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$\ce{0.007 M}$ 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 $\pu{2E-5}$

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  • $\begingroup$ Deleting obsolete comments. // Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // No MathJax in titles. $\endgroup$
    – Poutnik
    Commented Nov 22, 2023 at 20:09
  • $\begingroup$ You can write $\pu{2E-5}$ as $\pu{2E-5}$ and $\ce{PhNH3+}$ as $\ce{PhNH3+}$ $\endgroup$
    – Poutnik
    Commented Nov 22, 2023 at 21:59
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    $\begingroup$ It’s either aniline hydrochloride or anilinium chloride. Otherwise, you’d have too many hydrogen atoms. $\endgroup$
    – Karsten
    Commented Nov 23, 2023 at 1:06
  • $\begingroup$ @Karsten the question said it and thus the word 'anilinium hydrochloride'. Ig its either $\ce{PhNH2+ HCl-}$ or $\ce{PhNH3+ Cl-}$ $\endgroup$ Commented Nov 23, 2023 at 3:28
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    $\begingroup$ PhNH2.HCl=aniline hydrochloride, PhNH3+ Cl- = anilinium chloride, PhNH3+ . HCl = anilinium hydrochloride . The last one is wrong. $\endgroup$
    – Poutnik
    Commented Nov 23, 2023 at 3:56

2 Answers 2

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The molar conductivity $\Lambda_\ce{AHCl hydr}=\pu{119.4 S cm2 mol−1}$ is equivalent to the partially hydrolyzed hydrolyzed anilinium chloride, according to the hydrolysis constant.

$$\ce{PhNH3+(aq) + Cl-(aq) + H2O(l) <=>[K_\text{h}] PhNH32(aq) + Cl-(aq) + H3O(aq)}$$

$$K_\text{h} = \frac{[\ce{PhNH2}][\ce{H3O+}]}{[\ce{PhNH3+}]}$$

The molar conductivity $\Lambda_\ce{HCl}=\pu{413 S cm2 mol−1}$ is equivalent to the fully hydrolyzed anilinium chloride.

$$\ce{PhNH3+(aq) + Cl-(aq) + H2O(l) -> PhNH32(aq) + Cl-(aq) + H3O(aq)}$$

The molar conductivity $\Lambda_\ce{AHCl}=\pu{103 S cm2 mol−1}$ is equivalent to the not hydrolyzed hydrolyzed anilinium chloride, as free aniline suppresses hydrolysis by increasing pH and shifting the equilibrium.

$$\ce{PhNH3+(aq) + Cl-(aq) + H2O(l) <-[PhNH2] PhNH32(aq) + Cl-(aq) + H3O(aq)}$$

Therefore, the degree of hydrolysis

$$h= \frac{\Lambda_\ce{AHCl hydr} - \Lambda_\ce{AHCl} }{ \Lambda_\ce{HCl} - \Lambda_\ce{AHCl}} = \frac{\pu{119.4 S cm2 mol−1} - \pu{103 S cm2 mol−1} }{ \pu{413 S cm2 mol−1} - \pu{103 S cm2 mol−1}}\approx 0.0529$$

$$K_\text{h} = \frac{[\ce{A}][\ce{H3O+}]}{[\ce{AH+}]}\approx \frac{[\ce{A}]^2}{[\ce{AH+}]}=\frac{c_{\ce{AHCl}}^2\cdot h^2}{c_{\ce{AHCl}}(1-h)}=\frac{c_{\ce{AHCl}}\cdot h^2}{1-h}=\\ =\frac{\pu{0.007 mol L-1}\cdot 0.0529^2 }{ 1 - 0.0529}\approx\pu{2.07E-5 mol L-1}$$

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  • $\begingroup$ In summation, molar conductivity of hydrolysed - molar conductivity of non hydrolysed = how much of the anilinium hydrochloride is actually being hydrolysed, and the molar conductivity of HCl - molar conductivity of non-hydrolysed anilinium hydrochloride = concentration of the conjugate acid of aniline. Thus, you got $h$ from there and then went on to find $K_h$. Thank you so much @Poutnik! $\endgroup$ Commented Nov 23, 2023 at 1:13
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This answer is more about the chemistry and how the conductivity relates to the ions present, and does not show a numerical solution. For that, check the answer by Poutin.

Here is the acid/base reaction without the chloride counter ion:

$\ce{PhNH3+ <=> PhNH2 + H3O+}$

The number of ions on the left and on the right is the same. However, the hydronium ion is a much better conductor, so the dissociation reaction changes the conductivity.

Chloride, on the other hand, is present at the same concentration in all three solutions, so it is not responsible for the changes in conductivity.

The part of the question I did not get at first is the addition of aniline. Aniline is neutral, so its concentration does not change the conductivity directly. However, as you can see in the equation above, it can react with hydronium ions to form anilinium ion. This will lower the conductivity of the mixture (ignoring dilution).

If you assume that the addition of the aniline drops the hydronium concentration sufficiently that the conductivity reflects 100% anilinium and chloride, then the way Poutnik calculated the equilibrium constant is sound. You can check after the fact that the approximation is valid by calculating the pH and the ratio of anilinium to hydronium before and after the addition of aniline.

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    $\begingroup$ aniline is slightly basic in nature having $pK_b=9.38$ and thus on addition of acid it will accept the proton donated by HCl thus forming anilinium hydrochloride (that's what the question says). So by common ion effect the hydrolysis is suppressed. At first when Poutnik said it I realised this fact and this answer gives me more clarity into this. $\endgroup$ Commented Nov 23, 2023 at 3:26

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