Question

What was the percent by mass of aspirin in the tablets?

I dissolved four aspirin tablets with a total mass of 1.427 g in water and added 50.00 mL of 0.500 mol/L sodium hydroxide solution. Then I titrated the excess sodium hydroxide to a phenolphthalein end-point with 31.92 mL of 0.289 mol/L hydrochloric acid.

Answer 1

WARNING! Very long answer! The aspirin was 99.6 % pure.

Explanation:

Step 1. Start with the balanced equation.

`underbrace("C"_9"H"_8"O"_4)_color(red)("aspirin") + "2OH"^"-" → underbrace("C"_7"H"_5"O"_3^"-")_color(red)("salicylate") + underbrace("C"_2"H"_3"O"_2^"-")_color(red)("acetate") + "H"_2"O"`

For simplicity, let's rewrite the equation as

`"Asp + 2OH"^"-" → "Sal"^"-" + "Ac"^"-" + "H"_2"O"`

Step 2. Calculate the moles of `"NaOH"` used.

`"Moles NaOH" = "0.050 00" color(red)(cancel(color(black)("L NaOH"))) × "0.500 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.025 00 mol NaOH"`

Step 3. Calculate the moles of `"HCl"` used in the back-titration.

`"Moles of HCl" = "0.031 92" color(red)(cancel(color(black)("L HCl"))) × "0.289 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.009 225 mol HCl"`

Step 4. Calculate the moles of excess `"NaOH"`

The only base remaining after the reaction is the excess base that has not reacted with the aspirin.

You are doing a strong acid-strong base titration.

Once the NaOH is neutralized, the solution also contains the salts of the weak acids salicylic acid and acetic acid.

The salts of weak acids are stronger bases.

Thus phenolphthalein (which changes colour at pH 9) is a good choice for the indicator.

The equation for the titration is

`"NaOH" + "HCl" → "H"_2"O" + "NaCl"`

`"Moles of excess NaOH" = "0.009 225" color(red)(cancel(color(black)("mol HCl"))) × "1 mol NaOH"/(1 color(red)(cancel(color(black)("mol HCl")))) = "0.009 225 mol NaOH"`

Step 5. Calculate the moles of `"NaOH"` used in the reaction.

`"Original moles = moles reacted + excess moles"`

`"Moles reacted" = "original moles - excess moles" = "(0.02500 - 0.009 225 mol) NaOH" = "0.015 775 mol NaOH"`

Step 6. Calculate the moles of aspirin.

`"Moles of Asp" = "0.015 78" color(red)(cancel(color(black)("mol NaOH"))) × "1 mol Asp"/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.007 888 mol Asp"`

Step 7. Calculate the mass of aspirin.

`"Mass of Asp" = "0.007 888" color(red)(cancel(color(black)("mol Asp"))) × "180.16 g Asp"/(1 color(red)(cancel(color(black)("mol Asp")))) = "1.421 g Asp"`

Step 8. Finally (whew!), calculate the purity of the aspirin.

`"Percent by mass" = "mass of pure Asp"/"mass of impure Asp" × 100 % = (1.421 color(red)(cancel(color(black)("g"))))/(1.427 color(red)(cancel(color(black)("g")))) × 100 %`
`= 99.6 %`