Assignment 3 - Solutions

1.
 The concentration of sulfate (CO32-) anion in solution can be measured with an electrode made with ZnCO3.  An unknown solution of CO32- has a voltage of -1.270 V vs. the SCE reference electrode.  What is the concentration of CO32- in that solution?  For Ksp values, see Appendix 1.  For Eo values, see Appendix 4.  For reference electrode voltage, see Table 19.1.
 
Soln This is a 2-step question. 
First, you must solve the Eo value for the electrode being used (ZnCO3), then use that electrode to find the [CO32-]:

For the reduction of Zn4+,
 Zn2+ (aq)  +  2 e-  --->  Zn (s)      (Eocell = -0.763 V vs. SHE)
But in our reaction, the [Zn2+] is not 1 M ([CO32-] is).
Ksp = [Zn2+][CO32-] = 1.0 x 10-10

[Zn2+] = [CO32-]/Ksp = 1/1.0 x 10-10

EoZnCO3/Zn = -1.059


Now we know the Eo for the ZnCO3/Zn2+ couple, we can solve for the unknown [CO32-]:
[CO32-] = antilog(2(-1.059 + 1.270 -0.2444)/0.0592) = 0.0744 M
 
2.
 The standard addition method was used to determine the concentration of zirconium in the water near a mine.  The following data were obtained:
 
volume of water sample (mL)
volume of 0.00750 M Ni2+ (mL) 
potential (V)
20.0
0
 -1.092
20.0
5.0
-1.079

Determine the concentration of Zr4+ in the original solution.  Report the result as both molarity (mol/L) and parts per million (ppm = mg/L).  (For this question, do not assume Ej = 0)
 
Soln: for a cation:

Therefore, for sample #1,
log[Zr4+] = 4(E1 - L)/0/0592

and for sample #2,
log((20[Zr4+]+5*0.05)/25) = 4(E2 - L)/0/0592

Eq. #1 - Eq. #2:
25[Zr4+] - 20*0.1323[Zr4+] = 25*0.1323

[Zr4+] = 0.00148 M

Using the fact that ppm = mg/L, [Ni2+] = (0.00148 mol/L)(91.22 g/mol)(1000 mg / 1 g) = 135 ppm
 
3.
 Can Pb2+ and Sn2+ be separated by voltammetry (in concentration ranges from 10-4 to 10-2 M)?
 
Soln
 Using the Nernst equation, the range over which Pb2+ plates is -0.185 to -0.244
the range over which Sn2+ plates is -0.195 to -0.254

Since these ranges overlap, it is not possible to separate them,
 
4.
 Using linear-scan voltammetry, a solution that is 5.00 ppm in Mn2+ has a limiting current of 60.0 μA.  What is the concentration of a solution that has a limiting current of 43.8 μA?
 
Soln
 Il = kCA
Therefore, Il(1) = kCA(1)Il(2) / kCA(2) = (5.00)(43.8)/60 = 3.65 ppm


 

Last updated, April 4, 2005
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