Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on April 22, 2013, 06:30:41 PM

Not a difficult one, but definitely based on something else than all the previous ones.
Ultra pure water was put into a cell with electrodes both of exactly 2 cm^{2} surface and exactly 0.5 cm apart. At 283 K measured resistance was 8.772 MΩ. Estimate value of water ion product at this temperature, knowing that limiting molar conductivities are respectively
[tex]\Lambda^\infty_{H^+} = 275 \frac {cm^2} {\Omega ~mol}[/tex]
and
[tex]\Lambda^\infty_{OH^} = 140 \frac {cm^2} {\Omega ~mol}[/tex]

Given that resistance is 8.772 MΩ,
R = l/(kA) where l is the length and A is the area of the cell. k is the conductivity.
I get k= 2.85 x 10^{8} /Ωcm
So molar conductivity is Λ=k/c where c is the concentration. Here assuming 1g/cc as the density of water, I get c = (1/18) mol/cc.
This gives Λ = 51.3 x 10^{8} cm^{2}/Ωmol.
knowing that, I can find the degree of dissociation as α= Λ/Λ^{∞}
Λ^{∞} = 415 cm^{2}/Ωmol.
So α = 1.23 x 10^{9}.
Thus the water product is (c*α)^{2} which is 4.67 x 10^{15} mol^{2}/L^{2}.

Hard to deny, that's the expected answer.
Actually literature value for the water ion product at 20°C (http://www.chembuddy.com/?left=pHcalculation&right=waterionproduct) is 6.76×10^{15}. Apparently the measurements were not as accurate as they pretend to be ;)