3
$\begingroup$

This is something that may seem small, but it bothers me … I mean really, really bothers me. The following issue will be expressed through the Br��nsted-Lowry theory of acids and bases and specifically through the neutralistaion reaction of ammonia and hydrochloric acid. When I dissolve hydrochloric acid in water, the following equilibrium reaction takes place: $$\ce{HCl + H2O <=> Cl- + H3O+}\tag{1}$$ When I then add a base such as ammonia, $\ce{NH3}$: $$\ce{NH3 + H3O+ <=> H2O + NH4+}\tag{2}$$ Adding these 2 equations should give: $$\ce{NH3 + HCl <=> NH4Cl}\tag{3}$$

This implies that the first reaction produces hydroxonium ions which the second reaction uses up, hence the number of hydroxonium ions in the solution should remain the same as before the addition of the hydrochloric acid.

Now let's actually asses the implications of such a procedure. First of all, assume that the reaction has gone to completion. This implies that when I mix water with hydrochloric acid, I will get hydroxonium and chloride ions produced. Subsequently, when I add the same number of moles of ammonia as hydrochloric acid, this will react with all of the hydroxonium ions present in the solution, as is shown in reaction (2). One problem though; this isn't a full reaction. If I add the same number of moles of ammonia as hydrochloric acid, this time there will be more hydroxonium ions than there were in the solution before addition of the hydrogen chloride (this what what I guess will happen, and this is the center of my problem). So how do I become sure that the hydroxonium ions cancel out when the equations are added?

Improve this question
$\endgroup$
15
  • 3
    $\begingroup$ The first problem: you can't assume that "the reaction has gone to completion", because the ammonia clearly doesn't. Next, you're assuming that the balanced equation represents everything that is going on in the system. You know already that the system involves the reaction of species such as $\ce{H2O}$ and $\ce{H3O+}$, so don't expect an equation that doesn't include those species to necessarily be accurate. A more accurate equation would be something like $\ce{NH3 + HCl + H2O <=> $(1-x)$NH4+ + x NH3 + Cl- + x H3O+}$, however that's incredibly tedious and nobody likes writing that. $\endgroup$
    – orthocresol
    Commented Aug 17, 2016 at 5:24
  • $\begingroup$ Why would the real equation look like that? I'm interested. Please do tell me through a full answer. $\endgroup$
    – Mathematician
    Commented Aug 17, 2016 at 8:35
  • $\begingroup$ Simply because some of the $\ce{NH4+}$ ions will undergo hydrolysis: $\ce{NH4+ + H2O <=> NH3 + H3O+}$. The issue is that without consulting a thermodynamic data table (and I am not about to do so right now) you don't know exactly how much of it will do that, hence the variable $x$. Sorry, don't have time for lots of explanation. $\endgroup$
    – orthocresol
    Commented Aug 17, 2016 at 8:42
  • $\begingroup$ Also, I made the assumption that the reaction goes to completion just to shown you why I thought it's ok to cancel out some species in the case where the reaction is complete. However, since NO REACTION is ever complete, I also wanted to show that this whole cancelling thing which we learnt through Hess' law was actually wrong. I only noticed that it shouldn't work when I started studying acid- base equilibria (which I'm still on). Furthermore, since you just said they don't actually cancel, I'm doubting the truth of the Bronsted-Lowry theory. $\endgroup$
    – Mathematician
    Commented Aug 17, 2016 at 8:44
  • $\begingroup$ The reaction equation you have written is a simplification of the real thing. Yes strictly speaking that means it is "wrong", however as was mentioned in my comment (as well as the answer) the equation is not meant to be rigorously correct, it is just a tool that captures the essence of the reaction (from a glance one can see it is acid-base) and also keeps track of stoichiometry. I don't see how there is a problem with Bronsted-Lowry theory, which merely says that an acid is a proton donor and a base a proton acceptor - nothing less, nothing more. $\endgroup$
    – orthocresol
    Commented Aug 17, 2016 at 8:48

2 Answers 2

Reset to default
4
$\begingroup$

When you write a chemical equation, you don't care whether the reaction has gone to completion or not. On a deeper level, none of them do. There is always an equilibrium, and there is always some starting compound left. To find out how much is that "some", you need to know the equilibrium constant and make certain calculations.

The written equation does not tell you that each and every molecule in the sample should react exactly like that. This is simply not what it is all about. It does, though, tell you another thing: if you got such-and-such amounts of products, you must have lost such-and-such amounts of the starting compounds. This is simply the conservation of mass, and there is no going around it.

Improve this answer
$\endgroup$
1
$\begingroup$

(1) "H2O+HCl-> H3O$^+$ + Cl$^- has gone to completion" fair enough if the concentration of HCl is low.

(2) NH3+H2O <-> HN4OH never goes to completion. So, when you add 1 mole of HCl*H2O to 1 mole of NH3*H2O you actually add 1 mole of Cl$^-$+1mole of H3O$^+$+ x mole of OH$^-$+ x mole of Nh4$^+$ + 1-x moles of NH3*H2O. (0

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.