Calculations for Molarity, Dilution, and Titration (Solution Stoichiometry)

Stoichiometry is the mathematics of chemistry. While balancing equations tells you the ratios of atoms, solution stoichiometry allows you to predict exactly how much chemical is in a beaker, how to dilute a medicine to a safe level, or precisely when a reaction has finished. Whether you are analyzing a pharmaceutical compound or neutralizing an acid spill, mastery of molarity, dilution, and titration is non-negotiable.

This guide moves beyond simple definitions to applying these concepts to rigorous exam-style problems. We will cover calculating concentration, manipulating the dilution equation, and solving complex titration scenarios where the mole ratio is not always 1:1.

Molarity and Concentration Unit Conversions

The most common unit of concentration in chemistry is Molarity ($M$). It provides a bridge between the volume of a liquid we can measure and the number of molecules (moles) reacting in the solution.

Calculating Molarity from Mass

To find the molarity of a solution prepared from a solid, you must first convert the mass of the solute into moles using its molar mass ($g/mol$).

Illustrative Example:
Calculate the molarity of a solution prepared by dissolving $13.56 \text{ g}$ of sodium sulfate ($Na_2SO_4$) in enough water to make $287 \text{ mL}$ of solution.

1. Calculate Molar Mass: Sum the atomic masses of $Na_2SO_4$. $$ (2 \times 22.99) + 32.07 + (4 \times 16.00) = 142.05 \text{ g/mol} $$
2. Convert Mass to Moles: $$ \text{Moles} = \frac{13.56 \text{ g}}{142.05 \text{ g/mol}} \approx 0.0955 \text{ mol} $$
3. Convert Volume to Liters: $$ 287 \text{ mL} = 0.287 \text{ L} $$
4. Calculate Molarity: $$ M = \frac{0.0955 \text{ mol}}{0.287 \text{ L}} = \mathbf{0.333 \text{ M}} $$

Advanced Concept: Concentration of Specific Ions

Sometimes you need to find the concentration of a specific ion, not just the whole molecule. This often happens when mixing solutions containing a common ion.

Consider dissolving $2.50 \text{ g}$ of magnesium chloride ($MgCl_2$) into $150.0 \text{ mL}$ of $0.250 \text{ M}$ $NaCl$. What is the total concentration of chloride ions ($Cl^-$)?

• Step 1: Analyzed Dissociation.
  • $NaCl \rightarrow Na^+ + Cl^-$ (1 mol $NaCl$ gives 1 mol $Cl^-$)
  • $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$ (1 mol $MgCl_2$ gives 2 mol $Cl^-$)
• Step 2: Calculate moles from NaCl. $$ 0.150 \text{ L} \times 0.250 \text{ mol/L} = 0.0375 \text{ mol } Cl^- $$
• Step 3: Calculate moles from $MgCl_2$.
  Molar mass of $MgCl_2 \approx 95.21 \text{ g/mol}$. $$ \text{Moles } MgCl_2 = \frac{2.50 \text{ g}}{95.21 \text{ g/mol}} = 0.0263 \text{ mol} $$ Since each unit gives two chlorides: $$ 0.0263 \text{ mol } MgCl_2 \times 2 = 0.0525 \text{ mol } Cl^- $$
• Step 4: Combine and Solve. $$ \text{Total } Cl^- = 0.0375 + 0.0525 = 0.0900 \text{ mol} $$ $$ [Cl^-] = \frac{0.0900 \text{ mol}}{0.150 \text{ L}} = \mathbf{0.600 \text{ M}} $$

The Dilution Equation ($M_1V_1 = M_2V_2$)

Dilution involves adding solvent (usually water) to a stock solution to decrease its concentration. The key principle here is the conservation of moles: the number of moles of solute before dilution equals the number of moles after dilution.

Where $1$ represents the initial (concentrated) state and $2$ represents the final (diluted) state.

Practice Problem: Preparing a Standard Solution

Question: To make $50.0 \text{ mL}$ of a $0.100 \text{ M}$ NaOH solution, how much stock solution ($1.00 \text{ M}$ NaOH) is required?

Solution:

1. Identify variables: $M_1 = 1.00 \text{ M}$, $M_2 = 0.100 \text{ M}$, $V_2 = 50.0 \text{ mL}$. We need $V_1$.
2. Set up the equation: $$ (1.00 \text{ M})(V_1) = (0.100 \text{ M})(50.0 \text{ mL}) $$
3. Solve for $V_1$: $$ V_1 = \frac{5.0 \text{ mL}\cdot\text{M}}{1.00 \text{ M}} = \mathbf{5.00 \text{ mL}} $$

To prepare this, you would measure $5.00 \text{ mL}$ of the stock solution and add enough distilled water to reach the total volume of $50.0 \text{ mL}$ (adding roughly $45.0 \text{ mL}$ of water).

Multi-Step Dilution: The Serial Method

In analytical chemistry (like analyzing steroids or pollutants), concentrations are often extremely low, requiring serial dilutions.

