# Problem 4

- Page ID
- 302678

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)**Calculate the pH of a solution that is prepared by mixing 45 ml of 0.224 M 3-chlorobenzoic acid with 30 ml of 0.187 M ethylamine.**

Chlorobenzoic acid (Hcba) is a weak acid with a pK_{a} value of 3.824. Ethylamine is not in the table, but ethylammonium, its conjugate acid, is (pK_{a} = 10.63). Therefore ethylamine (EA) is a weak base (pK_{b} = 3.37). This solution consists of a mixture of a weak acid and a weak base.

What happens when we mix an acid with a base? From prior material we should know that an acid and a base react with each other in what is known as a **neutralization reaction**. The neutralization reaction between chlorobenzoic acid and ethylamine is shown below.

\[\ce{Hcba + EA \leftrightarrow cba- + EAH+} \hspace{60px} \mathrm{K_n}\nonumber\]

We can calculate initial amounts of Hcba and EA that exist in solution, but some of these will react according to the neutralization reaction. What we need to know is the extent of the neutralization reaction, in other words the value of K for this reaction. There are no tables of K_{n} values so what we need to do is see if there is a way to come up with the K_{n} expression by adding up a series of reactions that we do have K values for.

We do have reactions for Hcba and EA that we can look up in the table. These are as follows:

\[\begin{align}

&\ce{Hcba + H2O \leftrightarrow cba- + H3O+} \hspace{60px} &&\mathrm{K_a\: of\: Hcba}\nonumber\\

&\ce{EA + H_2O \leftrightarrow EAH+ + OH-} \hspace{60px} &&\mathrm{K_b\: of\: EA}\nonumber

\end{align}\nonumber\]

Adding these two together produces the following reaction:

\[\ce{Hcba + EA + 2H2O \leftrightarrow cba- + EAH+ + H3O+ + OH-} \hspace{40px} \mathrm{K = K_a(acid) \times K_b(base)}\nonumber\]

This almost looks like K_{n} but it is not exactly the same. The reactant side has two water molecules, and the product side has the hydronium and hydroxide ion. Note that these species do not show up in the neutralization reaction above. As tempting as it might be to say hydronium and hydroxide will react to produce the water molecules (thereby just cancelling these out and ignoring them), they are real species in the reaction that need to be accounted for in the final form of K_{n}. The way to eliminate these would be to add in the following reaction:

\[\ce{H3O+ + OH- \leftrightarrow 2H2O} \hspace{60px} \mathrm{K = \dfrac{1}{K_w}}\nonumber\]

This reaction is the reverse of K_{w}, a reaction we have seen before. If the direction of a reaction is reversed, its equilibrium constant is just the inverse or reciprocal value, 1/ K_{w} in this case.

The final expression to calculate the value of K_{n} then is the following:

\[\mathrm{K_n = \dfrac{K_a(acid)\times K_b(base)}{K_w} = K_a(acid)\times K_b(base) \times 10^{14}}\nonumber\]

If we evaluate the value of K_{n} for the reaction in the problem, we get the following value.

\[\begin{align}

&\mathrm{pK_a\: (Hcba)} = 3.824 &&\mathrm{K_a = 1.5\times 10^{-4}}\nonumber\\

&\mathrm{pK_b\: (EA)} = 3.37 &&\mathrm{K_b = 4.27\times 10^{-4}}\nonumber

\end{align}\nonumber\]

\[\mathrm{K_n = (1.5\times 10^{-4})( 4.27\times 10^{-4})(10^{14}) = 6.4\times 10^6}\nonumber\]

This K_{n} value of slightly more than six million is very large. That says that this reaction, for all practical purposes, will go to completion.

Before solving this problem, it would be worth a digression to examine more generally what we might expect for the value of K_{n}. Can we always expect K_{n} to be large such that neutralization reactions always go to completion? Or are there occasions when K_{n} might be relatively small such that the reaction will not go to completion?

One thing to keep in mind is a solution with excessively dilute concentrations of an acid or base. For example, suppose the concentrations of the acid and base are on the order of 10^{-10} M. In this case, there is so little acid and base, that even if the K_{n} value were large, the actual extent of reaction could still be small. It is not that common that we would encounter such solutions in a laboratory setting where we usually use much higher concentrations. But this could occur in environmental samples for some species.

Assuming solutions with appreciable concentrations of acid and base, would we ever have a small value of K_{n}? Recollecting back, we talked about weak acids as having K_{a} values on the order of, from strongest to weakest, 10^{-3}, 10^{-5}, 10^{-7}, and 10^{-9}. Similarly weak bases had K_{b} values from strongest to weakest on the same scale (10^{-3} to 10^{-9}). Remembering the equation for K_{n}:

\[\mathrm{K_n = K_a \times K_b \times 10^{14}}\nonumber\]

We can see that it will take a mixture of an excessively weak acid and base to get a small value for K_{n}. For example, mixing an extremely weak acid with a K_{a} of 10^{-9} with an extremely weak base with a K_{b} of 10^{-9} will give a K_{n} of 10^{-4}, a small number. This neutralization reaction would not proceed much at all. If the acid had a K_{a} value of 10^{-7} and the base a K_{b} value of 10^{-7}, the value of K_{n} would be 1, an intermediate value. This neutralization reaction would proceed to some extent.

