
Now, we need to calculate how much volume of methanol need to take for making 15% v/v methanol solution.
(Vmethanol/Vsolution) * 100 = 15
Vmethanol = 15* Vsolution/100 = 15 * 500 mL/100 = 75 mL
Therefore, to make our required buffer solution, we need to take 50 mL of 200 mmol/L sodium acetate, 75 mL of methanol (100%) and remaining 375 mL water (i.e; 500 - 50 - 75 = 375 mL).
b) we need to prepare one liter of 200 mmol/L sodium acetate.
Sodium acetate mol. wt = 83.02 g/mol
Molarity = (wt.in g/mol.wt) * (1/volume in L)
0.200 M = (wt/ 83.02)*(1/1L)
wt = 0.2 * 83.02 = 16.60 g
we need to take 16.60 g of sodium acetate and dissolve in 1 L solution.
c) Two liters of 100 mmol/L sodium acetate form NaC2H3O3.3H2O
Molar mass of NaC2H3O2*3H2O is 136.08 g/mol
Molarity = (wt.in g/mol.wt) * (1/volume in L)
0.100 M = (wt/ 136.08)*(1/2L)
wt = 0.1 * 136.08 * 2 = 27.22g
we need to take 27.22g g of sodium acetate (NaC2H3O2*3H2O) and dissolve in 2 L solution.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.