Prepare Aspirin Solutions m 1. 2. Record the concentration of the NaOH solution.
ID: 956357 • Letter: P
Question
Prepare Aspirin Solutions m 1. 2. Record the concentration of the NaOH solution. it has been standardized by your TA 0.051 Simultaneously prepare two of the following solutions obtain the mass of an aspirin tablet and add it toan Erlenmeyer flask . Obtain the mass of an aspirin tablet and ad it to an Erlenmeyer flask)o.b59 )03Ag 2)0M5 j 3. 4. Using a 5Oml volumetric pipette, add S0.00mL of NaOH to the tablet in the flask 5. Heat the flasks in a hot water bath (near boiling) for 5 minutes. (big beaker with water in it). While you are wating standardize the HCL Standardize HC OIMHCL t to anErlenmeyer flask. 0S 6. Obtain 0.0s to 0.1 g of Na,Co, (record the mass) and add it to an Erlenmeyer flask 7. Add approximately 25 ml of deionized water and 3 drops of phenol red to the flask to dissolve the Na CO, 8. Prepare a burette with the Ha solution-rinse burette and tip, fill burette, run bubbles out of tip, record initial volume. Init 50m 9. Slowly add the HCl to the flaks with the Na,cO, until the end point is reached. Record the final volume of HCL. 10. Repeat (add more HCI to burette if necessary before you begin). ·added 9.5ml HCI Odded G.1ml H o ci Titrate Aspirin Tablets 11. Refill the burette with the HCI solution 12. Add a few drops of phenol red to the flasks with the aspirin. 13. Titrate your aspirin samples in the flasks with the HCl solution. added 15.20 C 2) added 15. u HC
Explanation / Answer
Calculate the % by mass of aspirin in the tablet
For Trial-1
Weight of aspirin powder taken: 0.369g
Concentration of NaOH: 0.0951 M
Titration figure: 15.20 ml = 15.20 cm3
0.0951 mol dm3 * 0.0152 dm3 = 0.00144552 mol NaOH.
1 mol NaOH neutralises 1 mol acetylsalicylic acid.
0.369 g powder contains 0.00144552 mol aspirin.
mol = mass / molar mass so mass = mol * molar mass.
molar mass aspirin = 180 g mol-1 ( same as glucose!).
mass aspirin = 180 g mol-1 * 0.00144552 mol =0.2601 g.
% aspirin = (0.2601 / 0.369) * 100 = 70.48 %.
For Trial -2
Weight of aspirin powder taken: 0.365g
Concentration of NaOH: 0.0951 M
Titration figure: 15.40 ml = 15.40 cm3 ( as the image is not clear i am assuming that it is 15.40 ml )
0.0951 mol dm3 * 0.0154 dm3 = 0.00146454 mol NaOH.
1 mol NaOH neutralises 1 mol acetylsalicylic acid.
0.365 g powder contains 0.00146454 mol aspirin.
mol = mass / molar mass so mass = mol * molar mass.
molar mass aspirin = 180 g mol-1 ( same as glucose!).
mass aspirin = 180 g mol-1 * 0.00146454 mol =0.263 g.
% aspirin = (0.263 / 0.365) * 100 = 72.05 %.