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I have a question with $\ce{NaOH}$ initially present with given concentration and conductivity, then an equal volume of given concentration of HCl is added and the final conductivity is given, again a same volume of HCl of same concentration is added, the conductivity of which is again given. Now I need to find molar conductivity of $\ce{HCl ,NaOH, NaCl, H2O}$.

Molar conductivity($\lambda_m$), conductivity($\kappa$), Molarity($M$)

Using the formula $\lambda_m=\kappa/M$,firstly I found the molar conductivity of NaOH, afterwards when HCl was added, there will be only NaCl remaining, the rest being used up, so I again found molarity of NaCl and put in the formula and found molar conductivity, but in the last part I am confused, there is only HCl and NaCl remaining and I think one of these must be true: $$\kappa_{solution}=\kappa_{\ce{NaCl}}+\kappa_{\ce{HCl}}\\\lambda_{solution}=\lambda_{\ce{NaCl}}+\lambda_{\ce{HCl}}$$ Probably the second one due to Kolrauch's Law? Also I am not aware of what is going on here because I'm just using a pretty formula. Please explain in detail the physical situation and the processes involved.

Edit: The question is:

The conductivity of $0.1$ M $\ce{NaOH}$ is $0.0221\;\Omega^{-1}cm^{-1}$. When an equal volume of $0.1$ M $\ce{HCl}$ solution is added, the conductivity decreases to $0.0056\;\Omega^{-1}cm^{-1}$. A further addition of $\ce{HCl}$ solution, the volume of which is equal to that of first portion added, the conductivity increases to $0.017\;\Omega^{-1}cm^{-1}$. Calculate $\Lambda_m(\ce{NaOH}),\Lambda_m(\ce{NaCl}),\Lambda_m(\ce{HCl}),\Lambda_m(\ce{H^+,OH^-})$

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  • $\begingroup$ Your formula for molar conductivity is wrong, it is λ=κ/M. And you should better write the whole problem with all the data given. $\endgroup$
    – EJC
    Commented Sep 22, 2014 at 11:36
  • $\begingroup$ @Marko edited... $\endgroup$
    – RE60K
    Commented Sep 22, 2014 at 16:36

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As the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are equal, that concentration can be regarded as the "concentration" of the solution. If you divide both sides of your first equation (which includes $κ$) with that concentration you would get your second equation (which includes $λ$). So both equations are valid. The problem would arise if the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are different, you couldn't define the concentration of the solution, and you couldn't calculate it's molar conductivity, therefore only the first equation would be usable. Note that the Kolrauch's Law can also be expressed through $κ$.

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