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Concentration of water vs. equilibrium

  1. Mar 21, 2013 #1
    Hi,
    I understand from calculation that the molar concentration of Pure water is 55.5 moles/Liter
    Then how come in equilibrium reactions when calculation the dissociation constant, we say that the concentration of water is 1 Molar? This seems like a huge difference to me? What is it I need to understand here?
     
  2. jcsd
  3. Mar 21, 2013 #2

    DrDu

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    This is not correct.
    What enters the equilibrium constant is the ratio of the concentration of a substance c relative to some standard concentration ##c_0##, i.e. ##c/c_0##. For diluted substances this standard concentration is ##c_0=##1 mol/l (molarity) or 1 mol/ kg (molality) or the like (or more precisely the behaviour at infinite dilution extrapolated to a concentration of 1 mol/l). For solvents etc. we use as a standard state the pure substance, i.e. ##c_0=55,5## mol/l for water. In a dilute solution, the concentration ##c## of water is to an excellent extent equal to ##c_0## so that we can set the ratio equal to 1, at least for calculations with chemical precision.
     
  4. Mar 21, 2013 #3
    Okay, so I might understand what you are saying in 2 possible ways, which of the following two ways is the correct?

    So if we are measuring the equilibrium H2O + CO2 ⇔ H(+) + HCO3(-) which is a reaction taking place in the blood of a person, then the concentration of water is only 1Molar because a) we have a diluted solution, where water is the solvent so [H2O] =1M ?
    or b) the ratio between the initial concentration of water and final concentration ( at products) is almost the same so the ratio between the two concentrations is 1?
     
  5. Mar 21, 2013 #4

    DrDu

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    The equilibrium constant is a dimensionless quantity as it only depends on the ratios of concentrations to their respective standard concentrations.
    Specifically for your reaction

    ## K=(c_\mathrm{H^+}/1\mathrm{ mol/l})\cdot (c_\mathrm{HCO_3^-}/1\mathrm{ mol/l})/[(c_\mathrm{H_2O}/ 55,5\mathrm{ mol/l})\cdot (c_\mathrm{CO_2}/1\mathrm{ mol/l})]\approx (c_\mathrm{H^+}/1\mathrm{ mol/l})\cdot (c_\mathrm{HCO_3^-}/1\mathrm{ mol/l})/(c_\mathrm{CO_2}/1\mathrm{ mol/l})
    ##
     
  6. Mar 21, 2013 #5

    DrDu

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  7. Mar 21, 2013 #6
    That's exactly the part I don't understand: so you have 55.5M H2O in the denominator in your equation (which makes the fraction 55 times as small) and then you can say it's approximately the same as "≈" removing the 55.5M H2O? How come we can just do that?
     
  8. Mar 21, 2013 #7

    DrDu

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    The point is that c is to a very good approximation equal to c_0. So their ratio is 1.
    E.g. the concentration of water in a solution of CO2 containing 1 mol/l is still about c=(55,5-1) mol/l=54,5 mol/l, hence it differs very little from c_0
     
  9. Mar 21, 2013 #8
    PErfect, so the ratio between [H2O]_Start and [H2O]end is 1 :)
    Thank you.
     
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