Cut along the dotted borders to place at lab stations
STATION 1
An AP chemistry student mixes \(25.0\,\text{mL}\) of \(0.20\,\text{M}\) silver nitrate, \(AgNO_3\), with \(25.0\,\text{mL}\) of \(0.20\,\text{M}\) sodium carbonate, \(Na_2CO_3\). A bright yellow-white precipitate forms rapidly.
• Write the complete balanced molecular equation.
• Write the net ionic equation representing the physical change.
• State which spectator ions remain unchanged in solution.
STATION 2
In the lab, you mix \(50.0\,\text{mL}\) of \(1.0 \times 10^{-4}\,\text{M}\ Pb(NO_3)_2\) with \(50.0\,\text{mL}\) of \(2.0 \times 10^{-3}\,\text{M}\ NaCl\) at \(25^\circ\text{C}\). Given that the \(K_{sp}\) of \(PbCl_2\) is \(1.7 \times 10^{-5}\):
• Calculate the diluted concentrations of both \(Pb^{2+}\) and \(Cl^-\) ions after mixing.
• Formulate the expression for the reaction quotient \(Q\).
• Determine if a precipitate of \(PbCl_2(s)\) will form at this equilibrium point.
STATION 3
The solubility product constant (\(K_{sp}\)) for calcium sulfate, \(CaSO_4\), is \(2.4 \times 10^{-5}\) at room temperature.
• Calculate the molar solubility (\(s\)) of calcium sulfate in pure distilled water.
• Calculate the molar solubility of calcium sulfate in a \(0.10\,\text{M}\) sodium sulfate (\(Na_2SO_4\)) solution.
• Explain the physical reasoning behind this shift using Le Chatelier's Principle.
SOLUBILITY SHOWDOWN | PRINTABLE LAB CARDS (1-3) PAGE 3 OF 4
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STATION 4
Imagine mixing \(50.0\,\text{mL}\) of \(0.20\,\text{M}\) Iron(III) chloride, \(FeCl_3\), with \(50.0\,\text{mL}\) of \(0.30\,\text{M}\) sodium hydroxide, \(NaOH\). The Net Ionic reaction is: \(Fe^{3+}(aq) + 3\,OH^-(aq) \rightarrow Fe(OH)_3(s)\)
• Determine which reactant is fully consumed (the Limiting Reactant).
• Calculate the moles of remaining excess ions suspended in the final \(100\,\text{mL}\) solution.
• Translate this stoichiometry into a detailed particulate representation.
STATION 5
The hydroxide salt Magnesium Hydroxide, \(Mg(OH)_2\), is sparingly soluble with a \(K_{sp} = 1.8 \times 10^{-11}\). Instead of pure water, you dissolve it in a robust buffer solution maintained at \(pH = 9.00\).
• Calculate the equilibrium concentration of \(OH^-\) enforced by the buffer.
• Calculate the new molar solubility (\(s\)) of \(Mg(OH)_2\) in this buffered environment.
• Explain why lowering the pH below 9 would increase the solubility of this solid.
STATION 6
A solid alloy sample containing an unknown fraction of barium is analyzed. A \(1.50\,\text{g}\) alloy sample is dissolved, and excess sulfate solution is added to precipitate all barium as dry barium sulfate, \(BaSO_4\). You collect and dry \(0.466\,\text{g}\) of precipitate solid.
• Calculate the moles of barium sulfate precipitate collected.
• Find the mass of barium present in the original \(1.50\,\text{g}\) alloy sample.
• Determine the precise mass percent (\(\%\)) of barium in the original alloy.
SOLUBILITY SHOWDOWN | PRINTABLE LAB CARDS (4-6) PAGE 4 OF 4
• Spectator Ions (Unchanged):
\(Na^+(aq)\) and \(NO_3^-(aq)\)
STATION 2: Lead Lactate Mystery (\(Q\) vs. \(K_{sp}\)) Q vs Ksp Evaluation
The total mixed volume is \(V_t = 100.0\,\text{mL}\). Dilution math is required first:
• \([Pb^{2+}]_i = (1.0 \times 10^{-4}\,\text{M}) \times \frac{50.0\,\text{mL}}{100.0\,\text{mL}} = 5.0 \times 10^{-5}\,\text{M}\)
• \([Cl^-]_i = (2.0 \times 10^{-3}\,\text{M}) \times \frac{50.0\,\text{mL}}{100.0\,\text{mL}} = 1.0 \times 10^{-3}\,\text{M}\)
• Reaction Quotient:
\(Q = [Pb^{2+}][Cl^-]^2 = (5.0 \times 10^{-5})(1.0 \times 10^{-3})^2 = 5.0 \times 10^{-11}\)
• Conclusion: Since \(Q < K_{sp}\) (\(5.0 \times 10^{-11} < 1.7 \times 10^{-5}\)), the solution remains unsaturated. No precipitate of \(PbCl_2(s)\) forms.
