Determining Volume of Base for ASA Experiment

[Mary1910]

1. Homework Statement .
This question is based on an experiment to determine the mass of ASA.
Information from the experiment :
• ASA is a weak acid with a Ka value of 3.2 x 10^-4.
• An ASA tablet has an approximate mass of 500 g.
• The experiment used a titration with a standardized 0.100 mol/L NaOH.

The balanced equation is HC9H7O4(aq) + NaOH(aq) -> NaC9H7O4(aq) + H2O(l)

a) What volume of base will you substitute into the formula when you calculate the number of moles of base used?

2. Relevant equations.

n=m/M

v=n/c

3. The attempt at the solution

First determine molar mass of HC9H7O4
=180.17 g/mol

n=m/M
=(0.500g ) / (180.17g/mol)
=2.78 x 10^-3 mol

v=n/c
=(2.78 x 10^-3) / (0.1mol/L)
=0.0278 mL
=27.8 L

Could someone let me know if I determined the volume of the base correctly? Thanks.

 

[Borek]

=(2.78 x 10^-3) / (0.1mol/L)
=0.0278 mL
=27.8 L

You mixed up units in the last two lines, but - assuming you meant L first, mL second - your calculations look OK.

But it wouldn't hurt to write more precisely what you are calculating. I am guessing you are checking what should be approximate volume of the base required to titrate ASA from a single tablet, but it doesn't necessarily follow from your post.

 

[Mary1910]

Thanks, and sorry for my lack of clarity.