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Concentration of water vs. equilibrium |
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| Mar21-13, 06:54 AM | #1 |
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Concentration of water vs. equilibrium
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? |
| Mar21-13, 07:05 AM | #2 |
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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. |
| Mar21-13, 07:21 AM | #3 |
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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? |
| Mar21-13, 07:54 AM | #4 |
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Recognitions:
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Concentration of water vs. equilibrium
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}) ## |
| Mar21-13, 08:00 AM | #5 |
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| Mar21-13, 08:39 AM | #6 |
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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?
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| Mar21-13, 09:27 AM | #7 |
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Recognitions:
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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 |
| Mar21-13, 09:51 AM | #8 |
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PErfect, so the ratio between [H2O]_Start and [H2O]end is 1 :)
Thank you. |
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