Question

Chapter 4: Types of Chemical Reactions and Solutions During this Workshop you will: Gain familiarity with the major classes of chemical reactions. Solve chemistry problems in solutions. Quantitative Problem Solving: The steps for solving complex problems are: 1. Interpretation. 2. Guess. 3. Plan. 4. Look up. 5. Calculations 6. Check _: Answer the following: a. What is an aqueous solution? b. What is the difference between something that dissociates and something that dissolves? : A 500.0 mL aqueous solution of 5.89 x 10' M calcium hydroxide is mixed with 1.50 L of a 6.34 x 103 M aqueous solution of calcium hydroxide. What is the resulting molarity of calcium hydroxide in this solution?

Answer 1

Answers 1:-

1(a) Aqueous solution: The solution which is mainly prepared in water, known as an aqueous solution. For example, a solution of table salt, or sodium chloride (NaCl), in water can be shown as Na⁺(aq) + Cl⁻(aq). The word aq (comes from aqua) stands for water.

1(b) Dissociation:- When the ionic compounds separate in smaller ions (usually in a reversible manner) into the solution known as Dissociation. e.g., In the aqueous sodium chloride (NaCl) solution, the positive sodium ions (Na⁺) are attracted towards the negative pole of the water molecule (i.e. OH^-), while the negative chloride ions (Cl⁻) are attracted to the positive pole of the water molecule (i.e. H⁺).

Dissolve:- When the ionic crystal breaks down into its ions in water, known as Dissolution. For example, In aqueous NaCl, the water molecules surround the negative chloride ions (Cl^-) and positive sodium ions (Na⁺) and pull them away into the solution.

Answer 2:-

For the aquous solution 1 - Ca(OH)₂

volume = 0.5 Liter, Molarity = 5.89 x 10^-3 M

number of moles n₁ = 0.5 x 5.89 x 10^-3 = 2.945 x 10^-3

For the aquous solution 2 - Ca(OH)₂

volume = 1.5 Liter, Molarity = 6.34 x 10^-3 M

number of moles n₁ = 1.5 x 6.34 x 10^-3 = 9.51 x 10^-3

After Mixing of solution 1 and 2

total number of moles = n₁ + n₂ = 2.945 x 10^-3 +9.51 x 10^-3

                                                          = 12.455 x 10^-3

total volume of final solution = V₁ + V₂ = 0.5 + 1.5

                                                           = 2.0 Litr

Molarity of final solution = total number of moles / toal volume (in Litr)

                                     = 12.455 x 10^-3/ 2

                                     = 6.22 x 10^-3 M