[Borek]

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 cm² 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]

[Sunil Simha]

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 cm²/Ωmol.

knowing that, I can find the degree of dissociation as α= Λ/Λ^∞
Λ^∞ = 415 cm²/Ωmol.

So α = 1.23 x 10^-9.

Thus the water product is (c*α)² which is 4.67 x 10^-15 mol²/L².

[Borek]

Hard to deny, that's the expected answer.

Actually literature value for the water ion product at 20°C is 6.76×10^-15. Apparently the measurements were not as accurate as they pretend to be