[bamoida2]

Hello, I'm kinda stuck on how to solve problem that could come on my Analytical chemistry exam: (behold my bad translation)


Potentiometric Titration:  For the quantitative analysis of Cadmium a solution containing Cadmium gets titrated with Na2S and its endpoint gets determined via a Cd-selective ISE.


a)For a 50ml equivalent of the sample 2.5 ml of a 0.02 M Na2S was used.
What was the concentration of the sample?

That one is easy   (0.0025*0.02)/0.050 =  10^-3 M


b) How much (in % do does the concentration have to change and)  titrant solution do you have to add that the potential of the ISE changes by 2, 30 and 200 mV.    (T=298K)

Disregard the dilution of the sample and the solubility product.

(well there is a table that hints that you can calculate the volume of solution to add from the % the concentration needs to change.)

Here is my approach:  a≈c

ΔE=  R*T/(2*F) *ln (a[Cd 2+])


for 2mV:

0.002 = 8.3145*298/(2*96485) *ln(a [Cd ²⁺])

0.002/0.0129 = ln(a [Cd²⁺])

e^(0.002/0.0129) = a [Cd ²⁺]

=1.168

for the % (not sure if that is correct at all)
(1.168)/(1+1.168)=53.9%

2.5 ml * 0.539 =  1.35 mL.

I'm not sure if that is right at all (the end point detection confuses me as well, is this before or after the solution is fully titrated)


c)
Considering the result above how much does the mV value of the Potential at the Cd ISE change during  start and end of the titration if the detection limit of a Cd. Ise is 10^6 mol/L.

I have absolutely no idea on this one.


Please Help me solve b) and c), I hope my translation was understandable (its originally in German)

If you need any more info please don't hesitate to ask.


Thank you!

Edit:

this it what the template in b) looks like in german

http://imgur.com/Bpwt7pj

[mjc123]

a) Correct, but you make the implicit assumption that Cd²⁺ and Na₂S react in a 1:1 molar ratio. That is true in this case, but it is not true for all titration reactions, and you have to take the ratio into account. If this was an exam I would say explicitly "A and B react in a 1:1 molar ratio" so the examiner knows that you know it.

b) Starts OK, but you make a mistake. If ΔE = Δ(RT/nF*lnC), then ΔE*nF/RT = ln C - ln C₀ = ln (C/C₀)
e^(ΔE*nF/RT) = C/C₀
i.e 1.168 is not the difference between the two concentrations, but the ratio of the two concentrations.
Now during the titration you are reducing the concentration of Cd²⁺. The question says "potential changes by 2 mV", it doesn't say whether +2 or -2, but it must be negative as [Cd²⁺] is decreasing; so C/C₀ = e^(-0.002/0.129) = 1/1.168.
So the new concentration is 10^-3/1.168 = 8.56 x 10^-4 M. The change is -1.44 x 10^-4 M and the percentage change is 14.4%.
Can you now do the rest?

c) I'm not quite sure what it's asking for here, but I doubt if the detection limit of your ISE is 10⁶M.

[bamoida2]

Thank you that makes a lot of sense!

but doesn't the potential get bigger the less Cd²⁺ is in the solution?


to c):  the detection limit is 10^-6 ofc


could it be that the solution is it is just E= ±RT/(nF)*ln(10^-6)?

[mjc123]

The electrode potential of Cd²⁺/Cd will decrease (i.e. become more negative, as E° is negative relative to SHE) as [Cd²⁺] decreases. The cell potential will depend on the reference electrode, and depending whether the potential of the reference electrode is higher or lower than the cadmium electrode, the cell potential will either increase or decrease (if it is measured as simply a positive value, without saying which electrode is the anode and which the cathode).