Suppose a solution was too concentrated for an accurate reading with the spectrophotometer. The concentrated solution was diluted by placing 1.00mL of the concentrated solution in 4.00mL of water. The solution was then placed in the spectrophotometer, an absorbance was obtained, and after a few calculations the molar concentration was calculated to be 3.5 x 10-6 M. What was the concentration of the original stock?
I think that the M1V1 = M2V2 is a formula used to calculate some parts of the equation but I don't really know. I'm stuck because it says "before the dilution".
M1V1 = M2V2 is the correct equation. Here's why:
You need to find the concentration of the solution before it was diluted. You know that 1.00 ml of this undiluted solution was used to make a diluted solution. You also know that the volume of the diluted solution is 4 ml and that its concentration is 3.5 x 10^-6 M.
The number of moles of solute did not change when you diluted the solution, you simply added water.
Since molarity is moles solute/liters of solution, moles of solute = (molarity) x (liters of solution) or moles = MV.
This is where we get M1V1 = M2V2. If moles before diluting equals moles after diluting, then M x V before diluting equals M x V after diluting.
So M1, which you're trying to find, equals M2V2/V1
= (3.5 x 10^-6)(4 ml)/1 ml = 1.4 x 10^-5 moles per liter
You may have noticed that the volumes weren't converted to liters even though the concentration is in moles per liter. As long as the same volume unit is used for V1 and V2 it's unnecessary to convert. You would be multiplying both sides of the equation by the same conversion factor so it would cancel out.