1. Instrumental Analysis Ion selective electrodes
Tutorial 6

2. Objectives By the end of this session, the student should be able to:
Differentiate different types of ISE.
Define selectivity of ISE.
Calculate the E cell of ISE.
Differentiate between direct, indirect , Std addition and potentiometric titration.

3. Applying Nernst equation to Ion-selective Electrodes
where E is the potential of the galvanic cell containing the ion-selective electrode as an indicator electrode and a reference electrode. K is a constant specific for the ion selective electrode used.
This equation applies to any ion-selective electrode. If the analyte is anion, the sign for n is negative.

4. Evaluate the constant in the preceding equation.
Example 1
A cyanide ion-selective electrode obeys the equation: E = K – log [CN–]
The potential was 0.230 V when the electrode is immersed in 1.00 mM NaCN.
Evaluate the constant in the preceding equation.
Using the result of (a), find [CN–] if E = 0.300 V.
find [CN–] if E = 0.300 V.
Solution:
Evaluate the constant
E = K – log [CN–]
0.230 = K – log (1.00 x 10-3)
Constant, K = 0.407 V
0.300 = 0.407 – log [CN–] [CN–] = 1.55 x10-2 M

5. E ground water = k – (0.05916 / 1) log [F-]
Example 2
When a F ion-selective electrode was immersed in a sample of ground water, the potential was 40.0 mV more positive than the potential of tap water that maintains its fluorinated water at the recommended level of 1.00 mg F/L. what is the concentration of F mg/L in the groundwater.
Solution:
E = 40.0 mV, aF-(tap) = 1.00 mg/L
E ground water = k – ( / 1) log [F-]
Etap water = k – log 1.0
Subtract:
P.S. Note that we have /n if the potential is in Volt and 59.16/n if the potential measured in millivolt
We do not need to convert units .
E ground water - Etap water = k log [F-]- k log 1.0
0.04 = log [F-]
= log [F-] Shift log (-0.676) = 0.21
Conc of F- in ground water= mg/L

6. Selectivity Coefficient
No electrode responds exclusively to one kind of ion.
An electrode intends to measure ion A also may respond to ion X. The selectivity coefficient describes the relative response of the electrode to different species
Where kA,X is the selectivity coefficient of A over X. The smaller the selectivity coefficient, the more selective the electrode to ion A, i.e., the less the interference by foreign ion X.
A K+ ion-selective electrode has a selectivity coefficients:
kK+, Na+ = 1 x kK+, Cs+ = kK+, Rb+ = 2.8
Na+ hardly interferes Cs+ interferes to The electrode responds
with the measurement greater extent more to Rb+ than K+
For interfering ion with charge j, the response of the ion selective electrode can be described by the equation:
a A: conc of analyte
a X : conc of interfering ion
n: charge of analyte
j: charge of interfering ion

7. Example 3
One commercial sodium ion-selective electrode has a selectivity coefficient of kNa+, H+ = 36. When this electrode was immersed in 1.00 mM NaCl at pH = 8.00, a potential of 38 mV (versus SCE) was recorded.
Neglecting activity coefficient, calculate the potential if the electrode were immersed in a 5.00 mM NaCl solution at pH 8.00.
What would the potential be for 1.00 mM NaCl at pH 3.78?
Solution:
MNaCl = 0.001M, H+ = 10-8 M (pH =  log [H+])
a) E = constant + ( n) log (aNa+ + kNa+,H+ . a H+)
0.038 = constant + ( /1) log ( (36  10-8))
constant = V.
E = ( /1) log ( (36  10-8))
= 3.3 mV = V.

8. Example 4
A calcium ion-selective electrode obeys Nernst equation. The selectivity coefficients for several ions are listed.
Interfering ion Y k Ca2+,Y Mg Ba Zn
In a pure solution of 1.00x10-3 M Ca2+, the reading was mV.
a)What would be the voltage if the solution had the same calcium concentration plus Mg2+=1.00x10-3 M, Ba2+=1.00x10-3 M and Zn2+=5x10-4 M.
b) If the interfering ions are present at equal concentrations, which ion interferes the most with the Ca2+ electrode?

