Need solutions for #1, #2 Paragraph Styles 1. For the case of so issolving in wa

Question

Need solutions for #1, #2

Paragraph Styles 1. For the case of so issolving in water, what are so Hso the mole fractions of (aq), for SO s 5 ppm and pH of 5.0? (10 points) 2. What would be the pH change in a rain droplet the concentration of CO2 raises from if 400 to 700 ppm? (10 point). elected Henry's Law Constants and Equilibrium Constants at 298 K. Equilibrium H (mole L 1 atm 1) K (M) 1.0 x 10-14. CO2 H2O CO2. H20. 3.4 x 10-2. CO2. H2O HCO3 4.3 x 10 d. HCO3 CO32 4.7 x 10-11d. SO2 H2O SO2. H20. 1.23 1.32 x 10-2 HSO3 H+ SO32 6.42 x 10-8. NH3 H2O NH3. H20 62. NH3 H2O NH4+ OH 1.8 x 10 5. HNO3 H20 HNO3 H20. 2.1 x 105. HNO3 H2O H+ NO3 15.4 H2O2 H2O H2O2. H2O 7.45 x 104. 2.5 x 10-12 0.011

Explanation / Answer

1. for the case of so2 dissolving in water, what are the mole fractions of so2 (aq), HSO3-, SO32-, for
So2 = 5ppm and pH of 5.0?
PSO2 = (1 atm)(5 x 10-6) = 5 x 10-6 atm
[H2SO3] = KH.·PSO2
= (1.23 M/atm)(5 x 10-6 atm) = 6.15 x 10-6 M
Reaction:    H2SO3     H+ +    HSO3-   Ka2 = 0.0132

Ka2 = 0.0132 = [H+][HSO3-]/[H2SO3] = 10^-5 x [HSO3-]/6.15 x 10-6

[HSO3-] = 0.008 M

[H2SO3] = 6.15 x 10-6 M

HSO3-     H+ +    SO3-   Ka1 = 6.42 x 10-6
Ka1 = 6.42 x 10-6 = [H+][SO3-]/[HSO3-] = 10^-5 x [SO3-] /0.008

[SO3-] = 0.005 M

Total moles = moles of so2 (aq) + moles of HSO3- + moles of H2SO3 = 0.013 M

Mole fraction so2 (aq) = moles of so2 (aq)/ total moles = 6.15 x 10-6 M/0.013 M = 0.00047

Mole fraction HSO3- = moles of HSO3-/ total moles = 0.008 M/0.013 M = 0.61

Mole fraction H2SO3 = moles of H2SO3/ total moles = 0.005 M/0.013 M = 0.38

2. what would be the pH change in the rain droplet if the concentration of Co2 raises from 400 to 700 ppm

With the CO2 mixing ratio = 400 ppm, [CO2(aq)] = KH.PCO2 = (0.034 mol/L atm)(4.0 x 10-4 atm) = 1.36 x 10-5 M
Reaction:        CO2 + H2O     H+ +    HCO3-   Ka1 = 4.45 x 10-7

[H+] = [HCO3-] (assuming no other sources of H+) Ka1 = 4.3 x 10-7 = [H+][HCO3-]/[CO2]
                                                               = [H+]2/1.36 x 10-5 M
                                                               [H+]^2 = [(4.3 x 10-7)(1.36 x 10-5 M)]
                                                               =5.848 x 10^-12 M = 2.418 x 10-6
                                                               pH = 5.62
                                                              
   With CO2 mixing ratio = 700 ppm, [CO2(aq)] = 2.38 x 10-5 M,
                   [H+]^2 = [(4.3 x 10-7)(2.38 x 10-5 M)]
  
                   [H+]^2 = 10.234 x 10^-12
                   [H+]= 3.2 x 10^-6
       therefore; pH = 5.49
          
(so a decrease of 0.13 units).

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