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I'm reading about the electrical properties of solution where there is a problem like that:

The molar conductivity of $0.0250\ \mathrm M$ $\ce{HCOOH(aq)}$ is $4.61\ \mathrm{mS\ m^2\ mol^{-1}}$. Determine the $\mathrm pK_\mathrm a=-\log K_\mathrm a$ of the acid. (limiting ionic conductivity of $\ce{H+}=34.96\ \mathrm{mS\ m^2\ mol^{-1}}$ and limiting ionic conductivity of $\ce{OH-}=19.91\ \mathrm{mS\ m^2\ mol^{-1}}$)

I have to solve the problem using the equation below:

$$\frac1{\Lambda_\mathrm m}=\frac1{\Lambda_\mathrm m^0}+\frac{\Lambda_\mathrm mc}{K_\mathrm a\left(\Lambda_\mathrm m^0\right)^2}$$

Here which limiting ionic conductivity should I use to solve the problem? $\ce{H+}$ or $\ce{OH-}$?

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You need to add the limiting ionic conductivities for $\ce{H+}$ and $\ce{OH-}$ together to get the limiting ionic conductivity for all the ions in solution ($\Lambda_{0}$, which will replace $\Lambda^{0}_{\mathrm m}$ in your equation).

This arises from a simplification for calculating $\Lambda_{0}$ in weak electrolyte solutions (such as yours) according to Kohlrausch's Law in which it is stated:

Each ionic species makes a contribution to the conductivity of the solution that depends only on the nature of that particular ion, and is independent of the other ions present.

from which we can then estimate $\Lambda_{0}$ as:

$$\Lambda_{0} = \sum_{i}\lambda_{i,+}^{0} + \sum_{i}\lambda_{i,-}^{0}$$

For your problem:

$$\Lambda_{0} = \underbrace{(34.96 + 19.91)}_{54.87}\ \mathrm{mS\cdot m^{2}\cdot mol^{-1}}$$

we solve (I am omitting units here for clarity, and have confirmed via dimensional analysis that the answer is correct):

$${1\over 4.61} = {1\over 54.87} + {4.61\times 0.025\over K_{\mathrm a}\times (54.87)^{2}}$$

which simplifies to yield the answer (spoiler alert):

$$K_{\mathrm a} = 1.926\times 10^{-4} \implies \mathrm{p}K_{\mathrm a} = 3.72$$

and can be favorably compared to the value of 3.77 given here.

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