$\ce{0.007 M}$ aq. solution of anilinium hydrochloride had molar conductivity equal to $\pu{119.4 S cm2 mol-1}$, which became $\pu{103 S cm2 mol-1}$ when a few drops of aniline were added to the solution. Molar conductivity of $\ce{HCl}$ in the same ionic concentration was found to be $\pu{413 S cm2 mol-1}$. Find out the hydrolysis constant of anilinium hydrochloride.
I knew that $K_\text{h}=\dfrac{[\ce{PhNH2}][[\ce{H3O+}]}{[\ce{PhNH3+}]}$ and I tried to find out degree of hydrolysis $h=\dfrac{Λ_m^c}{Λ_m^∞}$ and hence $K_\text{h}$ but that attempt miserably failed. Can anyone please explain and give the correct method to go about this?
More details, from the comments:
I took the equilibrium,
$\ce{PhNH2 + HCl <=> PhNH3+ + Cl-}$
So I guessed that x=0.007 and found concentration, $K_h=0.007$. This was blatantly incorrect. But I am not able to proceed after writing down the reaction.
Further comment:
$h=\frac{Λ^c_m}{Λ^∞_m} = \frac {103}{119.4} = 0.86$ isn't it?
Thus $K_h = \frac {c \cdot h^2}{1-c} = \pu{3.7E-2}$
But this is incorrect... Answer is given as $\pu{2E-5}$
$\pu{2E-5}$
and $\ce{PhNH3+}$ as$\ce{PhNH3+}$
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