Aqueous Reaction Stoichiometry and Titration Notes

Introduction to Aqueous solution Stoichiometry

• Stoichiometry in aqueous solutions typically provides the volume of a reactant or product rather than its mass.

• To perform calculations, students must convert the volume of a solution to the number of moles of solute contained within that solution.

• Molarity ($M$) serves as the primary conversion factor between volume and moles.

Molarity and the Stoichiometry Flowchart

• Molarity is defined as the ratio of moles of solute per liter of solution:     $\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}$

• A comprehensive flowchart for solution stoichiometry includes the following levels:

  • Level 1: Volume of Solution A: The starting point when the amount of a substance is given in milliliters ($\text{mL}$) or liters ($\text{L}$).

  • Level 2: Moles of Substance A: Use the molarity of solution A as a conversion factor ($\frac{\text{moles}}{\text{L}}$) to find the moles of A.

  • Level 3: Moles of Substance B: Use the molar ratio from the balanced chemical equation to convert moles of A to moles of B.

  • Level 4: Desired Output for Substance B: Depending on the question, there are three common pathways from moles of B:

    1. Mass of B: Use the molar mass of substance B ($\frac{\text{grams}}{\text{mole}}$).

    2. Volume of B: Use the molarity of solution B as a conversion factor.

    3. Molarity of B: Divide the moles of B by the total liters of solution B to find the concentration.

Example 1: Calculating Volume of Reactant from Volume of Another Reactant

• Problem: What volume of $1.15\,M$ sodium sulfide ($Na_2S$) is required to react with $32\,\text{mL}$ of $0.75\,M$ tin-four chloride ($SnCl_4$)?

• Organizing Data:

  • $Na_2S$: $M = 1.15\,M$; Volume = unknown ($? \text{ mL}$).

  • $SnCl_4$: $M = 0.75\,M$; Volume = $32\,\text{mL}$.

• Conversion Factors as Fractions:

  • $SnCl_4$: $\frac{0.75\,\text{moles } SnCl_4}{1\,\text{L solution}}$

  • $Na_2S$: $\frac{1.15\,\text{moles } Na_2S}{1\,\text{L solution}}$

• Reaction Ratio: Based on the balanced equation, $1\,\text{mole}$ of $SnCl_4$ reacts with $2\,\text{moles}$ of $Na_2S$.

• The Plan: $\text{Volume } SnCl_4 \rightarrow \text{Moles } SnCl_4 \rightarrow \text{Moles } Na_2S \rightarrow \text{Volume } Na_2S$.

• Calculation Steps:

  1. Convert volume of $SnCl_4$ to moles using molarity (shortcut: use $1000\,\text{mL}$ instead of $1\,L$ on the bottom):         $32\,\text{mL } SnCl_4 \times \frac{0.75\,\text{moles } SnCl_4}{1000\,\text{mL } SnCl_4}$

  2. Apply the molar ratio:         $\dots \times \frac{2\,\text{moles } Na_2S}{1\,\text{mole } SnCl_4}$

  3. Convert moles of $Na_2S$ to volume using its molarity (shortcut: put $1000\,\text{mL}$ on top to get results in milliliters):         $\dots \times \frac{1000\,\text{mL } Na_2S}{1.15\,\text{moles } Na_2S}$

• Result: $41.7\,\text{mL}$ of $Na_2S$ solution is required.

Example 2: Calculating Mass of Product from Volume of Reactant

• Problem: How many grams of gallium chloride ($GaCl_3$) are formed by the reaction of $2.6\,\text{L}$ of $1.44\,M$ $HCl$?

• Organizing Data:

  • $HCl$: $2.6\,L$; $1.44\,M$.

  • $GaCl_3$: Find mass in grams.

• Molarity Fraction: $\frac{1.44\,\text{moles } HCl}{1\,\text{L HCl solution}}$.

• The Plan: $\text{Volume } HCl \rightarrow \text{Moles } HCl \rightarrow \text{Moles } GaCl_3 \rightarrow \text{Mass } GaCl_3$.

