-1
$\begingroup$

Consider the following Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 + 2H2O (Source: https://chemiday.com/en/reaction/3-1-0-440)

It appears to be double displacement reaction as CaCl2 is formed, but it’s not as NH4OH is not formed instead NH3 is formed

Why NH3 is formed instead of NH4OH ?

Is there intermediate step which transform NH4OH into NH3( I don’t think so but still)

$\endgroup$
1
$\begingroup$

It is postulated that NH4OH as a molecule does not exist in solution. A better way to write might be NH3(aq). As to the evolution of gaseous ammonia, this reaction would only proceed if you heat up (i) a moist paste of NH4Cl and Ca(OH)2 or (ii) strongly heat a solution, otherwise ammonia would remain dissolved in the solution.

$\endgroup$
1
$\begingroup$

$\ce{NH4OH <=>NH3 + H2O}$, that is, they are in equilibrium. Since the solution is weakly ionized, (Kb is ~1.8 * 10-5 at 293 K), there is little $\ce{NH4OH}$ present at a given instant, so ammonia water is only weakly alkaline.

One might say the answer $\ce{NH3 + H2O}$ is 99.998% correct, but $\ce{NH4OH}$ is still 0.002% correct, not entirely wrong.

See also this discussion on what species are present.

$\endgroup$
  • 1
    $\begingroup$ As the reaction takes place in water medium ,(More water on RHS) Why can’t we apply le’chatliiers principle and say equilibrium shifts to left hand side. Similar statement is made In the linked answer $\endgroup$ – user72730 Apr 9 at 8:08
  • 1
    $\begingroup$ I think the debatable and interesting part is the question: Does NH4OH as a molecule exist in solution? It was the same story as the hypothetical H2CO3 but people finally detected it. Not sure about the status of ammonium hydroxide. $\endgroup$ – M. Farooq Apr 9 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy