Question

How many grams of nitrogen and oxygen are dissolved in 125g of water at 20.0C when the water is saturated with air, in which Pnitrogen(2) equals 593 torr and Poxygen(2) equals 159 torr? At 1.00 atm pressure, the solubility of pure oxygen in water is 0.00430g O(2)/100.0g H(2)O, and the solubility of pure nitrogen in water is 0.00190g N(2)/100.0g H(2)O.

Answer 1

Answer on Question #50329, Chemistry, Inorganic Chemistry

How many grams of nitrogen and oxygen are dissolved in 125 g of water at 20°C when the water is saturated with air, in which Pnitrogen equals 593 torr and Poxygen equals 159 torr? At 1.00 atm pressure, the solubility of pure oxygen in water is 0.00430 g O₂/100.0 g H₂O, and the solubility of pure nitrogen in water is 0.00190 g N₂/100.0 g H₂O?

Solution:

Henry's law can be put into mathematical terms (at constant temperature) as

$$p = k _ {H} \times c$$

where p is the partial pressure of the gaseous solute above the solution,

c is the concentration of the dissolved gas and

k_H is a constant with the dimensions of pressure divided by concentration. The constant, known as the Henry's law constant, depends on the solute, the solvent and the temperature.

$$k _ {H} = \frac {p _ {0}}{c _ {0}} = \frac {1 \mathrm {a t m} \times 1 0 0 \mathrm {g}}{\mathrm {m} _ {\mathrm {g a s 0}}}$$ $$k _ {H} = \frac {p _ {\text {p a r t i a l}}}{c} = \frac {p _ {\text {p a r t i a l}}}{\frac {m _ {\text {g a s}}}{m _ {\text {w a t e r}}}} = \frac {p _ {\text {p a r t i a l}} \times m _ {\text {w a t e r}}}{m _ {\text {g a s}}}$$ $$\frac {1 \mathrm {a t m} \times 1 0 0 \mathrm {g}}{\mathrm {m} _ {\mathrm {g a s 0}}} = \frac {\mathrm {p} _ {\mathrm {p a r t i a l}} \times \mathrm {m} _ {\mathrm {w a t e r}}}{\mathrm {m} _ {\mathrm {g a s}}}$$ $$m _ {g a s} = \frac {p _ {p a r t i a l} \times m _ {w a t e r} \times m _ {g a s 0}}{1 \mathrm {a t m} \times 1 0 0 \mathrm {g}}$$

Convertion of torr into atm:

$$p _ {a t m} = \frac {p _ {t o r r} \times 1 a t m}{7 6 0 t o r r}$$

So:

$$m _ {g a s} = \frac {\frac {p _ {t o r r} \times 1 a t m}{7 6 0 t o r r} \times m _ {w a t e r} \times m _ {g a s 0}}{1 a t m \times 1 0 0 g} = \frac {p _ {t o r r} \times m _ {w a t e r} \times m _ {g a s 0}}{7 6 0 t o r r \times 1 0 0 g}$$

Oxygene:

$$m _ {O _ {2}} = \frac {1 5 9 \mathrm {t o r r} \times 1 2 5 \mathrm {g} \times 0 . 0 0 4 3 0 \mathrm {g}}{7 6 0 \mathrm {t o r r} \times 1 0 0 \mathrm {g}} = \frac {8 5 . 4 6 2 5 \mathrm {g}}{7 6 0 0 0} = 0. 0 0 1 1 2 4 5 \mathrm {g}$$

Nitrogen:

$$m _ {O _ {2}} = \frac {5 9 3 t o r r \times 1 2 5 g \times 0 . 0 0 1 9 0 g}{7 6 0 t o r r \times 1 0 0 g} = \frac {1 4 0 . 8 3 7 5 g}{7 6 0 0 0} = 0. 0 0 1 8 5 3 1 2 5 g$$

Answer:

0.0011245 g of Oxygene (O₂)

0.001853125 g of Nitrogen (N₂)

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