# Solutions that are Mixtures of Acids and Bases, Problem B

**5. Calculate the pH of a solution that is prepared by mixing 75 ml of 0.088 M aniline with 50 ml of 0.097 M 2-nitrophenol.**

From the table, we can determine that aniline (An) is a base and nitrophenol (HNp) is an acid. This solution consists of a mixture of an acid and a base, so the first thing we must consider is that a neutralization reaction takes place. In this case we also note that aniline is a very weak base (K_{b} = 3.94\(\times\)10^{-10}) and nitrophenol is a very weak acid (K_{a} = 5.83\(\times\)10^{-8}). The value of K_{n} is calculated below.

K_{n} = K_{a} \(\times\) K_{b}\(\times\)10^{14} = (5.83\(\times\)10^{-8})(3.94\(\times\)10^{-10})(10^{14}) = 2.3\(\times\)10^{-3}

This value is fairly small, so we cannot assume that this neutralization reaction will go to completion. Instead we anticipate that this reaction will go to a small extent. Since it goes to only a small extent, we can try making the assumption that x is small compared to the initial concentrations of the aniline (0.0528 \(\gg\) x) and nitrophenol (0.0388 \(\gg\) x).

\[\begin{align}

& &&\ce{An}\hspace{35px} + &&\ce{HNp} \hspace{25px}\leftrightarrow &&\ce{AnH+} \hspace{25px} + &&\ce{Np-}\\

&\ce{Initial} &&0.0528 &&0.0388 &&0 &&0\\

&\ce{Equilibrium} &&0.0528 - \ce{x} &&0.0388 - \ce{x} &&\ce{x} &&\ce{x}\\

&\ce{Assumption} &&0.0528 \gg \ce{x} &&0.0388 \gg \ce{x} && &&\\

&\ce{Approximation} &&0.0528 &&0.0388 &&\ce{x} &&\ce{x}

\end{align}\]

These values can be plugged into the K_{n} expression to solve for x:

\[\mathrm{K_n = \dfrac{[AnH^+][Np^-]}{[An][HNp]} = \dfrac{(x)(x)}{(0.0528)(0.0388)} = 2.3\times10^{-3}}\]

\[\mathrm{x = 2.17\times10^{-3}}\]

Now we could solve the two Henderson-Hasselbalch equations for each of the conjugate pairs, since we know the concentrations of both members of each pair.

\[\mathrm{pH = pK_a + \log\left(\dfrac{[An]}{[AnH^+]} \right) = 4.596 + \log\left(\dfrac{0.0528}{0.00217} \right ) = 5.98}\]

\[\mathrm{pH = pK_a + \log\left(\dfrac{[Np^-]}{[HNp]} \right) = 7.234 + \log\left(\dfrac{0.00217}{0.0388} \right ) = 5.98}\]

The two identical values suggest that the pH of the solution will be 5.98. It is interesting to check the assumptions that were used in calculating the values.

\[\dfrac{0.00217}{0.0388}\times100=5.6\% \hspace{60px} \dfrac{0.00217}{0.0528}\times 100 = 4.1\%\]

One does meet the 5% rule, the other is just a little over. This might suggest that solving a quadratic is in order, however, if you solve the quadratic and substitute in the values, you will still get a pH of 5.98.