Calculate the concentration of SO_2 in rainwater at 25^degree Celsius that is in

Question

Calculate the concentration of SO_2 in rainwater at 25^degree Celsius that is in equilibrium with polluted air where the SO_2 concentration is 5 ppm_v. The Henry 's law constant for SO_2 is 1.2 times 10^-2 mol/L-atm. Molecular weight of SO_2 = 64. The concentration of SO_2 in rainwater is mg/L (ppm) For reaction Cr_2O_7^2 + SH^+ + 3H_2C_2O_4 right arrow 2Cr^3+ + 7H_2O + 6CO_2. the net change of oxidation state for each carbon atom (C) from H_2C_2O_4 to CO_2 is (Do not input the sign, input numbers only)

Explanation / Answer

According to henry's law the amount of dissolved gas is proportional to its partial pressure.

H=x/p where H=henry's constant

x=concentration in water

c=partial pressure of gas

H=1.2*10^-2 mol/L-atm* 64 g/mol*1000mg/g=768 mg/L-atm=768 ppm/atm

x=H*p=768 mg/L-atm *5 *10^-6 atm=0.00384 mg/L or ppm

2)H2C2O4

Let oxidation state of C be x,then net o.s for the molecule

2*(+1)+2X+4*(-2)=0

-6+2X=0

X=+3

In CO2

Let o.s of C be Y

Y+2*(-2)=0

Y=+4

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