# Problem 2

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

1. Calculate the pH of a solution that is 0.332 M in anilinium iodide.

The anilinium ion is in the table of Ka values and is a weak acid (pKa = 4.596).  Anilinium iodide could be formed by the reaction between aniline (the conjugate base of anilinium) and hydrogen iodide, as shown below.

$\ce{An + HI ↔ AnH+I- }\nonumber$

In water, anilinium iodide will dissociate to produce the anilinium cation and the iodide anion.  What must now be assessed is whether either of these ions will react with water.  The two possible reactions that could occur are shown below.

$\ce{AnH+ + H2O ↔ An + H3O+}\nonumber$

$\ce{I- + H2O ↔ HI + OH^-}\nonumber$

The anilinium ion is behaving as an acid and since it has a pKa value in the table (4.596), this reaction will occur.  The iodide ion is acting as a base.  To see if this reaction occurs, we would need to look up hydroiodic acid (HI) in the table, and see that it is a strong acid.  The important feature of strong acids is that, for all practical purposes, strong acids go 100% to completion.  This means that HI in water will dissociate 100% to H3O+ and I.  Actually, some amount of undissociated HI must remain, but it is so small that we never need to consider it under normal circumstances in water.  Regarding I acting as a base, this means that it will all stay as I and no HI will form as shown above.  We can therefore solve the answer to this problem by only using the reaction of the anilinium ion.

The procedure is rather analogous to what we have already used in problems 1 and 2 above.  We ought to write a table for initial values, equilibrium values, and then examine whether any assumptions can be made.  If x is the amount of AnH+ that reacts, there are two important assumptions that can be made in this problem.  One is that, because Ka is so small, very little of the AnH+ reacts so that 0.332 >> x.  However, enough AnH+ reacts to produce a much larger concentration of H3O+ than was initially in solution such that x >> 10-7 M.

\begin{align} & &&\ce{AnH+}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{An} \hspace{25px} + &&\ce{H3O+}\nonumber\\ &\ce{Initial\: amount} &&0.332 &&0 &&10^{-7}\nonumber\\ &\ce{Equilibrium} &&0.332 - \ce{x} &&\ce{x} &&\ce{x}+10^{-7}\nonumber\\ &\ce{Assumption} &&0.332 >> \ce{x} && &&\ce{x}>>10^{-7}\nonumber\\ &\ce{Approximation} &&0.332 &&\ce{x} &&\ce{x}\nonumber \end{align}\nonumber

These values can now be plugged into the equilibrium constant expression for the reaction.

$\mathrm{K_a=\dfrac{[An][H_3O^+]}{[AnH^+]}=\dfrac{(x)(x)}{0.332}=2.54×10^{-5}}\nonumber$

$\mathrm{x = [H_3O^+] = 2.9×10^{-3}}\nonumber$

$\mathrm{pH = 2.54}\nonumber$

Both approximations must now be checked for validity.

$\mathrm{\dfrac{2.93×10^{-3}}{0.332}×100=0.88\% \qquad\qquad \dfrac{10^{-7}}{2.93×10^{-3}}×100=0.0034\%}\nonumber$

Both are okay, so the pH we calculated above is correct.