EXPERIMENT 9: ION-SELECTIVE ELECTRODES The determination of the concentration of
ID: 1043845 • Letter: E
Question
EXPERIMENT 9: ION-SELECTIVE ELECTRODES The determination of the concentration of both cations and anions in solu tions can take several forms, including spectrophotometry, complexation titrations, precipitation titrations, etc. For several potential analytes (like Na, K, Ca, Cl, NO,, F), the ion-selective electrode (ISE) represents one of the most attractive and straightforward means of determining solu tion concentrations. An ideal quantitative technique would involve placing a "probe" in the sample to be measured and then "reading out" the concentrations of all chemical species present. Potentiometric techniques approach this ideal situation. The probe (the ISE) is placed into a sample solution and is con nected to a potentiometer, a device for measuring the potential (voltage) differences between the ISE and a reference electrode. If the conditions of the measurement are well understood and controlled, the potential differ- ence between these two electrodes can be related to the concentration of a specific chemical species present in the sample. OBJECTIVES Educational Objectives A student who has successfully completed this experiment will be able to: . prepare a series of standards using a serial dilution, make voltage measurements using an ISE generate a calibration plot using voltage measurements, and determine the molarity of an ion from ISE measurementsExplanation / Answer
In Page 56, it is mentioned that you need to measure the concentration of Ca2+ ions
In Page 60, it is mentioned that you have to make at least 4 standard solutions in the concentration range 1.00 - 1.00 x 10-7 M.
In Page 60, it is also mentioned that you need to add ionic strength adjuster, 2 mL/100 mL, since you are going to measure Ca2+ ions, you can use 2M KCl Solution.
From the question, you will be provided with 2 M stock solution, in this case possibly 2 M CaCl2 / 2 M KCl as ionic strength adjustment.
Calculation for converting 2 M stock to 1.00 - 1.00 x 10-7 M solution, here showing the calculation for many dilutions you can choose any 4 or 5.
You can use the equation, M1V1 = M2V2, M1, V1 are the molarity and volume of the final solution, M2V2 molarity and volume of stock, Here V1 = 100 mL (since we are making a 100 mL solution), M1 = Concentration from 1.00 M to 0.0000001 M (as given in table), V2 = need to determine, M2 = 2 M (concentration of stock).
Once the first dilution is made you can re-dilute serially as below.
Stock
Standard Concentration
Volume stock
Volume of Ionic strength adjustment
Water
Total volume
Provided
2.00 M
-
-
-
-
1
1.00 M
50 mL of 2 M stock
2 mL
48 mL
100 mL
2
0.10 M
10 mL of Stock 1
2 mL
48 mL
100 mL
3
0.001 M
10 mL of Stock 2
2 mL
48 mL
100 mL
4
0.0001 M
10 mL of Stock 3
2 mL
48 mL
100 mL
5
0.00001 M
10 mL of Stock 4
2 mL
48 mL
100 mL
6
0.000001 M
10 mL of Stock 5
2 mL
48 mL
100 mL
7
0.0000001 M
10 mL of Stock 6
2 mL
48 mL
100 mL
Stock
Standard Concentration
Volume stock
Volume of Ionic strength adjustment
Water
Total volume
Provided
2.00 M
-
-
-
-
1
1.00 M
50 mL of 2 M stock
2 mL
48 mL
100 mL
2
0.10 M
10 mL of Stock 1
2 mL
48 mL
100 mL
3
0.001 M
10 mL of Stock 2
2 mL
48 mL
100 mL
4
0.0001 M
10 mL of Stock 3
2 mL
48 mL
100 mL
5
0.00001 M
10 mL of Stock 4
2 mL
48 mL
100 mL
6
0.000001 M
10 mL of Stock 5
2 mL
48 mL
100 mL
7
0.0000001 M
10 mL of Stock 6
2 mL
48 mL
100 mL
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