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What is the ratio of $\ce{MbO2/Mb}$ if the solution is saturated with air?

Details of the question: $$ \begin{align} \ce{Mb + O2 (aq) &-> MbO2}\\ \Delta G' &= -30.3~\mathrm{kJ/mol}\\ T &= 298~\mathrm{K}\\ \ce{pH} &= 7\\ \text{Henry's Constant for }\ce{O2} &= 42 \cdot 10^3~\mathrm{bar}\\ \text{Content of }\ce{O2}\text{ in dry air} &= 21\%\\ \end{align} $$

Attempt at Solution:

So, basically this is a two part question, as I see it. First I need to figure out the concentration of dissolved $\ce{O2}$ using Henry's Law. Then I can simply plug this value into the $$\Delta G' = -RT\ln K$$ equation where $$K = \frac{\ce{[MbO2]}}{\ce{[O2][Mb]}}.$$

Therefore: $$ \begin{align} P &= K\cdot m\\ 0.21~\mathrm{bar} &= (42 \cdot 10^3)\cdot m\\ m &= 5 \cdot 10^{-6} \pu{mol/kg}\\ -30.300~\mathrm{J/mol} &= -8.314 \cdot 298 \cdot \ln\left(\frac{\ce{[MbO2]}}{[5\cdot10^{-6}]\ce{[Mb]}}\right) \end{align} $$ After dividing and raising both sides to $e^x$ (to get rid of $\ln$) I get $$\frac{\ce{MbO2}}{\ce{Mb}} = 1.02.$$

However the answer sheet says it is $21.6$. I cannot figure out why this is so.

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  • $\begingroup$ I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. || For the ones unfamiliar (like me) with this kind of chemistry, it would be nice, if you could introduce your variables. What is $m$ or $P$? As a general remark: You should include units in every step. $\endgroup$ – Martin - マーチン Dec 18 '14 at 1:59
  • $\begingroup$ Thanks for the edit, I didn't know how to use the markup. Also units have been included where necessary (except for m which I have now included). I will, however, try to be more clear in the future $\endgroup$ – Dider Dec 18 '14 at 2:07
  • $\begingroup$ You are very welcome. As for the units, you should include them always and everywhere, they are usually the first hint that something went wrong, when they don't cancel correctly. $\endgroup$ – Martin - マーチン Dec 18 '14 at 2:31
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    $\begingroup$ I don't have time to answer properly, but I am inclined to think that the mistake is in the Henry's Law calculation. Usually when the constant is given as a pressure, the formula calculates a mole fraction rather than a molality. The answer you were given works back to a molarity of 1.1x10^-4M for oxygen, and water is 55 mol/L, so that works out nicely. @Martin It's the units that point me in that direction. $\endgroup$ – Jason Patterson Dec 18 '14 at 2:33
  • $\begingroup$ While I admit that there is an error in using molality instead of mole fractions, even after using mole fractions I am getting an answer of 2.78 x 10^-4 Mol/L for the O2 concentration $\endgroup$ – Dider Dec 18 '14 at 3:07

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