Determination of Aspirin by Spectrophotometric Analysis Pre-Laboratory Questions

Question

Determination of Aspirin by Spectrophotometric Analysis Pre-Laboratory Questions 1. A 0.40 g sample of sodium salicylate was dissolved and diluted to a final volume of 250 ml with deionized water. What is the concentration of this solution a. b. An 8.0 mL aliquot of the sodium salicylate solution above was diluted to a final volume of 100 mL with FeCl, solution. What is the concentration of this dilute sodium salicylate solution? · A calibration plot of absorbance vs. concentration was obtained, with the slope of the best-fit straight line as 1220 M-1. The absorbance of a dilute aspirin solution mixed with FeCl, was 0.72. What is the concentration of the dilute aspirin solution? a. A 5.0 mL aliquot of concentrated aspirin Brand X solution was diluted to a final volume of 100 mL. with FeCl3. This solutions concentration was determined to be 55 × 10-s M. w hat is the concentra- tion of the aspirin in the original aspirin Brand X solution? Prelab Questions: Determination of Aspirin by Spectrophotometric Analysis

Explanation / Answer

Q1a

[sodium salicylate] = mol of sodium salicylate / volume in ltier

mol of sodium salicylate = mass of sodium salicylate / MW of sodium salicylate = 0.40/160.11 = 0.002498

V = 250 mL = 0.25 L

[ sodium salicylate ] = 0.002498 /0.25 = 0.009992 M

b)

V = 8 mL are used..

diluted up to V = 100 mL

find concentation of diltued system

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

C1*V1 = C2*V2

C2 = C1*V1 /V2

C2 = (0.009992 )(8)/100

C = 0.00079936 mol per liter

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