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Chemical Thermodynamics: CO2 dissolves in water, find molality and pH.

  1. May 16, 2011 #1
    1. The problem statement, all variables and given/known data
    When carbon dioxide "dissolves" in water, essentially all of it reacts to form carbonic acid, H2CO3:

    CO2(s) + H2O(l) <--> H2CO3(aq).

    The carbonic acid can then dissociate into H+ and bicarbonate ions,
    H2CO3(aq) <--> H+(aq) + HCO3-(aq).

    Consider a body of otherwise pure water that is in equilibrium with the atmosphere near sea level, where the partial pressure of carbon dioxide is 3.4 x 10-5 bar (aka 340 ppm). Calculate the molality of carbonic acid and of bicarbonate ions in the water, and determine the pH of the solution. Note that even 'natural' precipitation is somewhat acidic.

    2. Relevant equations
    At 298 K and 1 bar:
    [tex]\Delta G[/tex] of [tex] H_{2}CO_{3}[/tex](aq) = -623.08 kJ
    [tex]\Delta G[/tex] of [tex] CO_{2}[/tex](g) = -394.36 kJ
    [tex]\Delta G[/tex] of [tex] H_{2}O[/tex](l) = -237.13 kJ
    [tex]\Delta G[/tex] of [tex] H^{+}[/tex](aq) = 0 kJ
    [tex]\Delta G[/tex] of [tex] HCO_{3}^{–} [/tex](aq) = -586.77 kJ

    3. The attempt at a solution

    Molalilty of carbonic acid:
    First reaction [tex]\Delta G = 8.41 kJ[/tex].
    [tex] m = (P/P_o)e^{-\Delta G / RT} [/tex]
    [tex] m = (3.4 x 10^(-4) bar)/(1 bar) * e^{\frac{-8.41 x 10^3 J}{(8.315 J/mol k)* 298 K}} [/tex]
    [tex] m = 1.141 x 10^{-5} [/tex] mol/kg

    Molality of Bicarbonate ions:
    Second reaction: [tex]\Delta G = 36.31 kJ[/tex].
    [tex] m_{H^{+}} m_{HCO_{3}^{–}} = e^{-\Delta G / RT} [/tex]
    [tex] m_{H^{+}} m_{HCO_{3}^{–}} = \displaystylee^{\frac{-36.31 x 10^3 J}{(8.315 J/mol k)*298 K}} [/tex]
    [tex] m_{H^{+}} m_{HCO_{3}^{–}} = 4.3247 x 10^{-7} [/tex]
    Since one of each ion comes out of every carbonic acid molecule, we must have:
    [tex]m_{H+}=m_{CO_{3}^{–}} = sqrt{(4.3247E{-7})} = 6.576E{-4} mol / kg [/tex].

    Then the pH is:
    [tex] pH = –log_{10}(m_{H^{+}}) = –log_{10}(6.576 x 10^{-4}) = 3.18 [/tex]

    This all seems well and good, but the final answer for pH that I'm getting is hugely acidic. And the problem says that we should get something just slightly acidic, since this is like rain water. I sense that the issue might be that when I calculate the molality of the bicarbonate ions, I'm not taking into account the fact that the carbonic acid was already really dilute. But I don't know how I would take that into account. Help please?

    Thanks!
     
    Last edited: May 16, 2011
  2. jcsd
  3. Mar 27, 2016 #2
    divide by molality in equation below \displaystylee is my guess.
     
  4. Mar 27, 2016 #3

    SteamKing

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    Please don't necropost to these old threads. The OP hasn't been seen on PF in more than 3 years and is long gone.
     
  5. Mar 27, 2016 #4
    what's the harm. someone might google this the way i did, it's not necessarily for OP.
     
  6. Mar 27, 2016 #5

    SteamKing

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    Yeah, but the folks running PF don't like having old threads dredged up.
     
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