5.R: Le Chatelier's Principle (Lab Report)
- Page ID
- 127150
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Name: ____________________________ Lab Partner: ________________________
Date: ________________________ Lab Section: __________________
Part A – Equilibrium and an Acid-Base Indicator
Equilibrium system:
\[\ce{ HA (aq) <=> H^{+} (aq) + A^{-} (aq)}\]
Observations
Record your results upon completing each of the following steps:
| Step 1: Color of bromothymol blue in distilled water | |
|---|---|
| Step 2: Name of reagent “A” causing color change when added | |
| Step 3: Name of reagent “B” causing a return to original color |
Analysis
- Complete the following:
The acidic form of the bromothymol blue indicator, \(\ce{HA}\) (aq), is _______________ in color.
The basic form of the bromothymol blue indicator, \(\ce{A^{–}}\) (aq), is _______________ in color.
- Explain why reagent A (in Step 2) caused the color change observed.
- Explain why reagent B (in Step 3) caused the color change observed.
Part B – Solubility Equilibrium and \(K_{sp}\)
Equilibrium system:
\[\ce{PbCl2 (s) <=> Pb^{2+} (aq) + 2 Cl^{–} (aq)}\]
Observations
| Step 4: Observations upon addition of just 1.0 mL of \(\ce{HCl}\) to the \(\ce{Pb(NO3)3}\) solution | |
|---|---|
| Step 5: Total volume of \(\ce{HCl}\) required for noticeable precipitation | mL |
| Step 6: Observations upon placing the test tube with precipitate in hot water | |
| Step 7: Observations upon placing the test tube with precipitate in cold water | |
| Step 9: Volume of water added to just dissolve \(\ce{PbCl2}\) precipitate | mL |
| Step 10: Total solution volume upon completion | mL |
Analysis
- Why didn’t any solid \(\ce{PbCl2}\) form immediately upon addition of 1 mL of \(\ce{HCl}\) (aq) in Step 4? What condition must be met by \([\ce{Pb^{2+}}]\) and \([\ce{Cl^{–}}]\) if solid \(\ce{PbCl2}\) is to form?
- Consider your observation in hot water in Step 6:
In which direction did the equilibrium shift? ____________________
Did the value of \(K_{sp}\) get smaller or larger? ____________________
Is the dissolution of \(\ce{PbCl2}\) (s) exothermic or endothermic? ____________________
Explain below.
- Explain why the solid \(\ce{PbCl2}\) dissolved when water was added to it in Step 9. What was the effect of this water on \([\ce{Pb^{2+}}]\), \([\ce{Cl^{–}}]\), and \(Q_{sp}\)? In which direction would such a change drive the equilibrium system?
- The point at which the \(\ce{PbCl2}\) precipitate just dissolves in Step 9 can be used to determine the value of \(K_{sp}\) for this equilibrium system, where \(K_{sp} = [\ce{Pb^{2+}}][\ce{Cl^{–}}]^{2}\). Calculate \([\ce{Pb^{2+}}]\) and \([\ce{Cl^{–}}]\) in the final solution (consider the “dilution effect”). Then use these equilibrium concentrations to determine the value of \(K_{sp}\) for this system. Show all work below.
Part C – Complex Ion Equilibria
Equilibrium system:
\[\underbrace{\ce{Co(H2O)6^{2+}(aq) }}_{\text{Pink}} + \ce{4Cl^{-} (aq) <=> } \underbrace{\ce{CoCl4^{2-}(aq) }}_{\text{Blue}} + \ce{6 H2O (l) }\]
Observations
| Step 2: Color of solution in 12 M HCl | |
|---|---|
| Step 3: Color of solution upon addition of water | |
| Step 4: Color of solution in hot water | |
| Step 5: Color of solution in cold water |
Analysis
- What form of the complex ion, \(\ce{Co(H2O)6^{2+}}\) (aq) or \(\ce{CoCl4^{2–}}\) (aq), is predominate in:
The 12 M \(\ce{HCl}\) (aq) __________________
The diluted solution __________________
The heated solution __________________
- Explain why you obtained the observed color in 12 M \(\ce{HCl}\) (aq) (Step 2).
- Explain the observed color change that occurred when water was added to the solution in Step 3. Consider how water affects the ion concentrations and \(Q\) in this system.
- Consider your observations in the hot water bath in Step 4.
In which direction did the equilibrium shift? ______________________
Did the value of K get smaller or larger? ______________________
Is the reaction (as written) exothermic or endothermic? ______________________
Explain.
Part D – Dissolving Insoluble Solids
Equilibrium system:
\[\ce{Zn(OH)2 (s) <=> Zn^{2+} (aq) + 2 OH^{-} (aq)}\quad K_{sp} << 1 \nonumber\]
Observations
| Step 1: Adding 1 drop of \(\ce{NaOH}\) (aq) to \(\ce{Zn(NO3)2}\) (aq) | |
|---|---|
| Step 2: Tube A: Effect when \(\ce{HCl}\) (aq) is added | |
| Step 3: Tube B: Effect when \(\ce{NaOH}\) (aq) is added | |
| Step 4: Tube C: Effect when \(\ce{NH3}\) (aq) is added |
Analysis
- Explain your observation upon addition of \(\ce{HCl}\) (aq) to the precipitate in Tube A. You must consider the various equilibria that are occurring in solution and the effect of \(\ce{HCl}\) on \([\ce{OH^{–}}]\).
- Explain your observations upon addition of \(\ce{NaOH}\) (aq) to the precipitate in Tube B. Consider the various equilibria that are occurring in solution and remember that \(\ce{Zn^{2+}}\) forms stable complex ions with \(\ce{OH^{–}}\) at sufficiently high concentrations.
- Explain your observations upon addition of \(\ce{NH3}\) (aq) to the precipitate in Tube C. Consider the various equilibria that are occurring in solution and remember that \(\ce{Zn^{2+}}\) forms stable complex ions with \(\ce{NH3}\).
Part E
Equilibrium system:
\[\ce{Mg(OH)2 (s) <=> Mg^{2+} (aq) + 2 OH^{-} (aq)}\quad K_{sp} \ll 1\]
Observations
| Step 1 Adding 1 drop of \(\ce{NaOH}\) (aq) to \(\ce{Mg(NO3)2}\) (aq) | |
|---|---|
| Step 2: Tube A: Effect when \(\ce{HCl}\) (aq) is added | |
| Step 3: Tube B: Effect when \(\ce{NaOH}\) (aq) is added | |
| Step 4: Tube C: Effect when \(\ce{NH3}\) (aq) is added |
Analysis
- Based on your observations in Steps 3 and 4 do you think that \(\ce{Mg^{2+}}\) forms stable complex ions? Explain your reasoning.