Astudent obtained a 0.4513g sample containing aspirin. this sample was analyzed through a titration with naoh and phenolphthalein was used as the indicator. the endpoint (ph around 9) for the reaction was reached after the addition of 15.22 ml of 0.1105m naoh. molar mass for aspirin = 180.16g/mol.
calculate the % purity for the sample. show all your work
*hint: aspirin reacts with naoh in a 1: 1 ratio.
% purity = (actual moles of aspirin in sample based on titration÷ maximum moles of aspirin in sample) × 100%

Answer

The purity of the sample is 67.14 %

Explanation:

The titration reaction is as follows:

NaOH + aspirin-H → Na⁺ + aspirin⁻ + H₂O

When no more aspirin-H is left, the addition of more NaOH raises the pH and the color of the indicator turns, in this case to a pink color. This is the reaction at the endpoint that indicates that no more aspirin-H is left:

NaOH + aspirin⁻ → Na⁺ + aspirin⁻ + OH⁻

Then, the moles of NaOH added until the turn of the indicator must be equal to the number of moles of aspirin present in the solution since NaOH reacts with aspirin in a 1:1 ratio.

Then:

moles of aspirin in the solution = moles of added NaOH

moles of aspirin in the solution = Concentration of NaOH * volume

moles of aspirin in the solution = 0.1105 mol/l * 0.01522 l = 1.682 x 10⁻³ mol

Knowing the molar mass of aspirin, we can calculate the mass of aspirin present in the solution:

1.682 x 10⁻³ mol aspirin *(180.16 g / mol) = 0.3030 g.

Since the sample contained 0.4513 g, the percent of aspirin in the sample will be: 0.3030 g * (100 % / 0.4513 g) = 67.14 %

We will get the same result if we convert the mass of the sample to mol and calculate the purity using moles instead of mass:

moles of aspirin in the sample:

0.4513 g * ( 1 mol / 180.16g) = 2.505 x 10⁻³ mol aspirin

The purity will be then:

1.682 x 10⁻³ mol * ( 100 % / 2.505 x 10⁻³ mol) = 67.14 %

it is stable as it is in group 0 which have full outer shells of electrons