At $\pu{25 ^\circ C}$, $\lambda_\infty (\ce{H+}) = \pu{3.5 \times 10^{-2} S m2 mol-1}$ and $\lambda_\infty (\ce{OH-}) = \pu{2 \times 10^{-2} S m2 mol-1}$. Determine $\mathrm{pH}$ and $K_\mathrm{w}$.

Given: Specific conductance $= \pu{5.5 \times 10^{-6} Sm-1}$ for $\ce{H2O}$

My Attempt:

I know I need to find the concentration of $\ce{H+}$ or $\ce{OH-}$ for $\mathrm{pH}$. $K_\mathrm{w}$ can be found subsequently. However, I have no clue as to how do I find that with the given information.

I am also aware of the relation between conductivity, Specific conductance, and molarity. It is easy to find the molarity from the given information, but that would be of no use as we need the concentration of the individual ions rather than the molarity of the solution.

Any help would be appreciated.

  • $\begingroup$ Is your second value 𝜆∞ valid for OH- ? $\endgroup$ – Maurice Jan 20 at 13:12
  • $\begingroup$ Yes, I forgot to include that. My bad! $\endgroup$ – Tony Jan 20 at 13:22

We know,

$$\lambda_c = \frac{(\text{Specific Conductivity})}{1000 \times c}$$

$$\alpha\lambda_\infty = \lambda_c$$

(where $c$ is concentration in moles per litres of the electrolyte and $\alpha$ is degree of dissociation of the electrolyte)

$\lambda_{\infty}$ of water (the electrolyte) is the sum of $\lambda_{\infty}$ of $\ce{H+}$ and $\ce{OH-}$.

From equations above, one can easily find the the value of $c\alpha$ is $10^{-7}$, which is the concentration of $\ce{H+}$.

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  • $\begingroup$ @Tony could you verify the answer calculated based on this answer and report us until we get a second opinion? $\endgroup$ – Sir Arthur7 Jan 20 at 14:14
  • $\begingroup$ I do get the correct answer based on this explanation. $\endgroup$ – Tony Jan 22 at 19:03

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