• Calculation Steps:

  1. $2.6\,\text{L } HCl \times \frac{1.44\,\text{moles } HCl}{1\,\text{L } HCl}$

  2. Using the balanced equation ratio ($6\,\text{moles } HCl : 2\,\text{moles } GaCl_3$):         $\dots \times \frac{2\,\text{moles } GaCl_3}{6\,\text{moles } HCl}$

  3. Using the molar mass of $GaCl_3$ ($176.07\,\text{g/mol}$ from the periodic table):         $\dots \times \frac{176.07\,\text{g } GaCl_3}{1\,\text{mole } GaCl_3}$

• Result: $220\,\text{grams}$ of gallium chloride are formed.

Example 3: Calculating Molarity of a Reactant from Mass of Another Reactant

• Problem: $15.8\,\text{mL}$ of acetic acid ($HC_2H_3O_2$) is required to react with $2\,\text{grams}$ of sodium carbonate ($Na_2CO_3$). What is the molarity of the acetic acid?

• Organizing Data:

  • $HC_2H_3O_2$: $15.8\,\text{mL}$ ($0.0158\,L$); find Molarity ($M$).

  • $Na_2CO_3$: $2\,\text{grams}$.

• The Plan: $\text{Mass } Na_2CO_3 \rightarrow \text{Moles } Na_2CO_3 \rightarrow \text{Moles } HC_2H_3O_2 \rightarrow \text{Molarity } HC_2H_3O_2$.

• Calculation Steps:

  1. Convert mass to moles using molar mass of $Na_2CO_3$ ($105.99\,\text{g/mol}$):         $2\,\text{g } Na_2CO_3 \times \frac{1\,\text{mole } Na_2CO_3}{105.99\,\text{g } Na_2CO_3}$

  2. Using the balanced equation ratio ($1\,\text{mole } Na_2CO_3 : 2\,\text{moles } HC_2H_3O_2$):         $\dots \times \frac{2\,\text{moles } HC_2H_3O_2}{1\,\text{mole } Na_2CO_3}$

  3. Calculate molarity by dividing the moles obtained by the volume in liters:         $\text{Molarity} = \frac{\text{Result of Step 2}}{0.0158\,\text{L } HC_2H_3O_2}$

• Result: $2.39\,M$ acetic acid solution.

Titration Concepts

• Definition: Titration is a laboratory procedure where a solution of known concentration is used to determine the concentration of an unknown solution.

• Titrant: The solution with the known concentration.

• Analyte: The solution with the unknown concentration.

• Process: The titrant is added slowly to the analyte until the reaction is complete (neutralization in acid-base titrations). The volume of titrant required is recorded to calculate the analyte's concentration using stoichiometry.

Example 4: Titration Stoichiometry (Neutralization)

• Problem: $50\,\text{mL}$ of $HCl$ requires $35.23\,\text{mL}$ of $0.25\,M$ $NaOH$ to neutralize it. What is the molarity of the $HCl$ solution?

• Organizing Data:

  • $NaOH$: $35.23\,\text{mL}$; $0.25\,M$.

  • $HCl$: $50\,\text{mL}$ ($0.05\,L$); find Concentration ($M$).

• Molarity Fraction: $\frac{0.25\,\text{moles } NaOH}{1\,\text{L solution}}$.

• The Plan: $\text{Volume } NaOH \rightarrow \text{Moles } NaOH \rightarrow \text{Moles } HCl \rightarrow \text{Molarity } HCl$.

• Calculation Steps:

  1. $35.23\,\text{mL } NaOH \times \frac{0.25\,\text{moles } NaOH}{1000\,\text{mL } NaOH}$

  2. Using the balanced equation ratio ($1\,\text{mole } NaOH : 1\,\text{mole } HCl$):         $\dots \times \frac{1\,\text{mole } HCl}{1\,\text{mole } NaOH}$

  3. Divide by the volume of $HCl$ in liters ($0.05\,L$):         $\text{Molarity} = \frac{\text{Result of Step 2}}{0.05\,\text{L } HCl}$

• Result: $0.176\,M$ $HCl$ solution.

Critical Warning: Dilution Equation vs. Stoichiometry

• Dilution Equation: $M_1V_1 = M_2V_2$.

• Prohibition: Do not use the dilution equation for stoichiometry problems.

• Reasons:

  • Stoichiometry involves a chemical change (a reaction between two different substances) and requires the molar ratio from a balanced equation.

  • Dilution is a physical change where only solvent is added to one substance, lowering its concentration without changing the identity of the solute.

  • The dilution equation does not account for molar ratios that are not $1:1$.