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Why is $\ce{CH3CH2NH3Cl}$ acidic?

I know that it has something to do with how it is formed from a weak alkaline, but I cannot identify which acid and base it is formed from.

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  • 1
    $\begingroup$ Its a salt of weak base and strong acid so on hydrolysis it forms acidic solution $\endgroup$ – JM97 Sep 1 '16 at 13:56
  • $\begingroup$ Have you tried drawing it? What do you think ought to be the cation and anion parts when dissociating in water? $\endgroup$ – SCH Sep 1 '16 at 14:03
  • $\begingroup$ As, mentioned its a salt formed by reacting Ethyl amine with HCl, $\endgroup$ – Khan Sep 1 '16 at 14:18
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In short

$\ce{CH3CH2NH3Cl}$ is acidic because it is formed from a strong acid ($\ce{HCl}$) and a weak base ($\ce{CH3CH2NH2}$).


Elaborated

Notice that this compound is ionic, comprising $\ce{CH3CH2NH3+}$ cation and $\ce{Cl-}$ anion.

I mentioned that $\ce{CH3CH2NH2}$ is weak, meaning that this reaction is reversible:

$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH-}\tag1$$

In an acidic environment, we can rewrite it as:

$$\ce{CH3CH2NH2 + H3O+ <=> CH3CH2NH3+ + H2O}\tag2$$

Therefore, when $\ce{CH3CH2NH3Cl}$ dissolves in water, there would be initially an equal amount of $\ce{CH3CH2NH3+}$ and $\ce{Cl-}$ in terms of number of moles.

Using equation $(2)$, we know that $\ce{CH3CH2NH3+}$ will react with water reaching an acidic equilibrium.

Note that there is no $\ce{OH-}$ to start with (very little in reality), so equation $(1)$ is not applicable.

Therefore, the following species would be present:

  • $\ce{CH3CH2NH3+}$
  • $\ce{CH3CH2NH2}$
  • $\ce{H2O}$
  • $\ce{H3O+}$
  • $\ce{Cl-}$

Now it is apparent that $\ce{H3O+}$ makes it acidic.

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  • 3
    $\begingroup$ No mistakes. I wish there were more first answers like this! Have my upvote! $\endgroup$ – Jan Sep 1 '16 at 22:12

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