Imagine you have a stock solution of a steroid with a concentration of $9.6 \times 10^{-5} \text{ M}$. If you take a $500.0 \text{ }\mu\text{L}$ aliquot (portion) and dilute it to $500.0 \text{ mL}$, what is the final concentration?

1. Convert units to match: $500.0 \text{ }\mu\text{L} = 0.000500 \text{ L}$. Final volume = $0.500 \text{ L}$.
2. Calculate moles transferred: $$ n = (9.6 \times 10^{-5} \text{ mol/L}) \times (0.000500 \text{ L}) = 4.8 \times 10^{-8} \text{ mol} $$
3. Calculate new Molarity: $$ M_{final} = \frac{4.8 \times 10^{-8} \text{ mol}}{0.500 \text{ L}} = 9.6 \times 10^{-8} \text{ M} $$

Titration Stoichiometry: Equivalence Point Calculations

Titration determines the concentration of an unknown solution by reacting it with a solution of known concentration (the titrant). The goal is to find the equivalence point.

Endpoint vs. Equivalence Point:
Students often confuse these terms. The equivalence point is the theoretical point where moles of acid equal moles of base (stoichiometrically). The endpoint is the physical point where the indicator changes color. In a well-designed experiment, the endpoint occurs immediately after the equivalence point.

The Golden Rule of Titration Calculations

Do not blindly use $M_1V_1 = M_2V_2$ for titrations! This only works for 1:1 ratios. Instead, follow this logic:

1. Balanced Equation: Determine the mole ratio.
2. Moles of Known: Calculate $n = M \times V$ for the substance with known concentration.
3. Stoichiometry: Convert moles of known to moles of unknown.
4. Solve: Calculate the unknown Molarity ($n/V$).

Case Study: Diprotic Acids (The 1:2 Ratio)

Problem: It takes $50.0 \text{ mL}$ of $0.50 \text{ M}$ KOH to completely neutralize $125 \text{ mL}$ of sulfuric acid ($H_2SO_4$). What is the concentration of the acid?

Step-by-Step Solution:

1. Reaction: Notice that sulfuric acid has two protons ($H^+$). $$ H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O $$ Ratio: 1 mole acid reacts with 2 moles base.
2. Moles of Base (KOH): $$ 0.50 \text{ mol/L} \times 0.050 \text{ L} = 0.025 \text{ mol KOH} $$
3. Moles of Acid ($H_2SO_4$): $$ 0.025 \text{ mol KOH} \times \frac{1 \text{ mol } H_2SO_4}{2 \text{ mol KOH}} = 0.0125 \text{ mol } H_2SO_4 $$
4. Concentration of Acid: $$ M = \frac{0.0125 \text{ mol}}{0.125 \text{ L}} = \mathbf{0.10 \text{ M}} $$

Acid-Base Titrations and pH at Equivalence

For strong acid-strong base titrations, calculating the volume required to reach the equivalence point and the resulting pH is straightforward if you track the ions.

Finding the Required Volume

Problem: A $25.0 \text{ mL}$ sample of $0.200 \text{ M}$ $HNO_3$ is titrated with $0.150 \text{ M}$ $Ca(OH)_2$. What volume of the base is needed to reach the equivalence point?

Note the stoichiometry: $Ca(OH)_2$ provides two hydroxide ions per molecule.

1. Calculate moles of $H^+$: $$ 0.0250 \text{ L} \times 0.200 \text{ mol/L} = 0.00500 \text{ mol } HNO_3 \rightarrow 0.00500 \text{ mol } H^+ $$
2. Determine moles of $OH^-$ needed: At equivalence, moles $H^+ =$ moles $OH^-$, so we need $0.00500 \text{ mol } OH^-$.
3. Determine moles of $Ca(OH)_2$: $$ 0.00500 \text{ mol } OH^- \times \frac{1 \text{ mol } Ca(OH)_2}{2 \text{ mol } OH^-} = 0.00250 \text{ mol } Ca(OH)_2 $$
4. Calculate Volume: $$ V = \frac{n}{M} = \frac{0.00250 \text{ mol}}{0.150 \text{ M}} = 0.0167 \text{ L} = \mathbf{16.7 \text{ mL}} $$

Calculating pH at Equivalence

In the titration of a Strong Acid (like $HCl$, $HNO_3$) with a Strong Base (like $NaOH$, $RbOH$), the net ionic equation is simply:

At the equivalence point, the solution contains only water and "spectator ions" (like $Rb^+$ and $Cl^-$) which do not react with water. Therefore, the solution is neutral.

Result: For any Strong Acid + Strong Base titration, the pH at the equivalence point (at $25^\circ C$) is 7.00.

Key Takeaways for Stoichiometry Exams

• Convert to Liters: Always convert milliliters to liters ($L = mL / 1000$) before using Molarity formulas.
• Check the Ratio: Never assume a 1:1 reaction ratio. Write the balanced chemical equation first.
• Conservation of Moles: In dilution, $M \times V$ remains constant. In titration, $M \times V$ allows you to bridge from one reactant to another via the stoichiometric ratio.
• Unknown vs. Unknown: You cannot titrate a solution of unknown concentration with another unknown. One solution must be a "standard" with a known Molarity.