STATION 3: Common Ion Crises (Equilibrium Shifts) Common Ion Effect
• Pure Water Molar Solubility (\(s\)):
\(K_{sp} = [Ca^{2+}][SO_4^{2-}] = s^2 \implies s = \sqrt{2.4 \times 10^{-5}} = 4.9 \times 10^{-3}\,\text{M}\)
• Solubility in \(0.10\,\text{M}\ Na_2SO_4\):
\(K_{sp} = [Ca^{2+}][SO_4^{2-}] = (s)(0.10 + s) \approx 0.10s\)
\(2.4 \times 10^{-5} = 0.10s \implies s = 2.4 \times 10^{-4}\,\text{M}\)
Le Chatelier explanation: The high initial \([SO_4^{2-}]\) shifts the equilibrium to the left, decreasing molar solubility by an order of magnitude.
SOLUBILITY SHOWDOWN | EXEMPLARY ANSWER KEY PAGE 2 OF 3
STATIONS 4–6 SOLUTIONS
STATION 4: Particulate Precision (Drawing Guide) Stoichiometry
• Initial Moles: \(Fe^{3+} = 50\,\text{mL} \times 0.20\,\text{M} = 10.0\,\text{mmol}\); \(OH^- = 50\,\text{mL} \times 0.30\,\text{M} = 15.0\,\text{mmol}\).
• Reaction Stoichiometry: \(Fe^{3+} + 3\,OH^- \rightarrow Fe(OH)_3(s)\)
• Limiting Reactant: \(OH^-\) (runs out first). \(15.0\,\text{mmol}\ OH^-\) reacts with \(5.0\,\text{mmol}\ Fe^{3+}\).
• Remaining Excess Ions (in \(100\,\text{mL}\)):
\(Fe^{3+} = 5.0\,\text{mmol}\ (0.050\,\text{M})\), \(Cl^- = 30.0\,\text{mmol}\ (0.30\,\text{M})\), \(Na^+ = 15.0\,\text{mmol}\ (0.15\,\text{M})\).
Particulate Drawing Verification: The bottom solid should be alternating \(Fe^{3+}\) and \(OH^-\) ions (in a \(1:3\) ratio). Suspended randomly above must be separate \(Na^+\), \(Cl^-\), and excess \(Fe^{3+}\) ions.
STATION 5: Barium Barrier (pH-Driven Dissolution) pH Math
• \(pH = 9.00 \implies pOH = 5.00 \implies [OH^-] = 1.0 \times 10^{-5}\,\text{M}\)
• \(K_{sp} = [Mg^{2+}][OH^-]^2 \implies 1.8 \times 10^{-11} = (s)(1.0 \times 10^{-5})^2\)
• \(s = \frac{1.8 \times 10^{-11}}{1.0 \times 10^{-10}} = 0.18\,\text{M}\)
Solubility Contrast: In pure water, \(s = \sqrt[3]{K_{sp}/4} = 1.65 \times 10^{-4}\,\text{M}\). The low-pH buffer dramatically increases solubility because added acid depletes \(OH^-\) products, shifting equilibrium to the right.
STATION 6: Gravimetric Analysis Challenge Stoichiometry Steps
• \(Moles\ of\ BaSO_4 = \frac{0.466\,\text{g}}{233.38\,\text{g/mol}} = 2.00 \times 10^{-3}\,\text{mol}\)
• \(Moles\ of\ Ba^{2+} = Moles\ of\ BaSO_4 = 2.00 \times 10^{-3}\,\text{mol}\)
• \(Mass\ of\ Ba = 2.00 \times 10^{-3}\,\text{mol} \times 137.33\,\text{g/mol} = 0.275\,\text{g}\)
• \(Mass\ \%\ of\ Barium = \left(\frac{0.275\,\text{g}}{1.50\,\text{g}}\right) \times 100\% = 18.3\%\)
SOLUBILITY SHOWDOWN | EXEMPLARY ANSWER KEY PAGE 3 OF 3
