[Twickel]

Hi

I am having major problems with the calculation the slope of the standard solution.
When I plot this graph out, I have a positive slope, I know the slope should be negative.
The following calculation is used for my standard curve:

mathod, add 0.1mL of Fluoride to a standard of 100mL with 5mL of buffers, Repeat, keep adding 0.1mL take reading.

C1V1=C2V2
 After the first fluorine addition we have:

C1= 10000ppm
V1=0.1
C2= x
V2= 105.1

This equation is used for the remaining additons with V2 increasing by 0.1mL, everything else is constant.
Below is the result of my standard addition.

Please see attached file for a better understanding of what is going on.

[Borek]

Why do you think slope should be positive?

If you add 0.1 mL each time, doesn't it mean V1 changes as well?

[Twickel]

I meant it should be negative.
I do not know if V1 changes or V2 changes....

[Borek]

OK. I see what the problem with the slope is. You calculated it from the data, I just skimmed the table doing some assumptions. Ignore it for now, you have a simple problem earlier.

What is V₁ and what is V₂? Define them, explain what they are, elaborate how they are used and what for.

[Twickel]

V2 is the voume of what I am adding my fluoride ions to, with buffer ( 100mL of water with 5mL of buffer)
V1 is the volume of fluoride I add in each step ( 0.1)
I am using c1v1=c2v2

I am guessing V1 should increase while V2 stays the same?  since the concentration should be going up.

I.e. for step 1 its 10000x0.1= x x 105
          step 2 its  10000x0.2= x x 105


Or is it
step 1 its 10000x0.1= x 105
  step 2 its  10000x0.2= x x 105.1

Do I include the buffer in the solution of V2 by the way?

[Twickel]

F- conc. (ppm) in treated unknown (from standard addition graph)      22.9364
      
now how do I figure out the mass of NaF given to me?

I had an unknown of unknown voume given to me, I made it up to 500mL with distilled water, pipetted 100mL of that 500mL and that was what was analysed.

So I convert ppm to g/L thats the concentration which is 0.02291023. I mutiply that by 5? since dil factor = 1/5 th

0.11455115 = c [f-] n=cxv = 0.11455 x 0.5L = 0.05725 moles of F- ( now is the volume 0.5L or 0.1L)

m= n x M(NaF) = 0.05725x 41.98871 = 2.403 g?

For you information I used method 2 of the c1v1=c2v2 I showed you.

Edit: now I am not sure if it should be called a dilution factor. Since the unknown I had was given to me in an unknown volume ( not marked or referred to in the guide), its simply the fraction of solution used. I am so stuck on this for some reason.

[Borek]

V2 is the voume of what I am adding my fluoride ions to, with buffer ( 100mL of water with 5mL of buffer)

No, V₂ is teh final volume of the solution.

V1 is the volume of fluoride I add in each step ( 0.1)

No, V₁ is the total volume of the fluoride solution that was added.

I am guessing V1 should increase while V2 stays the same?

Almost. V₂ is growing as well, but 0.1mL added to 100 mL can be neglected, it changes the result by 1/1000th.

I.e. for step 1 its 10000x0.1= x x 105
          step 2 its  10000x0.2= x x 105


Or is it
step 1 its 10000x0.1= x 105
  step 2 its  10000x0.2= x x 105.1

If I understand what you did correctly, you took 100 mL, you added 5mL of buffer, and you added 0.1 mL of the fluoride solution. If so, final volume (V₂) is 100+5+0.1=105.1. As explained above, the decimal part you can ignore.

[Twickel]

But I have to calculate the concentration at each step, so add 0.1 calculate concentration, add another 0.1ml calculate concentration etc.

So v1 becomes the total amount of fluoride added so far...

So should it not be 0.1x10000= 105 x x, then 0.2x10000= 105.1 x x then 0.3x10000-105.2 x x etc etc?

How about the rest of my calculations, do they make sense regarding the dilution factors/

[Borek]

So should it not be 0.1x10000= 105 x x, then 0.2x10000= 105.1 x x then 0.3x10000-105.2 x x etc etc?

How much is 100+5+0.1? Is this not the volume after the first addition of 0.1 mL?

[Twickel]

I thought that for the first time I add 0.1mL of fluoride, the volume of the sample is only 100+5 ml, because no fluoride has been added to the sample yet.
But for the second step, the second time I add F, the sample already has 0.1mL of fluoride fromm step 1.  So it is
1) 0.1x 10000 = 105 x x
2)  0.2x 10000 = 105.1 x x

Do you  see  what I mean?

What about the second part of the calculations are they right ( figuring out the concentration in the original amount, then mass of NaF)

[Borek]

I thought that for the first time I add 0.1mL of fluoride, the volume of the sample is only 100+5 ml, because no fluoride has been added to the sample yet.

Perhaps it is just a matter of miscommunication, but for me it sounds like "After I add 0.1 mL of fluoride solution volume is 105 mL because I have not added 0.1 mL of fluoride solution yet".

But as I wrote earlier it is not that important. More important part was about using correct numbers for V₁. Has the sign of the slope changed? Because your original calculations were completely off.

If I understand you correctly you measured concentration of the solution prepared by diluting the original sample to 0.5L, if so, total amount of substance in the original sample is concentration times this volume.

[Twickel]

Yes it is now negative which is what it should be, however the book says correct electrode operation should indicate a slope of -25 I got 21. The meter fluctuated so much even after waiting for 3 minutes, it kept going up and down by +/- 1.3. I did not know which number to take.

I multiply the concentration by 5 as well right?

so 22 ppm x 5 = x
 n= x multiply by 0.5L like what I did?

[Twickel]

Every method ive shown her was in correct. The mass ended up being 0.0xxxx. I will put the method I did up soon, let me know i its right

Cocentrated fluoride in treated unknown = 22.9688 ppm.
Concentrate of fluoride in 100ml = 22.96x10^-3 x 105/100= 0.024117

Concentration of Naf is concentration of fluoride x mol weight of Naf/ atomic mass of fluoride= 0.024117 x41.988/18.99 = 0.05332 g/L

therefore in unknown = 0.05332 x 0.5 = 0.02666 g

so mass of NaF is 0.02666g