Problem 5
- Page ID
- 302681
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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 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.
Contributors and Attributions
- Thomas Wenzel, Bates College (twenzel@bates.edu)
- Sourced from the Analytical Sciences Digital Library