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# I need help with a titration problem, thanks so much?

Aspirin (C9H8O4) is an acid which can be titrated with a base to determine purity. If an aspirin tablet weighing 0.732 g is titrated with standardized 0.1345 M KOH, the endpoint is reached after 28.24 mL of KOH have been added. What is the percentage of aspirin in the tablet? The acid/base ratio is 1:1.

### 3 Answers

- pisgahchemistLv 76 months agoFavorite Answer
Aspirin titration .....

Mass of tablet: 0.732g .... which includes aspirin and binder

HC9H7O4 + OH- --> C9H7O4^- + HOH

???g ........ 28.24mL

.................. 0.1345M

To find the mass of aspirin in the tablet, use the unit-factor method for a 1-line setup and one chain calculation.

0.02824L x (0.1345 mol OH- / 1L) x (1 mol HC9H7O4 / 1 mol OH-) x (180.158g HC9H7O4 / 1 mol HC9H7O4) = 0.6843g HC9H7O4

Percent aspirin by mass = 0.6843g / 0.732g x 100 = 93.5% ............ the answer is expressed to three significant digits

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- Dr WLv 76 months ago
use dimensional analysis...

.. 28.24mL KOH... 0.1345 mmol KOH.. .1 mmol aspirin.... 180.2mg aspirin.. ... 1g

----- ----- ---- ---- x ---- ----- ----- ---- ---- x ---- ---- ----- ----- x --- --- ---- ---- ---- x ----- ---- x 100%

.. 0.732g tablet... ... .. 1 mL KOH... ... ... 1 mmol KOH.. ... . .1 mmol aspirin.. 1000mg

.. = 93.5%

just enter

.. 28.24 / 0.732 * 180.2 / 1000 * 100 =

in your calculator and round to 3 sig figs

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- davidLv 76 months ago
28.24 X 0.1345 /1000 = 0.003798 mole aspirin

0.003798 X 180 g/mole = 0.6837 g aspirin

0.6837/0.732 X 100 = 93.4%

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