4.1.6 Solution Stoichiometry

This topic explains how to perform stoichiometric calculations involving solutions. It covers molarity, solution stoichiometry, and how to use volumes and concentrations of solutions in chemical reactions.

Definition

Solution Stoichiometry: Stoichiometric calculations in which volumes and concentrations of solutions are used instead of masses of reactants and products.

Molarity (M): The number of moles of solute dissolved per dm³ of solution.

Main Idea

In solution stoichiometry, we use volumes of solutions of known concentration instead of masses of reactants and products. The molarity of a solution is used to determine the number of moles of solute present. Then, the mole ratio from the balanced chemical equation is used to calculate the moles of other substances involved in the reaction.

Core Concept

Molarity (M) = Moles of solutedm³ of solution

Key relationships: Moles = Molarity × Volume (in dm³)

Moles = Molarity × Volume (in cm³)1000

Detailed Explanation

What is Molarity?

• In solutions, the most common concentration unit is molarity.
• The molarity of a solution is simply the numerical value of its concentration in mol/dm³.
• Molarity is defined as the number of moles of solute dissolved per dm³ of solution.

Mathematically: M = moles of solutedm³ of solution

Understanding Concentrated Solutions

If you read the label on a bottle of concentrated H₂SO₄, you will notice 98% H₂SO₄ by mass and also 18 M H₂SO₄.

• 18 M means there are 18 moles of H₂SO₄ in each dm³ of solution.
• Similarly, concentrated HCl is 37% and 12.1 M HCl. This means there are 12.1 moles of HCl in each dm³ of solution.

We can express the concentration in terms of moles of solute in the given volume of solution.

Example 4.8: Molarity of Urea Solution

Problem: Urea (NH₂CONH₂) is a white solid used as fertilizer and starting material for synthetic plastic. A solution contains 40 g urea dissolved in 500 cm³ of solution. Calculate the molarity of this solution.

Solution:

• Mass of urea = 40 g
• Volume of solution = 500 cm³ = 0.5 dm³

Required: - Molarity of solution

Step 1: Calculate molar mass of urea (NH₂CONH₂) Molar mass = 14 + 1 × 2 + 12 + 16 + 14 + 1 × 2 = 60 g/mol

Step 2: Calculate moles of urea Moles = 4060 = 0.667 mol

Step 3: Calculate molarity M = molesvolume in dm³ = 0.6670.5 = 1.334 M

Final Answer: Molarity of urea solution = 1.334 M

Solution Stoichiometry

In solution stoichiometry, we use volumes of solutions of known concentration instead of masses of reactants and products, and their molarity is used to determine its moles.

Example 4.9: Mass of AgCl Formed

Problem: When 250 cm³ of 0.1 M AgNO₃ solution is added to an excess of NaCl solution, what mass of AgCl will be formed?

Chemical Equation: AgNO_3(aq) + NaCl_(aq) → AgCl_(s) + NaNO_3(aq)

Solution:

Step 1: Find moles of silver nitrate present in the solution

As the solution is 0.1 M: 1000 cm³ of solution contain = 0.1 mol of AgNO₃

250 cm³ of solution contain = 0.1 × 2501000 = 0.025 mol of AgNO₃

Step 2: Use mole ratio

From the equation: 1 mol of AgNO₃ produces = 1 mol of AgCl

0.025 mol of AgNO₃ will produce = 0.025 mol of AgCl

Step 3: Calculate mass of AgCl

Mass of AgCl = moles × molar mass \] \[ = 0.025 × 143.5 \] \[ = 3.59 g

Final Answer: Mass of AgCl formed = 3.59 g

Mixing Two Solutions

When two solutions are mixed, we proceed as follows:

Moles of reactant 1 = M₁ V₁

Moles of reactant 2 = M₂ V₂

The ratio between moles of reactants is the same as given by the balanced chemical equation.

M₁ V₁ : M₂ V₂ = n₁ : n₂

Where n₁ and n₂ are moles of reactants in the chemical reaction.

M₁ V₁M₂ V₂ = n₁n₂

In this expression, the ratio of two volumes remains the same whether you use volumes in dm³ or cm³.

Intermediate Level

Molarity

M = moles of solutedm³ of solution

Key relationships: Moles = M × V (in dm³) Moles = M × V (in cm³)1000

Example 4.8: Molarity of Urea Solution

• Mass of urea = 40 g
• Volume = 500 cm³ = 0.5 dm³
• Molar mass of urea = 60 g/mol

Solution: Moles = 4060 = 0.667 mol M = 0.6670.5 = 1.334 M

Final Answer: 1.334 M

Example 4.9: Mass of AgCl Formed

• Volume of AgNO₃ = 250 cm³
• Molarity of AgNO₃ = 0.1 M

Required: - Mass of AgCl

Equation: AgNO₃ + NaCl → AgCl + NaNO₃

Solution: Moles of AgNO₃ = 0.1 × 2501000 = 0.025 mol Moles of AgCl = 0.025 mol Mass of AgCl = 0.025 × 143.5 = 3.59 g

Final Answer: 3.59 g

Mixing Two Solutions

M₁ V₁M₂ V₂ = n₁n₂

Important Points

• 1. Molarity = moles of solute / dm³ of solution.
• 2. Moles = Molarity × Volume (in dm³).
• 3. Moles = Molarity × Volume in cm³1000.
• 4. Solution stoichiometry uses volumes and concentrations instead of masses.
• 5. Example: 250 cm³ of 0.1 M AgNO₃ = 0.025 moles.
• 6. Mass of AgCl = 0.025 × 143.5 = 3.59 g.
• 7. Mixing two solutions: M₁ V₁M₂ V₂ = n₁n₂.

Remember

• - Molarity: M = nV
• - Moles: n = M × V
• - Volume in cm³: n = M × V1000
• - **Example:** 250 cm³ of 0.1 M AgNO₃ = 0.025 mol
• - **Mass of AgCl:** 0.025 × 143.5 = 3.59 g

Common Mistakes

• 1. Forgetting to convert cm³ to dm³: Divide by 1000.
• 2. Mixing up molarity and moles: Molarity is concentration; moles is amount.
• 3. Forgetting the mole ratio: Use the balanced equation.
• 4. Using the wrong molar mass: Check the periodic table.

Quick Revision

• Molarity = moles of solute / dm³ of solution.
• Moles = Molarity × Volume (in dm³).
• Moles = Molarity × Volume in cm³1000.
• Solution stoichiometry uses volumes and concentrations instead of masses.
• Example: 250 cm³ of 0.1 M AgNO₃ = 0.025 moles.
• Mass of AgCl = 0.025 × 143.5 = 3.59 g.
• Mixing two solutions: M₁ V₁M₂ V₂ = n₁n₂.

60-Second Revision

• Molarity: M = nV
• Moles: n = M × V
• Volume in cm³: n = M × V1000
• **Example:** 250 cm³ of 0.1 M AgNO₃ = 0.025 mol
• **Mass of AgCl:** 0.025 × 143.5 = 3.59 g

QUALITY VALIDATION

• Topic: 4.1.6 Solution Stoichiometry
• Chapter: 4 (Stoichiometry)

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