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Problem 5

1. 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 (Kb = 3.94$$\times$$10-10) and nitrophenol is a very weak acid (Ka = 5.83$$\times$$10-8). The value of Kn is calculated below.

$\mathrm{K_n = K_a \times K_b \times 10^{14} = (5.83 \times 10^{-8})(3.94 \times 10^{-10})(10^{14}) = 2.3 \times 10^{-3}}\nonumber$

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 >> x) and nitrophenol (0.0388 >> x).

\begin{align} & &&\ce{An}\hspace{35px} + &&\ce{HNp} \hspace{25px}\leftrightarrow &&\ce{AnH+} \hspace{25px} + &&\ce{Np-}\nonumber\\ &\ce{Initial} &&0.0528 &&0.0388 &&0 &&0\nonumber\\ &\ce{Equilibrium} &&0.0528 - \ce{x} &&0.0388 - \ce{x} &&\ce{x} &&\ce{x}\nonumber\\ &\ce{Assumption} &&0.0528 >> \ce{x} &&0.0388 >> \ce{x} && &&\nonumber\\ &\ce{Approximation} &&0.0528 &&0.0388 &&\ce{x} &&\ce{x}\nonumber \end{align}\nonumber

These values can be plugged into the Kn expression to solve for x:

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

$\mathrm{x = 2.17\times10^{-3}}\nonumber$

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.

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

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

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.

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

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.

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