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I was watching a video in Khan academy about acids and bases. In the video, they asked the $\mathrm{pH}$ of a solution that is $\pu{0.15M}$ of $\ce{NH3}$ ($K_\mathrm{b}=1.8\times10^{-5}$) and $\pu{0.35M}$ of $\ce{NH4NO3}$. So, it was obvious to me that the first reaction is going to be that of a weak base because the $K_\mathrm{b}$ is given. But what wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.

Since this problem is presented in a beginners learning video I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base according to the information given in the video.

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closed as unclear what you're asking by Mithoron, Todd Minehardt, pentavalentcarbon, Tyberius, M.A.R. ಠ_ಠ Mar 17 '18 at 0:35

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tl;dr

What wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.

Remember the rule: "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution"


I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base

Realize that all such types of species of the form $\ce{A+B-}$ are ionic salts. If you try draw out the Lewis structure of $\ce{NH4NO3}$, you'll quickly realize it to be very different from that of, say, water or acetic acid.

In ammonium nitrate, you have two separate ions - ammonium cation and nitrate ion - held together by strong electrostatic forces of attraction (often called an "ionic bond"). Water, having a high dielectric constant, is able to break these electrostatic forces easily and dissolve the salt in 100% concentration.

Hence, we say that "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution" i.e. when dissolved in water, they'll dissociate completely. It seems you already knew the latter part, while the part for ionic salts is new for you.

As I said before, note that ionic salts are different from weak acids or weak bases like $\ce{CH3COOH}$. The $\ce{O-H}$ bond here is covalent and breaking it is an altogether different job as compared to breaking ionic bonds. Here, the reaction (or "equilibrium") has to favor the dissociation of acetic acid. This is ensured by the high stability of the conjugate base. If you've taken GOC classes, you would quickly realize that the conjugate base $\ce{CH3COO-}$ is much weaker than say $\ce{HSO3-}$ or $\ce{NO3-}$ (guess which acids the latter two are? ;) )

Again, realize that the rule is actually not completely correct. Later down the equilibrium road, you'll come across the equilibrium constant $K_\mathrm{sp}$ that defines exactly how much soluble certain salts are in aqueous solutions. Certainly, many ionic salts would be in that list, and that too with very low solubilities instead! Again, the way these people simplify calculations at an elementary level is to implicitly assume that "All ionic salts whose $K_\mathrm{sp}$ is provided are only partially soluble, while those whose $K_\mathrm{sp}$ is not provided are completely soluble (i.e. 100% dissociated)"

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The solution which is $0.15$ M in $\ce{NH_3}$ and $\ce{0.35 M}$ in $\ce{NH_4NO_3}$, acts as nothing but a basic buffer solution.
Applying the Henderson-Haselbach equation for basic buffers, $$\ce{pOH = pK_b + log({[salt]/[base]})}$$ Here the salt ( which is actually the salt formed by the base after reacting with strong acid) is $\ce{NH_4NO_3}$ because it is formed by the reaction of weak base $\ce{NH_3}$ with strong acid $\ce{HNO_3}$ and the salt concentration is $0.35$ M.
$\ce{pK_b}$ = $\ce{-log(1.8* 10^{-5}})$ =$4.745$.
So, the $\ce{pH}$ will be ,$$\ce{pH= 14 - pOH =14 -4.745 -log(0.35/0.15) = 8.887}$$ So, the pH can be determined simply by considering the buffer solution.
For your query, about the ionic property of $\ce{NH_4NO_3}$ , the $\ce{NH_4^+}$ ions and $\ce{NO_3^-}$ ions are simply held by pure electrostatic attraction. So, when you dissolve them in a solution they will purely dissociate, because there is no other force ( like covalent bonding ) to hold them back or dissociate partially. They have to fully dissociate in the solution in $\ce{NH_4^+}$ and $\ce{NO_3^-}$.

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