Question

For the following analytical scenario below which involved initially diluting 20 ml of the original water sample into a 1000 ml volumetric flask. Subsequently, the sample was prepared and analyzed as depicted below:


Volemetric flask (500ml) #1
Sample Composition: 5 ml of sample + 0 mL of 300 ppm Na 0.5% LiCl2
Absorbance: 0.12


Volemetric flask (500ml) #2
Sample Composition: 5 ml of sample + 1 ml of 300 ppm Na + 0.5% LiCl2
Absorbance: 0.17


Volemetric flask (500ml) #3
Sample Composition: 5 ml sample + 2 ml of 300 ppm Na + 0.5% LiCl2
Absorbance: 0.21


I. If all the flasks were diluted to a final volume of 500 ml with distilled water calculate the final concentration of sodium in the sample through a graphical and mathematical solution.

II. Why Lanthanum and Cesium or Lithium are added to solutions to be analysed for Calcium and Sodium respectively?

Answer 1

Answer on the Question #65323, Chemistry / Other

For the following analytical scenario below which involved initially diluting $20\mathrm{ml}$ of the original water sample into a $1000\mathrm{ml}$ volumetric flask. Subsequently, the sample was prepared and analyzed as depicted below:

Volumetric flask (500ml) #1

Sample Composition: 5 ml of sample + 0 mL of 300 ppm Na 0.5% LiCl $_2$

Absorbance: 0.12

Volumetric flask (500ml) #2

Sample Composition: 5 ml of sample + 1 ml of 300 ppm Na + 0.5% LiCl $_2$

Absorbance: 0.17

Volumetric flask (500ml) #3

Sample Composition: 5 ml sample + 2 ml of 300 ppm Na + 0.5% LiCl $_2$

Absorbance: 0.21

I. If all the flasks were diluted to a final volume of $500\mathrm{ml}$ with distilled water calculate the final concentration of sodium in the sample through a graphical and mathematical solution.

II. Why Lanthanum and Cesium or Lithium are added to solutions to be analyzed for Calcium and Sodium respectively?

Solution:

To solve this task the Beer-Lambert Law needed.

$$A = \varepsilon l c$$

where $A$ is absorbance, $\varepsilon$ is extinction coefficient, $l$ is absorption path length in cm, $c$ is concentration of solution in mol/l.

Conversion of concentration of standard solution from ppm to mol/l:

1 ppm is 0.001 g/l

300 ppm is $0.3\mathrm{g / l}$

0.3 g/l is 0.013 mol/l

Concentration of standard solution in volumetric flask (500 ml):

$$c _ {s t} = \frac {V _ {s t} \cdot c _ {s t}}{V _ {f l a s k}}$$

Input data:

Mathematical solution:

$$\left\{ \begin{array}{c} A _ {1} = \varepsilon l c _ {x} \\ A _ {2} = \varepsilon l (c _ {x} + c _ {s t}) \\ A _ {3} = \varepsilon l (c _ {x} + c _ {s t}) \end{array} \right.$$ $$\frac {A _ {1}}{A _ {2}} = \frac {\varepsilon l c _ {x}}{\varepsilon l (c _ {x} + c _ {s t})} = \frac {c _ {x}}{c _ {x} + c _ {s t}}$$ $$\frac {0 . 1 2}{0 . 1 7} = \frac {c _ {x}}{c _ {x} + 2 . 6 \cdot 1 0 ^ {- 5}}$$ $$c _ {x} = 6. 2 4 \cdot 1 0 ^ {- 5} \mathrm{mol} / \mathrm{l}$$ $$\frac {A _ {1}}{A _ {3}} = \frac {\varepsilon l c _ {x}}{\varepsilon l (c _ {x} + c _ {s t})} = \frac {c _ {x}}{c _ {x} + c _ {s t}}$$ $$\frac {0 . 1 7}{0 . 2 1} = \frac {c _ {x}}{c _ {x} + 5 . 2 \cdot 1 0 ^ {- 5}}$$ $$c _ {x} = 6. 9 0 \cdot 1 0 ^ {- 5} \mathrm{mol} / \mathrm{l}$$ $$\bar {c} _ {x} = 6. 5 7 \cdot 1 0 ^ {- 5} \mathrm{mol} / \mathrm{l}$$

Concentration of sodium in original sample of water:

$$\bar {c} _ {x} = 6. 5 7 \cdot 1 0 ^ {- 5} \frac {\mathrm{mol}}{\mathrm{l}} \cdot 5 0 0 0 = 0. 3 3 \frac {\mathrm{mol}}{\mathrm{l}}$$

Graphical solution:

$$c_x = \frac{\text{Intercept}}{\text{Slope}} = \frac{0.12167}{0.01731} = 7.03 \cdot 10^{-5} \text{mol/l}$$ $$c_x = 7.03 \cdot 10^{-5} \frac{\text{mol}}{\text{l}} \cdot 5000 = 0.35 \frac{\text{mol}}{\text{l}}$$

II. Lanthanum and Cesium or Lithium are added to solutions to be analyzed for Calcium and Sodium respectively like ionization suppression solution.

Answer provided by www.AssignmentExpert.com