Lecture Notes: Solution Stoichiometry and Titration

Solution Stoichiometry and Concentration

  • Definition of Solution Stoichiometry: Incorporating information from a balanced chemical equation into calculations involving aqueous solutions.

  • Solution Components: A solution is a homogeneous mixture composed of two primary parts:

    • Solute: The substance being dissolved (often a solid in lab settings).

    • Solvent: The substance doing the dissolving (commonly water).

  • Concentration: A measure of the amount of solute present in a given quantity of solvent or solution.

  • Molarity (MM): The standard measure of concentration, defined as the ratio of moles of solute to the volume of the solution in liters.

    • Formula for Molarity:     M=moles of solutevolume of solution in litersM = \frac{\text{moles of solute}}{\text{volume of solution in liters}}

    • Units: Molarity can be expressed as uppercase MM or as moles per liter\text{moles per liter} (mol/L\text{mol/L}). Using mol/L\text{mol/L} is often preferred in calculations to facilitate dimensional analysis and unit cancellation.

  • Derived Formulas:

    • Finding Moles: moles=M×volume in liters\text{moles} = M \times \text{volume in liters}

    • Finding Volume: volume in liters=molesM\text{volume in liters} = \frac{\text{moles}}{M}

Laboratory Preparation of Solutions

  • Preparation Process for a Solution of Known Concentration:

    • Calculate the required moles of solute based on the desired molarity and final volume.

    • Convert moles to grams using the molar mass of the substance.

    • Weigh the specific mass of the solute using an analytical balance.

    • Transfer the solute into a volumetric flask.

    • Add a small amount of solvent (e.g., water) and shake or stir until the solid is completely dissolved.

    • Add more solvent until the solution level reaches the calibration mark on the flask's neck, ensuring the bottom of the meniscus touches the line.

Molarity Calculation Examples

  • Example 1: Calculating Mass Needed for a Solution

    • Problem: How many grams of potassium dichromate (K2Cr2O7K_2Cr_2O_7) are required to prepare a 250 ml250\,ml solution with a concentration of 2.16 M2.16\,M?

    • Step 1: Convert Volume to Liters     250 ml×1 L1000 ml=0.25 L250\,ml \times \frac{1\,L}{1000\,ml} = 0.25\,L

    • Step 2: Calculate Moles     moles=2.16 mol/L×0.25 L=0.54 mol K2Cr2O7\text{moles} = 2.16\,mol/L \times 0.25\,L = 0.54\,mol\,K_2Cr_2O_7

    • Step 3: Convert Moles to Grams (Molar Mass = 294.2 g/mol294.2\,g/mol)     0.54 mol×294.2 g/mol=159 g K2Cr2O70.54\,mol \times 294.2\,g/mol = 159\,g\,K_2Cr_2O_7

    • Conclusion: Dissolving 159 g159\,g of solute in water to reach a total volume of 250 ml250\,ml produces a 2.16 M2.16\,M solution.

  • Example 2: Calculating Volume Required for a Specific Mass

    • Problem: Calculate the volume in milliliters of a 2.53 M2.53\,M glucose (C6H12O6C_6H_{12}O_6) solution needed to provide 3.81 g3.81\,g of glucose.

    • Step 1: Calculate Molar Mass of Glucose     (6×12.01)+(12×1.008)+(6×16.00)=180.2 g/mol(6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.2\,g/mol

    • Step 2: Convert Mass to Moles     3.81 g180.2 g/mol=2.114×10−2 mol\frac{3.81\,g}{180.2\,g/mol} = 2.114 \times 10^{-2}\,mol

    • Step 3: Calculate Volume in Liters     volume=2.114×10−2 mol2.53 mol/L=0.00836 L\text{volume} = \frac{2.114 \times 10^{-2}\,mol}{2.53\,mol/L} = 0.00836\,L

    • Step 4: Convert to Milliliters     0.00836 L×1000 ml/L=8.36 ml0.00836\,L \times 1000\,ml/L = 8.36\,ml

Solution Dilution

  • Definition: The process of preparing a less concentrated solution from a more concentrated "stock" solution by adding solvent.

  • Stock Solutions: Highly concentrated solutions stored for efficiency to avoid preparing fresh solutions from solids repeatedly.

  • Dilution Formula:     MiVi=MfVfM_i V_i = M_f V_f

    • Where MiM_i and ViV_i are the initial concentration and volume of the stock solution.

    • Where MfM_f and VfV_f are the final concentration and volume of the diluted solution.

  • Units in Dilution: In this specific formula, the volume units do not necessarily have to be in liters, provided the same units are used on both sides of the equation (as they cancel out).

  • Dilution Example Problem:

    • Task: Describe how to prepare 5.00×102 ml5.00 \times 10^{2}\,ml of a 1.75 M1.75\,M sulfuric acid (H2SO4H_2SO_4) solution starting with an 8.61 M8.61\,M stock solution.

    • Calculation of ViV_i:     Vi=1.75 M×500 ml8.61 M=101.6 mlV_i = \frac{1.75\,M \times 500\,ml}{8.61\,M} = 101.6\,ml

    • Interpretation and Procedure:

      1. Measure exactly 101.6 ml101.6\,ml of the 8.61 M8.61\,M stock solution.

      2. Transfer to a clean container.

      3. Add the solvent (deionized water) until the total volume reaches 500 ml500\,ml.

      4. The amount of water added is calculated as: 500 ml−101.6 ml=398.4 ml500\,ml - 101.6\,ml = 398.4\,ml.

Titration Concepts

  • Titration: An analytical technique where a solution of accurately known concentration (the titrant) is added gradually to another solution of unknown concentration (the analyte) until the chemical reaction between the two is complete.

