Instrumental Analysis Ion selective electrodes Tutorial 6

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.

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.

Evaluate the constant in the preceding equation. Example 1 A cyanide ion-selective electrode obeys the equation: E = K – 0.05916 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 – 0.05916 log [CN–] 0.230 = K – 0.05916 log (1.00 x 10-3) Constant, K = 0.407 V 0.300 = 0.407 – 0.05916 log [CN–] [CN–] = 1.55 x10-2 M

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 – (0.05916 / 1) log [F-] Etap water = k – 0.05916 log 1.0 Subtract: P.S. Note that we have 0.05916/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- 0.05916 log [F-]- k +0.05916 log 1.0 0.04 = - 0.05916 log [F-] -0.676 = log [F-] Shift log (-0.676) = 0.21 Conc of F- in ground water= 0.21 mg/L

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 10-5 kK+, Cs+ = 0.44 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

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 + (0.05916n) log (aNa+ + kNa+,H+ . a H+) 0.038 = constant + (0.05916/1) log (0.001+ (36  10-8)) constant = 0.1393 V. E = 0.1393 + (0.05916/1) log (0.005 + (36  10-8)) = 3.3 mV = 0.003 V.

Example 4 A calcium ion-selective electrode obeys Nernst equation. The selectivity coefficients for several ions are listed. Interfering ion Y k Ca2+,Y Mg2+ 0.040 Ba2+ 0.021 Zn2+ 0.081 In a pure solution of 1.00x10-3 M Ca2+, the reading was +300.0 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?

For the pure  Ca2+ solution (without interference) we can write E = constant + (0.05916 / 2) log (1.00 10-3) Putting in E = 0.300 V gives constant = 0.389 V. For the solution containing the interfering ions, we can say that … E= 0.389 + (0.05916 / 2) log ((1.00x103) + (0.040)(1.00x10-3)+(0.021)(1.00x10-3) +(0.081)(5.00x10-4 )) = 0.3015 V At equal concentrations, Zn2+ interferes the most because it has the largest selectivity coefficient.

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 = 29.052 x  373.54 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 = – 373.54+ 29.052 x x = 2.504 [Ca2+] = antilog (2.504) = 319 ppm

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 (0.05916/n).

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. 5.00 mL of a 0.006 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 = 207.2 g/mol conc. of Pb = 4.93 x 10-4 X 207.2 = 0.1023 g/L conc. of Pb = 0.1023 g/L X 1000 = 102.3 mg/L

Exercise 1 (Instrumental Analysis (Skoog) The following cell was found to have a potential of 0.124 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 0.105 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 = 0.241 V) (solution pCrO4 = 6.76) Exercise 3 (Instrumental Analysis (Skoog) A glass /calomel electrode system was found to develop a potential of 0.0412 V when used in a buffer of pH 6.00; with an unknown solution the potential was observed to be 0.2004 V. Calculate the pH and [H+] of the unknown. (solution pH = 8.69, [H+] = 2.05 x 10-9 M)

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 0.2714 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 0.1901 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 0.1823 V. Addition of 5.0 mL of a solution containing 0.00107 mg F/mL caused the potential to change to 0.2446 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, 15-40. Harris text book, p342-345