I I termination of Sodium and Potassium Amounts by Flame Photometry lI Pre-lab W

Question

I I termination of Sodium and Potassium Amounts by Flame Photometry lI Pre-lab Worksheet: Determination of Sodium um and Potassium Read the experiment, and then answer the following questions additional pages as needed questions. Show all calculations; atoch 1, Calculate the masses of Naci and KCl that are required to that contains Na and K, each present at 1000 ppm ww to prepare 250.0 mL of a standard at 1000 germ uiyed o prepare 250.0 ml of a standard w/v Narne the amount (ml)ot 1000pm ww NaK nedel to prepare 250 mil of 100 ppm wiv Na/k. 2. Calculate the amount (mL) of 1000 ppm w/v Na/K 3. Calculate how many ml of 100 ppm wiv Na/K are needed to prepare 100 ml each of standards containing 1,2, 3,4 and 5 ppm wiv Na and K 4. Calculate the Na and K' concentrations if you used 0.8500 g of aCfand 03959 of K g o instead of the values you calculated in Question 1, to prepare 250 mL of solution. What the would be the concentrations of the diluted solutions in Questions 2 and 3? behind how flame photometry is used to measure Na, and K 5. Briefly explain the theory behind how flame If a river water sample containing several through the flame photometer, howw being measured and not any other metal, everal diflferent metals at varying concentrations is put containingan we be certain that only one element, such as Na, is 6. certain that only ng

Explanation / Answer

NaCl = 0.85 g KCl =0.399 g

1. conc of Sodium ion

amount of sodium in NaCl = 23 g in 58.5 g of NaCl

so in 0.85g we will have = 0.3341 g of sodium

0.3341 g in 250mL = 1.3364 g /L = 1336.4 ppm or mg / L

2. conc of potassium ion

amount of sodium in KCl = 39 g in 74.5 g of KCl

so in 0.399g we will have = 0.2088 g of potassium

0.2088 g in 250mL = 0 .8352 g /L = 835.2 ppm or mg / L

...........

we can calculate mL of solution needed for each of the desired conc as

1. 1ppm

M1V1 = M2V2

1 X 100 = 1336.4 X V2

V2 =0.074 mL

2.2ppm

M1V1 = M2V2

2 X 100 = 1336.4 X V2

V2 = 0.149 mL

3.3ppm

M1V1 = M2V2

3 X 100 = 1336.4 X V2

V2 = 0.222mL

4.4ppm

M1V1 = M2V2

4 X 100 = 1336.4 X V2

V2 = .296mL

5. 5ppm

M1V1 = M2V2

5 X 100 = 1336.4 X V2

V2 = 0.37mL respectively

For potassium

1. 1ppm

M1V1 = M2V2

1 X 100 = 835.2 X V2

V2 =0.119 mL

2.2ppm

M1V1 = M2V2

2 X 100 = 835.2 X V2

V2 = 0.238 mL

3.3ppm

M1V1 = M2V2

3 X 100 = 835.2 X V2

V2 = 0.357mL

4.4ppm

M1V1 = M2V2

4 X 100 = 835.2 X V2

V2 = .476mL

5. 5ppm

M1V1 = M2V2

5 X 100 = 835.2 X V2

V2 = 0.595mL respectively

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