Im confused on how to answer 2 questions based on the experimental step number 9
ID: 557178 • Letter: I
Question
Im confused on how to answer 2 questions based on the experimental step number 9.
The first question is: why is it not necessary to test the supernatant from both samples for the excess reactant in procedural step number 9?
The second is what indication in the test tube test will be given if Na3PO4 is the limiting reagent? What if BaCl2 is the limiting reagent?
Figure 6.1 Opened filter paper to place into filter funnel. 7. Stir up your reaction vigorously before filtering. Decanting the liquid into the filter paper first will actually result in most of the solid flowing through the pores of the paper. 8. Filter the first sample through the first filter paper. Transfer all of the precipitate by using a rubber policeman (rubber or Teflon™ spatula) and no more than three 5 mL aliquots of the warmed wash water. Be sure to collect as much of the precipitate as possible. When the filtration is complete, carefully remove the filter paper to a clean watch glass for drying in a safe place until the next laboratory period. 9. Use the plastic transfer pipet to half-fill each of the two centrifuge tubes with the liquid filtrate. Label the tubes I and Il. Add 2 drops of BaCl2 test reagent to tube I, and 2 drops of Na3PO4 test reagent to tube Il. You should see an obvious precipitation reaction in only oneExplanation / Answer
Please check you have not scanned point no 9 of your lab procedure completely. i am answering the question based on your typed questions.
you are asked to carry out same experiment twice ( by preparing two duplicate samples), to ascertain repeatibility of the procedure.
i.e, you prepare two exact samples by mixing each time 20mL BaCl2 with 20 mL Na3PO4 and diluting the mixture with 100 mL Deionised water. Out of the two reagents that you mix, either BaCl2 is in excess( in which case Na3PO4 is the limiting reagent) or Na3PO4 is in excess ( In which case, BaCl2 is the limiting reagent). As you have taken same volume of the reactants in preparing the two duplicate samples, the excess reagent is same in the supernatant liquids obtained from the two samples. Thus you need to test the supernatant solution from any one sample only.
If Centrifuge tube I + BaCl2 solution = Precipitate
This means that the clear filtrate obtained after filtration still contains unreacted Na3PO4 , which again forms precipitate of Ba3(PO4)2 on addition of BaCl2. This tells you that BaCl2 was the limiting reagent in the sample.
If Centrifuge tube II + Na3PO4 solution = Precipitate
This will mean that the clear filtrate obtained after filtration still contains some unreacted BaCl2, which will again form precipitate of Ba3(PO4)2 on addition of Na3PO4. This tells you that Na3PO4 was the limiting reagent in the sample.
Note: Limiting reagent is the one which get completely consumed in the reaction.