  • Equivalence Point: The point in a titration where the reaction between the titrant and analyte is stoichiometricially complete.

  • Indicator: A substance that changes color at or near the equivalence point, providing a visual signal to stop the titration.

  • Lab Equipment Setup:

    • Burette: Contains the solution of known concentration (often a base like NaOHNaOH).

    • Erlenmeyer Flask: Contains the analyte (unknown concentration) and a few drops of indicator.

  • Standard Procedure: Add titrant dropwise until a permanent, faint color change (e.g., faint pink with phenolphthalein) is observed. Intense color indicates the equivalence point has been exceeded.

Titration Calculation Methodology

  • General Strategy: Every titration problem essentially requires using the molarity formula twice:

    1. Use the known concentration and volume (measured during titration) to find the moles of the known substance.

    2. Use the stoichiometric ratio from the balanced chemical equation to find the moles of the unknown substance.

    3. Use the moles and the initial volume of the unknown to calculate its concentration (or volume as required).

  • Acid-Base Titration Example:

    • Problem: How many milliliters of 0.61 MNaOH0.61\,M NaOH are needed to neutralize 20 ml20\,ml of 0.245 MH2SO40.245\,M H_2SO_4?

    • Balanced Equation: H2SO4+2NaOH→Na2SO4+2H2OH_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O

    • Step 1: Moles of Known (H2SO4H_2SO_4)     0.245 mol/L×0.020 L=4.9×10−3 mol H2SO40.245\,mol/L \times 0.020\,L = 4.9 \times 10^{-3}\,mol\,H_2SO_4

    • Step 2: Stoichiometry (1:21:2 ratio)     4.9×10−3 mol H2SO4×2 mol NaOH1 mol H2SO4=9.8×10−3 mol NaOH4.9 \times 10^{-3}\,mol\,H_2SO_4 \times \frac{2\,mol\,NaOH}{1\,mol\,H_2SO_4} = 9.8 \times 10^{-3}\,mol\,NaOH

    • Step 3: Calculate Volume of NaOHNaOH     volume=9.8×10−3 mol0.61 mol/L=0.0161 L=16.1 ml\text{volume} = \frac{9.8 \times 10^{-3}\,mol}{0.61\,mol/L} = 0.0161\,L = 16.1\,ml

  • Redox Titration Example:

    • Problem: A 16.42 ml16.42\,ml volume of 0.1327 MKMnO40.1327\,M KMnO_4 is needed to oxidize 25 ml25\,ml of FeSO4FeSO_4 in an acidic medium. Find the concentration of FeSO4FeSO_4.

    • Net Ionic Equation: 5Fe2++MnO4−+8H+→5Fe3++Mn2++4H2O5Fe^{2+} + MnO_4^{-} + 8H^{+} \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O

    • Step 1: Moles of Known (MnO4−MnO_4^{-})     0.1327 mol/L×0.01642 L=2.179×10−3 mol MnO4−0.1327\,mol/L \times 0.01642\,L = 2.179 \times 10^{-3}\,mol\,MnO_4^{-}

    • Step 2: Stoichiometry (5:15:1 ratio)     2.179×10−3 mol×5=1.09×10−2 mol Fe2+2.179 \times 10^{-3}\,mol \times 5 = 1.09 \times 10^{-2}\,mol\,Fe^{2+}

    • Step 3: Calculate Concentration of FeSO4FeSO_4     Molarity=1.09×10−2 mol0.025 L=0.436 M\text{Molarity} = \frac{1.09 \times 10^{-2}\,mol}{0.025\,L} = 0.436\,M

Gravimetric Analysis

  • Gravimetric Analysis: An analytical method based on the measurement of mass.

  • Process Overview (Example: Phosphorus in Fertilizer):

    • Dissolve the unknown sample in water.

    • React the analyte with a specific reagent to form a solid precipitate.

    • Filter the mixture to isolate the precipitate from the aqueous solution.

    • Dry and weigh the precipitate.

    • Calculate the mass and percentage of the element of interest based on the precipitate's formula and mass.

Chemistry in Action: Metal Extraction

  • Real-World Application: Extracting Magnesium (MgMg) from seawater.

  • Significance: Magnesium is used in industries to form alloys, batteries, and in chemical synthesis.

  • Reaction Steps:

    1. Precipitation: Seawater is treated to precipitate magnesium as solid magnesium hydroxide, Mg(OH)2Mg(OH)_2.

    2. Acid-Base Reaction: Mg(OH)2Mg(OH)_2 reacts with hydrochloric acid (HClHCl) to form aqueous magnesium chloride (MgCl2MgCl_2).

    3. Redox Reaction: Electrolysis is performed on the magnesium chloride where magnesium ions gain electrons to form liquid magnesium metal (Mg(l)Mg_{(l)}), and chlorine ions lose electrons to form chlorine gas (Cl2(g)Cl_{2(g)}).

Questions & Discussion

  • Question: Why do we use mol/Lmol/L instead of just MM in calculations?

  • Response: Using mol/Lmol/L allows students to visually track how units cancel out via dimensional analysis, helping to prevent mathematical errors in the final result.

  • Question: What happens if the titration turns intense pink instead of faint pink?

  • Response: An intense color indicates that one has gone past the equivalence point, meaning too much titrant was added, making the resulting data inaccurate.

  • Question: In dilution, do I always subtract the initial volume from the final volume?

  • Response: Yes, if you need to determine the specific quantity of solvent (water) to add to the measured stock solution to reach the desired final volume.