4. (30pts; 20,10) A groundwater contains 0.132 mM hydrogen sulfide (H2S) and 2.9

Question

4. (30pts; 20,10) A groundwater contains 0.132 mM hydrogen sulfide (H2S) and 2.93x10 pM radioactive Radon-222 (Rn- 222). Recall the prefix p indicates pico, which is a factor of 101 What percent removal of H2S will be achieved at pH 7.04 if the following conditions and constants apply? P(H,S, clean air)-1.0x10 atm; K(HS,25°c)-102.3mM/atm; pK,t 7.04 for reaction H,S(aqe->H'(aq)+HS (aq) at 25°C; neglect K,z at pH7. a. b. What percent removal of Rn-222 will be achieved if the following conditions and constants apply? P(Rn-222, clean air)=2.8x10 18atm; KH(Rn-222, 25°C)-9.4m M/atm

Explanation / Answer

By means of air stripping, unwanted gases such as H2S and radon are removed. The Henry's law constant and the partial pressure of the gases in the fresh air blown to the groundwater is given.

The Henry's law constant of H2S indicates that 102.3 mM of the gas is aerated per unit atm pressure of air. The partial pressure of H2S in the air is 1.0 x 10-7 atm. So, the percentage amount of H2S that is removed:

[(102.3 x 10-7) / (0.132)] x 100 = 0.0077%

The Henry's law constant of Rn-222 indicates that 9.4 mM of the gas is aerated per unit atm pressure of air. The partial pressure of Rn-222 in the air is 2.8 x 10-18 atm. So, the percentage amount of Rn-222 that is removed:

[(9.4 x 2.8 x 10-18 x 109 ) / (2.93 x 10-4)] x 100 = 0.0089%

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