If we considered a neutralization reaction in which either the acid or base was strong (a strong acid or base might have a K_{a} or K_{b} value on the order of 10^{6} or higher), you would need the other species to have a K value of 10^{-20} or lower to get a small value of K_{n}. Since this is an unreasonably low value for the weak acid or base, any acid-base reaction that involves either a strong acid or a strong base will go to completion.

The first step is to calculate the initial concentrations of Hcba and EA, remembering that mixing the two solutions dilutes each of the species.

\[\textrm{[Hcba]: } \mathrm{(0.224\:M)\times\dfrac{45\:ml}{75\:ml} = 0.1344\:M}\nonumber\]

\[\textrm{[EA]: } \mathrm{(0.187\:M)\times\dfrac{30\:ml}{75\:ml} = 0.0748\:M}\nonumber\]

The next step, since K_{n} is large, is to allow the reaction to go to completion. This is a one-to-one reaction, so the species with the lower concentration will be used up and limit the amount of product that forms.

\[\begin{align}

& &&\ce{Hcba}\hspace{25px} + &&\ce{EA} \hspace{25px}\leftrightarrow &&\ce{cba-} \hspace{25px} + &&\ce{EAH+}\nonumber\\

&\ce{Initial} &&0.1344 &&0.0748 &&0 &&0\nonumber\\

&\ce{Completion} &&0.0596 &&0 &&0.0748 &&0.0748\nonumber

\end{align}\nonumber\]

Of course, the amount of EA cannot really be zero, since K_{n} is a finite value and there needs to be some finite amount of EA. The next step in this problem is to think that some small amount of back reaction occurs.

\[\begin{align}

& &&\ce{Hcba}\hspace{25px} + &&\ce{EA} \hspace{25px}\leftrightarrow &&\ce{cba-} \hspace{25px} + &&\ce{EAH+}\nonumber\\

&\ce{Initial} &&0.1344 &&0.0748 &&0 &&0\nonumber\\

&\ce{Completion} &&0.0596 &&0 &&0.0748 &&0.0748\nonumber\\

&\ce{Back\:reaction} &&0.0596 + \ce{x} &&\ce{x} &&0.0748 - \ce{x} &&0.0748 - \ce{x}\nonumber

\end{align}\nonumber\]

And we can now consider whether there are any approximations that can be made. Considering that K_{n} is so large, the extent of back reaction is very small. This means that it is likely that x is very small compared to 0.0596 and 0.0748, such that 0.0596 >> x and 0.0748 >> x.

\[\begin{align}

& &&\ce{Hcba}\hspace{25px} + &&\ce{EA} \hspace{25px}\leftrightarrow &&\ce{cba-} \hspace{25px} + &&\ce{EAH+}\nonumber\\

&\ce{Initial} &&0.1344 &&0.0748 &&0 &&0\nonumber\\

&\ce{Completion} &&0.0596 &&0 &&0.0748 &&0.0748\nonumber\\

&\ce{Back\:reaction} &&0.0596 + \ce{x} &&\ce{x} &&0.0748 - \ce{x} &&0.0748 - \ce{x}\nonumber\\

&\ce{Assumption} &&0.0596 >> \ce{x} && &&0.0748 >> \ce{x} &&0.0748 >> \ce{x}\nonumber\\

&\ce{Approximation} &&0.0596 &&\ce{x} &&0.0748 &&0.0748\nonumber

\end{align}\nonumber\]

Before we go on, it is worth examining these final concentrations. One interesting thing to note is that we have appreciable quantities of Hcba and cba^{–}. These two are conjugate pairs, and we know that a solution with appreciable quantities of both members of a conjugate pair represents a buffer. We can therefore use the appropriate Henderson-Hasselbalch equation for chlorobenzoic acid to solve for the pH of this solution.

\[\mathrm{pH = pK_a + \log \left(\dfrac{[cba^-]}{[Hcba]}\right) = 3.824 + \log\left(\dfrac{0.0748}{0.0596}\right)=3.92}\nonumber\]

Before assuming that this answer is the correct one, we ought to check our assumptions. Using the K_{n} expression, we can calculate the value of x.

\[\mathrm{K_n = \dfrac{[cba^-][EAH^+]}{[Hcba][EA]} = \dfrac{(0.0748)(0.0748)}{(0.0596)(x)} = 6.4 \times 10^6}\nonumber\]

\[\mathrm{x = 1.46\times 10^{-8}}\nonumber\]

This number is very small and obviously less than 5% of 0.0596 and 0.0748. If we assume that 1.46\(\times\)10^{-8} is the final value of EA after back reaction, and that 0.0748 is the final value of EAH^{+}, we have final values for both members of a conjugate pair. If we substitute these into the Henderson-Hasselbalch equation of ethylammonium we ought to get the same pH as above. (Note, the EA and EAH^{+} are not a buffer since the amount of EA is not appreciable. But if you know the concentration of both members of a conjugate pair, you can still use the Henderson-Hasselbalch equation to solve for the pH, since it is just a rearrangement of the K_{a} expression.)

\[\mathrm{pH = pK_a + \log\left(\dfrac{[EA]}{[EAH^+]} \right ) = 10.63 +\log\left(\dfrac{1.46\times10^{-8}}{0.0748} \right ) = 3.92}\nonumber\]

It should not be a surprise that the two values are identical.

## Contributors and Attributions

- Thomas Wenzel, Bates College (twenzel@bates.edu)
- Sourced from the Analytical Sciences Digital Library