Question

Lab 5_Liquid-Liquidem 129A Part 2: Multiple Extraction and Determination o Between Ether and Water action and Determination of the Distribution Coefficient for Propionic Acid Table 2: Titration of Propionic Acid Solution after Multiple Extractions to ter Multiple Extractions (3 x 50 mL) with Ether Volume PA Aliquot (mLVolume NaOH (ml) Molarity NaOH (M) 10 2.0m 0.3190 MM in coefficient you calculated in part A ns. I might think in fter 3. 50 mL extractions extraction se the partition coefficient von calentated in art. Question 6. to calculate the weight of PA you would expect to obtain in the combined sther extracts after 3, 50 ml extractions. I might is of the water lever how many prams P will remain in the water layer after the extraction This will be the initial value of grams of PA in the extraction. Now calculate the grams PA emaining in the water layer after the extraction? Then the 3" The mass that is not in the - aqueous layer must be in the combined ether extracts. 9. Now you will calculate the actual weight of PA obtained in your combined ether extracts a. Using data from Table 2, calculate the weight of PA left in the water layer after your 3 ether extractions. b. Calculate the weight of PA extracted into the combined ether extracts (3 x 50 mL = 150 mL).

question 9! parts a and b!
table shows data after 3 extractions!!!

Answer 1

a) The titration you performed was based on the reaction between propionic acid and NaOH:

$CH_{3}CH_{2}COOH+NaOH\rightarrow CH_{3}CH_{2}COO^{-}+H_{2}O+Na^{+}$

As you can see, one mole of acid reacts with one mole of NaOH. We can calculate the amount of moles of NaOH you added in the titration using:

$n=M\cdot V(L)=0.3130\frac{mol}{L}\cdot 0.002L=6.26x10^{-4}moles$

So this is also the number of moles of propionic acid that were present in the 10 mL you used for the titration. They are equivalent to a mass of:

$m=n\cdot m_{molar}=6.26x10^{-4}moles\cdot 74.08\frac{g}{mol}=0.0464g$

I am missing here the total volume of water you have extracted the acid from. This is important because we know there are 0.0464 g of acid in 10 mL of water, but if there was more water volume, the total mass would be higher. If there were 50 mL of water, for example, the mass of acid in the whole volume would be 5 times the one we have calculated, because the original volume is 5 times higher than the volume used for the titration.

b) for this part, the initial mass is necessary, since the mass in the ether extracts will be equal to the initial mass - mass calculated in item a.

If you need it, post the missing data in the comments and we can calculate the results.