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This question already has an answer here:

A book which I am reading has this topic on hydrolysis of salts.
Suppose $\ce{NH4Cl}$ is dissolved in water. It is a salt of a strong acid and a weak base. $\ce{NH4+}$ ions being the conjugate acid of a weak base is strong and reacts with $\ce{H2O}$ to give $\ce{H+}$ thus making the solution acidic. $$\ce{NH4+ + H2O <=> NH4OH + H+}$$ Now for calculating degree of hydrolysis $h$ it takes the initial concentration to be $C$. This means $C(1-h)$ of $\ce{NH4+}$ remains and $Ch$ each of $\ce{NH4OH}$ and $\ce{H^+}$ is formed. Therefore, $K_h = \frac{Ch^2}{1-h}$. In the next step $1-h$ has been approximated as $1$. While calculating degree of dissociation of weak electrolyte we do a similar thing but that is because degree of dissociation is too small compared to 1 but here, I cannot understand how can degree of hydrolysis be very small compared to 1 when $\ce{NH4+}$ is a strong acid.

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marked as duplicate by Jan, pentavalentcarbon, ron, airhuff, Todd Minehardt Nov 26 '17 at 19:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I am aware that the question looks like it’s asking something very different, but I believe that the core of your problem is the one answered in the question I am about to link as a duplicate. Or, to put it in one sentence: what in the world makes you think that ammonium be a strong acid? $\endgroup$ – Jan Nov 26 '17 at 14:36
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I cannot understand how can degree of hydrolysis be very small compared to 1 when $\ce{NH4+}$ is a strong acid.

That is wrong.

You also wrote:

It is a salt of a strong acid and a weak base.

Well, the salt of a strong acid and a weak base, i.e. the conjugated acid of that weak base is a weak acid.

In the case of the ammonium ion, $pK_a=9.25$, very weak indeed.

Also, the hydolysis reaction equilibrium is (because free $\ce{H+}$ don't exist in water and $\ce{NH4OH}$ is old notation, long obsolete):

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$

$h$ is negligibly small because this equilibrium leans very strongly to the left, so $[\ce{H3O+}]\approx 0$.

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