Posted by Gweedo8 on Wednesday, July 1, 2009 at 6:27pm.
Online chemistry lab preparing standards help?
I was told to prepare some standard for a chemistry lab and here where the instructions:
Obtain a solution of known phosphate concentration. In this lab we will use (1.e-002M) potassium dihydrogen phosphate, KH2PO4. Note the concentration on your data sheet.
Pipet 0.50mL of the standard phosphate solution into a 100mL volumetric flask. Using a graduated cylinder, add 10.0mL of the ammonium molybdate solution to the flask. Then dilute to the mark with distilled water. Make sure you label this volumetric with the concentration of phosphate. (McVc=MdVd) Do this by choosing label from the menu (or right clicking)
Mc = concentration of the original standard phosphate solution.
Vc = volume of concentrated standard phosphate solution TRANSFERRED to the volumetric.
Md = new diluted concentration.
Vd = total volume of the diluted solution.
The Part I do not understand is filling in the McVc=MdVd.
Here are my guesses on some but I am not 100% sure and could use some help:
- chemistry - GK, Wednesday, July 1, 2009 at 7:19pm
0.50 mLs of KH2PO4 was diluted to 100 mLs in the volumetric flask.
The concentration of the diluted KH2PO4 in the 100.00 mL volumetric flask is:
(0.50mLs /100mLs)(Original concentration of KH2PO4). Not sure where the decimal point is in the original phosphate concentration.
- chemistry - DrBob222, Wednesday, July 1, 2009 at 7:31pm
I share GK's dilemma about the original concentration. What is 1.e-002M?
Your question is about McVc = MdVd.
Mc = molarity of the standard solution. That is the 1.e-002M whatever that is.
Vc is the 0.50 mL. Md is the new molarity and is what you are to calculate. Vd is the 100 mL. So the set up looks like this (and you will need to change the concn to whatever 1.e-002 M means.
1.e-002M*0.50 = Md*100.
Solve for Md.
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