2. A fluoride electrode is used to determine fluoride in a water sample. Standards and samples are diluted 1:10 with TISAB solution. For a 1.00 x10-3M(before dilution) standard, the potential reading relative to the reference electrode is -211.3mV and for a 4.00 x10-3M standard it is -238.6mV.The reading with the unknown is -226.5mV.What is the concentration of fluoride in the sample?
Well, this question sure seems serious! But fear not, I'm here to sprinkle some humor into it. Let's dive in, shall we?
To determine the concentration of fluoride in the water sample, we can use the heartthrobbing Nernst equation. It goes like this:
E = E° - (0.0592/n) * log C,
where E is the electrode potential, E° is the standard electrode potential, n is the number of electrons transferred, and C is the concentration.
Now, keeping that in mind, let's put on our detective hats and solve this equation. We have a few data points to work with:
For the 1.00 x 10^-3 M standard, the potential is -211.3 mV.
For the 4.00 x 10^-3 M standard, the potential is -238.6 mV.
And for the unknown sample, the potential is -226.5 mV.
Now, let me whip out my trusty clown calculator. *honk honk*
First, we need to find the standard electrode potential (E°). Sadly, I don't have that information from you. But if you provide it, I can calculate it faster than a cheetah chasing its prey!
So, please share E° with me, and we'll continue this chemical escapade together.
To determine the concentration of fluoride in the sample, we need to compare the potential reading of the unknown sample with the potential readings of the standards.
1. Calculate the potential difference (E) between the two standard solutions:
E1 = -211.3 mV
E2 = -238.6 mV
ΔE = E2 - E1
= -238.6 mV - (-211.3 mV)
= -27.3 mV
2. Calculate the concentration difference (ΔC) between the two standard solutions:
C1 = 1.00 x 10^(-3) M (before dilution)
C2 = 4.00 x 10^(-3) M (before dilution)
ΔC = C2 - C1
= 4.00 x 10^(-3) M - 1.00 x 10^(-3) M
= 3.00 x 10^(-3) M
3. Calculate the calibration factor (K):
K = ΔE / ΔC
= -27.3 mV / (3.00 x 10^(-3) M)
= -9.10 mV/M
4. Calculate the concentration of fluoride in the sample (C_sample):
E_sample = -226.5 mV (reading with the unknown)
C_sample = (E_sample - E1) / K
= (-226.5 mV - (-211.3 mV)) / -9.10 mV/M
Finally, plug in the values:
C_sample = -15.2 mV / -9.10 mV/M
≈ 1.67 x 10^(-3) M
Therefore, the concentration of fluoride in the water sample is approximately 1.67 x 10^(-3) M.
To determine the concentration of fluoride in the sample, we can use the Nernst equation, which relates the potential difference to the concentration of the analyte. The Nernst equation for the fluoride electrode is given by:
E = E° + (0.0592/n) * log([Fluoride])
Where:
E is the potential reading relative to the reference electrode,
E° is the standard potential,
[Fluoride] is the concentration of fluoride in the solution,
0.0592 is the value representing the conversion from natural logarithm to base 10 logarithm,
and n is the number of electrons transferred in the reaction (for fluoride, it is one).
First, we need to calculate the standard potential (E°):
E° = E2 - E1
Where E2 is the potential reading for the 4.00 x 10^-3 M standard (-238.6 mV) and E1 is the potential reading for the 1.00 x 10^-3 M standard (-211.3 mV):
E° = -238.6 mV - (-211.3 mV) = -27.3 mV
Now we can use the Nernst equation to find the concentration of fluoride in the sample:
-226.5 mV = -27.3 mV + (0.0592/1) * log([Fluoride])
Simplifying the equation:
-226.5 mV + 27.3 mV = 0.0592 * log([Fluoride])
-199.2 mV = 0.0592 * log([Fluoride])
Taking the antilogarithm of both sides:
10^(-199.2 mV/0.0592) = [Fluoride]
Calculating the concentration of fluoride:
[Fluoride] = 10^(-3368.918/0.0592)
[Fluoride] ≈ 1.16 x 10^-10 M
Therefore, the concentration of fluoride in the sample is approximately 1.16 x 10^-10 M.
I think the easy way to do this is to graph it as mv on the y axis and M on the x axis. If you do that you can estimate just on scratch paper, that it is about 2.7 x 10^-3 M for the unknown. If you want to do it without graphing you can try this
You have -211.3 mv for 1.00E-3M and -238.6. mv for 4.00E-3. So that's (238.6-211.3 = 27.3 mv for a difference in M of 4.00E-3 - 1.00E-3 = 3.00E-3 or
3.00E-3/27.3 = 1.09E-4M (and you may want to carry that out more) for each 1 mv.
(I've ignored the - sign for the voltage. You can keep it straight in yur head.)
So how far are you from the 1E-3 standard. That's 226.5-211.3 = 15.2 mv away and at 1.09M/mv x 15.2 mv = about 1.67E-3M and that + 1.00E-3 from the starting point makes about 2.67M for the unknown. We should get close to that if we try the top end and come down. You can do that yourself if you wish but it's something like this.
Difference is 238.5-226.5 = 12.10 x 1.09E-4 = 1.32M so we should be down that far from the 4.00E-3.
4.00E-3-1.32E-3 = about 2.68E-3 for the unknown. If you will put this on a graph and determine the slope of the line and the intercept for y = mx+b you can calculate exactly what it should be for the unknown.