Prior to performing a titration, three drops of phenolphthalein were added to 15.0mL of a tartaric acid solution. The solution was then titrated to the end point with 24.5 mL of 0.2297 M NaOH. What was the molarity of the tartaric aid solution?
I don't want just an answer please... I need to know how to do it!]
Thanks.
2007-01-30 06:53:36 · 2 answers · asked by woventemple 1 in Science & Mathematics ➔ Chemistry
Convert the volume of NaOH into liters, and multiply that by the molarity...That gives you moles of NaOH. Tartaric acid has 2 carboxyl groups, doesn't it? (You'll need to check that, if you don't know for sure...). Assuming that it does, then 2 moles of NaOH will be required to react with 1 mole of tartaric acid. So, dividing the moles of NaOH by 2 gives you the moles of tartaric acid. Finally, divide the moles of tartaric acid by the volume you used (15 mL) also expressed in Liters, and you have the molarity of the tartaric acid solution.
2007-01-30 07:11:41 · answer #1 · answered by hcbiochem 7 · 0⤊ 0⤋
at endpoint, moles of acid = moles of base
u have concentration of base so u can find moles of acid and conc. of acid from that
moles of base =conc. of NaOH * volume of NaOH.
using the moles of base, divide by volume of acid to get Molarity of tartaric acid.
M= moles/Volume
0.2297mol/1L * 0.0245L = 0.0056 mols
0.0056 mols / 0.015L = 0.37M
this should be correct or
2007-01-30 07:09:25 · answer #2 · answered by Anonymous · 0⤊ 0⤋