9. For the pure  Ca2+ solution (without interference) we can write
E = constant + ( / 2) log (1.00 10-3)
Putting in E = V gives constant = V.
For the solution containing the interfering ions, we can say that …
E= ( / 2) log ((1.00x103) + (0.040)(1.00x10-3)+(0.021)(1.00x10-3) (0.081)(5.00x10-4 )) = V
At equal concentrations, Zn2+ interferes the most because it has the largest selectivity coefficient.

10. Applications of potentiometry
1- Direct potentiometry
Example 5
In the potentiometric determination of Ca2+ in solution, the following calibration data was collected. A Ca2+ analyte solution yielded a measured potential of 300.8 mV. Find the concentration of Ca2+ in the unknown solution.
[Ca2+], ppm
log [Ca2+]
Emeas, mV 15
1.176
-338.5 35
1.544
-329.8 89
1.949
-316.5
150
2.176
-312.2
230
2.362
-303.7
400
2.602
-296.4
500
2.699
-295.5
650
2.813
-292.5
Calibration curve equation:
y = m x + a = x 
where m (29.052) is the slope of the Nernst equation (theoretically 59.16/2=29.58) and a (373.54) is the intercept that represents the constant value.
Substituting in the above equation by the potential measured for the unknown Ca2+, we get:
300.8 = – x
x = [Ca2+] = antilog (2.504)
= 319 ppm

11. 2- Standard addition Potentiometry
Basis of Calculations (single point standard addition)
The formula for the Standard Addition calculation is:
Where,
Cu = concentration in the unknown sample.
Cs = concentration in the standard.
Vs = volume of standard.
Vu = volume of sample
E1 = electrode potential (mV) in the pure sample solution.
E2 = electrode potential after the addition of standard.
S = the electrode slope ( /n).

12. Example 6
The wastewater from an industrial processing plant was routinely analyzed for lead as required by EPA. When a Pb ion-selective electrode and a SCE reference were immersed in 50.0 mL of the sample, the cell potential was found to be –0.118 volts mL of a M solution of Pb2+ standard solution was added to the above sample and the measurement repeated; the cell potential was changed to –0.109 volts. What is the approximate concentration of Pb in the wastewater in mg/L?
Solution
cu = 4.93 x 10-4 mol/L
atomic mass of Pb = g/mol
conc. of Pb = 4.93 x 10-4 X = g/L
conc. of Pb = g/L X 1000 = mg/L

13. Exercise 1 (Instrumental Analysis (Skoog)
The following cell was found to have a potential of V:
SCE // Cu2+ (3.25x10-3 M) / membrane electrode for Cu2+
When the solution of known copper activity was replaced with unknown solution, the potential was found to be V. What is the pCu of this unknown solution? (solution pCu =  log [Cu2+] =3.14)
Exercise 2 (Instrumental Analysis (Skoog)
The following cell was employed for the determination of pCrO4:
SCE// CrO42- (x M) / Ag2CrO4(s) / Ag(s)
Find pCrO4 if the cell potential is 0.402 V. (EAg/Ag2CrO4 = 0.361 V, ESCE = V)
(solution pCrO4 = 6.76)
Exercise 3 (Instrumental Analysis (Skoog)
A glass /calomel electrode system was found to develop a potential of  V when used in a buffer of pH 6.00; with an unknown solution the potential was observed to be  V. Calculate the pH and [H+] of the unknown (solution pH = 8.69, [H+] = 2.05 x 10-9 M)

14. SCE // Mg2+ (a = 3.32 x 10-3 M / membrane electrode for Mg2+
Exercise 4 (Instrumental Analysis (Skoog)
The following cell was found to have a potential of V:
SCE // Mg2+ (a = 3.32 x 10-3 M / membrane electrode for Mg2+
When the solution of known magnesium activity was replaced with an unknown solution, the potential was found to be V. What was the pMg of this unknown solution?
(solution pMg = 5.226)
Exercise 5 (Instrumental Analysis (Skoog)
A 0.40 g sample of toothpaste was boiled with a 50 mL solution containing a citrate buffer and NaCl to extract the fluoride ion. After cooling, the solution was diluted to exactly 100 mL. The potential of a F selective ion/calomel system in a 25.0 mL aliquot of the sample is found to be  V. Addition of 5.0 mL of a solution containing mg F/mL caused the potential to change to  V. Calculate the weight percent of F in the sample (solution % F in toothpaste = 4.25x10-4 %)
Try to solve
Problems 15-1, 15-8, 15-28, 15-29, 15-30, 15-34,
Harris text book, p