{"id":15924,"date":"2016-03-20T19:34:04","date_gmt":"2016-03-20T19:34:04","guid":{"rendered":"http:\/\/www.ch.imperial.ac.uk\/rzepa\/blog\/?p=15924"},"modified":"2017-09-25T07:45:31","modified_gmt":"2017-09-25T06:45:31","slug":"how-many-water-molecules-does-it-take-to-form-ammonium-hydroxide-from-ammonia-and-water","status":"publish","type":"post","link":"https:\/\/www.ch.ic.ac.uk\/rzepa\/blog\/?p=15924","title":{"rendered":"How many water molecules does it take to form ammonium hydroxide from ammonia and water?"},"content":{"rendered":"
\n

This is a corollary to the previous post<\/a>\u2021<\/sup> exploring how many molecules are needed to ionise HCl. Here I am asking how many water molecules are required to form the ionic ammonium hydroxide from ammonia and water.<\/p>\n

As Wikipedia will inform you<\/a>, “it is actually impossible to isolate samples of NH4<\/sub>OH (more formally the ion-pair NH4<\/sub>+<\/sup>OH–<\/sup>\u00a0) as these ions do not comprise a significant fraction of the total amount of ammonia except in extremely dilute solutions\u00a0(<\/em>my italics)”. In fact, the ionization constant K<\/em>b<\/sub>\u00a0= [NH4<\/sub>+<\/sup>][OH–<\/sup>]\/[NH3][H2<\/sub>O]\u00a0is ~1.8 x 10-5 <\/sup>(pK<\/em>b\u00a04.75) equivalent to a free energy difference of \u00a0~6.5\u00a0kcal\/mol between the two forms.\u2020<\/sup> This is in stark contrast to solutions of e.g.<\/em> HCl in water, where essentially all of the HCl is ionised to hydronium chloride or\u00a0H3<\/sub>O+<\/sup>Cl– <\/sup>by addition of just ~4-5 water molecules.\u00a0So what is the water model\u00a0required\u00a0to compute this known behaviour of\u00a0ammonia? Here, this will be \u03c9B97Xd\/Def2-TZVPPD\/SCRF=water, i.e.<\/em>\u00a0a continuum water model\u00a0is already included and we add n<\/strong> further discrete water molecules to enhance it.<\/p>\n

For n=0 or 2,\u00a0the ion-pair is not an explicit minimum (although it appears to be a “hidden intermediate”[1]<\/a><\/span>).\u00a0Values of\u00a0e.g.<\/em> n=4,6,8\u00a0allow\u00a0the formation of two or three “bridges” comprising two or three water molecules connecting\u00a0the N and O atoms by hydrogen bonds\u00a0and\u00a0this additional solvation enables location of a transition state for proton transfer between O and N. This implies an equilibrium can be established as\u00a0NH3<\/sub> + H2<\/sub>O \u21cc*<\/sup>\u00a0NH4<\/sub>+<\/sup>.OH–<\/sup>\u00a0with the ion-pair now a\u00a0genuine minimum\u00a0stabilized by those ion-pair bridges. Note in particular how the hydrogen bond lengths involving the water salt-bridge in the ion-pair are shorter than for the neutral water-ammonia complex.<\/p>\n

\"NH3-8\"<\/p>\n

The contact ion-pair is nevertheless\u00a0a very shallow minimum when surrounded by 4 or more explicit waters, the barrier from proton transfer from N being less than a vibrational quantum, and so the lifetime of the contact ion-pair is very much\u00a0defined by the proton dynamics of the system..<\/p>\n

\"4\"<\/p>\n

\"8\"<\/a><\/p>\n

For n=8, the dipole moment changes along the\u00a0IRC for proton transfer between N and O as might be expected for the collapse of a contact ion-pair.<\/p>\n

\"8dm\"<\/a><\/p>\n

The relative free energies of the ion-pair and the un-ionized pair are shown below, the former being the higher. The values correspond approximately to the known ionization constant. As more explicit\u00a0water molecules are added, there is a hint that the proportion of ion-pairs might actually decrease\u00a0relative to neutral ammonia.\u00a0However, these calculations\u00a0are for a contact ion-pair and not a solvent-separated ion pair, the latter form possibly being\u00a0the more appropriate form for\u00a0extremely dilute solutions <\/em>(see above). Modelling the latter type of ion-pair is not as straightforward as the contact variety; \u00a0as the ion separation increases, so the propensity for barrierless proton transfers increases, ultimately leading back to the contact form. So to understand if it is correct that in\u00a0extremely dilute solutions <\/em>there is no remaining neutral ammonia, probably only a full molecular dynamics treatment of such a system is likely to give further insights.<\/p>\n\n\n\n\n
n<\/th>\n4<\/th>\n6<\/th>\n8<\/th>\n<\/tr>\n
\u0394\u0394G298<\/sub><\/td>\n6.4, DOI: 10.14469\/ch\/191950<\/a><\/td>\n5.9, DOI: 10.14469\/ch\/191957<\/a><\/td>\n7.0, DOI: 10.14469\/ch\/191946<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

To summarise, in order to compute the formation of the ammonium hydroxide ion pair from ammonia and water, one has to include an additional four or more explicit water molecules in the calculation. This model confirms that in the resulting equilibrium, only a tiny proportion of the ammonia becomes ionised. With such a base model in place, one can now proceed to investigate how addition of substituents on the nitrogen might impact upon such ionisation, i.e.<\/em> to form a stronger or a weaker base.<\/p>\n

\n

\u2021<\/sup>A more complete analysis followed.[2]<\/a><\/span> *<\/sup>If you are wondering how to produce a reversible arrow, see here<\/a>. \u2020<\/sup>This is only approximate, since the concentration of water needs renormalising.<\/p>\n

References<\/h2>\n
  1. https:\/\/doi.org\/<\/a>\n\n<\/li>\n
  2. A. Vargas\u2010Caamal, J.L. Cabellos, F. Ortiz\u2010Chi, H.S. Rzepa, A. Restrepo, and G. Merino, \"How Many Water Molecules Does it Take to Dissociate HCl?\", Chemistry \u2013 A European Journal<\/i>, vol. 22, pp. 2812-2818, 2016. https:\/\/doi.org\/10.1002\/chem.201504016<\/a>\n\n<\/li>\n<\/ol>\n\n<\/div> ","protected":false},"excerpt":{"rendered":"

    This is a corollary to the previous post\u2021 exploring how many molecules are needed to ionise HCl. Here I am asking how many water molecules are required to form the ionic ammonium hydroxide from ammonia and water. 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