# Quick quantitative analysis problem

Xelb

## Homework Statement

A 1.00 × 10^-4 M NaF solution can be prepared in several ways. Two methods are listed below. The formula weight of NaF is 41.9984 ± 0.0003.

For each method use propagation of error methods to determine the absolute and relative uncertainty in the final concentration.

Method 1
0.0042 ± 0.0001 g of NaF is placed in a 1000.0 ± 0.3 mL volumetric flask. The flask is filled to the mark with water

Method 2
0.0420 ± 0.0001 g of NaF is placed in a 100.00 ± 0.08 mL volumetric flask. The flask is filled to the mark with water. This solution is diluted 100:1 by pipeting 1.000± 0.003 mL of the solution to another 100.00 ± 0.08 mL volumetric flask and then filling to the mark with water.

For each method use propagation of error to determine the absolute and relative uncertainty in the final concentration.

Which method has the most uncertainty? Which individual measurement introduces the most uncertainty?

## Homework Equations

Molarity = mol/L

moles = mass/MW

Final Concentration = (Initial Concentration)*(Initial Volume)/(Final Volume)

## The Attempt at a Solution

Note: When writing the absolute uncertainties, I purposely write "2." because it needed to be just one significant figure, and writing down just "2" means that there are no significant figures.

For method one, I took the grams of NaF and converted it to moles, which turned out to be 1.0*10^-4 ± 2. * 10^-6 when propagation of uncertainty and the correct significant figures were used. Once I found moles, I then converted to molarity using the following: (1.0*10^-4 ± 2. * 10^-6 )/(1.0*10^3 ± 0.3L) and arrived at 1.0 * 10^-7 ± 2. * 10^-9 M.

Is 1.0 * 10^-7 ± 2. * 10^-9 M the final concentration in this case? I know the problem says it needs to be 1.0 * 10 ^-4 M, but it wanted the percent relative uncertainty, which ended up being 2%. I'm not really sure what else to do regarding method one.

For method two, it seemed a bit easier, but I'm not even sure I did it right. I made use of the following formula: Final Concentration = (Initial Concentration)*(Initial Volume)/(Final Volume).

Since I had found the initial concentration from method one, I just plugged that in and multiplied it by the pipet volume and divided it by the volume of the flask. I arrived at a final concentration, again different from the value that the problem originally stated, of 1.0*10^-9 ± 2. * 10^-11 M with a relative uncertainty of 2% as well. Not sure what else to do...