8.2: Pre-lab
- Page ID
- 537577
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this pre-laboratory exercise, you will determine how dilution or the addition of a strong acid or base affects the pH of a buffer.
Suppose you need to prepare a buffer solution with a pH of 4.00 and a total buffer concentration of 0.10 M using benzoic acid, C6H5COOH, (pKa = 4.20) and sodium benzoate, C6H5COONa. The equilibrium involved in this buffer is
C6H5COOH(aq) + H2O(l) ⇌ C6H5COO-(aq) + H3O+(aq)
To simply the equation, let HA represent the acid and A- represent the benzoate ion.
HA(aq) + H2O(l) ⇌ A-(aq) + H3O+(aq)
Using this representation, the Henderson-Hasselbalch equation can be written as
pH = pKa + log [A-]/[HA] = 4.20 + log [A-]/[HA]
Since the total buffer concentration must be 0.10 M, then
[HA] + [A-] = 0.10 M
or [A-] = 0.10 - [HA]
Since we want a solution with a pH of 4.00, the Henderson-Hasselbalch equation can be written as follows:
4.00 = 4.20 + log (0.10-[HA])/[HA] or -0.20 = log (0.10-[HA])/[HA]
Using the definition of logarithms, we can write the above equation as
10-0.20 = (0.10-[HA])/[HA] or 0.63 = (0.10-[HA])/[HA]
Solving this equation gives
[HA] = 0.061 M
and [A-] = 0.039 M.
Therefore, a solution with a benzoic acid concentration of 0.061 M and a sodium benzoate concentration of 0.039 M should have a pH of 4.00. You should be prepared to do a calculation similar to this when you come to the laboratory session.
The initial pH water and this buffer is recorded in the table below. When 1.0 M HCl is added to the beaker containing water and the beaker containing the buffer solution, the pH changes as recorded below.
|
mL 1.0 M HCl added |
0.0 mL |
1.0 mL |
2.0 mL |
3.0 mL |
5.0 mL |
30 mL |
|
pH water |
6.65 |
2.15 |
1.97 |
1.88 |
1.78 |
1.50 |
|
pH buffer |
3.80 |
3.27 |
3.00 |
2.90 |
2.00 |
1.50 |
When 1.0 M NaOH is added to the beaker containing water and the beaker containing the buffer solution pH changes as recorded in the following table.
|
mL 1.0 M NaOH added |
0.0 mL |
1.0 mL |
2.0 mL |
3.0 mL |
5.0 mL |
30 mL |
|
pH water |
6.55 |
11.21 |
11.47 |
11.63 |
11.77 |
12.03 |
|
pH buffer |
3.75 |
4.39 |
4.84 |
5.47 |
11.81 |
12.09 |
When the indicated amount of pure water is added to the beaker containing water and the beaker containing the buffer solution, the pH changes are recorded in the table below.
|
mL of pure water added |
0.0 mL |
1.0 mL |
2.0 mL |
3.0 mL |
5.0 mL |
30 mL |
|
pH water |
4.92 |
4.88 |
4.90 |
4.93 |
4.95 |
4.91 |
|
pH buffer |
3.87 |
3.89 |
3.88 |
3.87 |
3.87 |
3.87 |
1. Upon adding a small amount of strong acid, how does the pH change for the buffer solution compare to that for water?
2. Upon adding a small amount of strong base, how does the pH change for the buffer solution compare to that for water?
3. Can a buffer indefinitely resist a significant change in pH? What observations do you have to support your answer?
4. How does dilution affect the pH of a buffer solution? Use the Henderson-Hasselbalch equation to explain your answer.