SCHA 511/711 Standard Additions for F- problem.
The surest way to make something that is simple complicated is to try to make it simple.
I have tried in the past to make the Chapter 23 problem simple by giving you a formula. What a mistake that was. Let’s look at standard additions in this case. Standard additions is a simple concept but the algebra will get us every time.
Let’s look at our friend the Nernst Equation for this electrode. The value of n is 1 so we have a simple expression.
But for this problem we do not know the concentration of fluoride or the Eo but we do know the starting potential. In the second part of the experiment we add a known amount (mass or moles) of fluoride and the potential changes. In each case we know or can readily determine the volume of solution. The only real unknown in the initial number of mgs that was in the 25.0 mL aliquot analyzed. [ ] usually denotes Molarity but let’s just use mg/mL here.
For solution 1 we have 25.0 mL, a potential of –0.1823V and x mgs of Fluoride.
Since we add 5.0 mL of a 0.00107 mg /mL solution we have added 0.00535 mg to the second solutions which gives use the Nernst equation for the second solution of
Now we have two equations with two unknowns, E0 and x. Dust off the algebra skills.
Subtract the second equation from the first and this will give us
Factor out the 0.0592 and divide both sides by the 0.0592 and we get
Combine the log terms, remember those rules!!!!
Clear the log term by taking in the antilog of both sides.
x = 4.271 x 10-4 mg and since this was 1/4 of the total solution that was prepared we had a total of 4 times as much in the original solution.
1.708 x 10-3 mg in the 100 mL.
So the percent present was 1.708 x 10-3 mg * 1.000 g / 1000 mg divided by the 0.4000 grams * 100 =
4.27 x 10-4 %