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 [Cd
2+])
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