How much of solution to reach the endpoint of titration?

[ammora313]

Homework Statement

You prepare your standardization solution by weighing out 0.0750 grams of standard ascorbic acid (176.12 g/mol) and dissolving it in 50.00 mL of water. You then add ten drops of starch indicator. How many milliliters of 0.015M triiodide solution will you need to reach the endpoint of the titration?

Homework Equations

M1V1=M2V2

The Attempt at a Solution

[(0.0750 g X 176.12 g/mol) / (50 mL water) V1]= [(0.015 M) (V2)]

How do I solve this, seeing as I have 2 unknowns (V1 and V2)? Am I doing it wrong?

 

[Borek]

Your problem is in applying blindly M1V1=M2V2 equation. It is not needed here, have you read the page on the titration calculation I linked to in another thread?

 

[ammora313]

The formula for the reaction that takes place is:
C₆H₈O₆ + I₃^- +H₂O $\leftrightarrow$ C₆H₆O₆ +3I^- +2H₃O⁺

So relevant equations are C=n/V or n=CV

0.0750 g x 1mol/176.12 g/mol = 4.25846x10^-4 mol ascorbic acid

4.25846x10^-4 mol ascorbic acid/ 0.050 L water = 8.51692x10^-3 M abscorbic acid.

I'm not sure where to go from here.

 

[Borek]

You don't need concentration of the ascorbic acid. You have to calculate how many moles of triiodide will react with the ascorbic acid. Usually you will use n=CV to calculate number of moles of ascorbic acid, but you are given mass, so it is enough to convert it to number of moles (which you already did).

Next step is a simple stoichiometry - how many moles of triiodide will react with the 4.258x10^-4 moles of ascorbic acid?

After that - what volume of triiodide solution contain this amount of triiodide?

 

[ammora313]

Ok I figured it out, thanks for being patient with me.
So since the stoichiometry is one to one, it takes 4.258 x 10-4 mol of triiodide to react with the same number of moles of ascorbic acid. then using v=n/c, it takes 28.3 mL of triiodide.

 

[Borek]

I got 28.4 mL, but it is close enough to show you are on the right track.