Question

Which of the following mixtures contain less number of ions? (1) \(2 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4} ; 3 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (2) \(2 \mathrm{M} \mathrm{NaC} 1 ; 2 \mathrm{M} \mathrm{KC} 1\) (3) \(2 \mathrm{M} \mathrm{MnSO}_{4} ; 2 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\) (4) \(2 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4} ; 2 \mathrm{M}\left(\mathrm{NII}_{4}\right)_{2} \mathrm{SO}_{4}\)

Short answer

Mixture 2 contains the least number of ions.

Step-by-step solution

- List the components for each mixture

Identify the components of each mixture and their concentrations.(1) Mixture 1: 2 M K2SO4 and 3 M Na2CO3(2) Mixture 2: 2 M NaCl and 2 M KCl(3) Mixture 3: 2 M MnSO4 and 2 M K2SO4(4) Mixture 4: 2 M K2SO4 and 2 M (NH4)2SO4

- Dissociate each compound into ions

Write the dissociation equation for each compound in water:(1) K2SO4 → 2K⁺ + SO4²⁻Na2CO3 → 2Na⁺ + CO3²⁻(2) NaCl → Na⁺ + Cl⁻KCl → K⁺ + Cl⁻(3) MnSO4 → Mn²⁺ + SO4²⁻K2SO4 → 2K⁺ + SO4²⁻(4) K2SO4 → 2K⁺ + SO4²⁻(NH4)2SO4 → 2NH4⁺ + SO4²⁻

- Calculate the total number of ions for each mixture

Calculate the number of ions produced by each compound in the mixture and sum them up:(1) K2SO4: 2 M produces 3 ions (2K⁺ and 1 SO4²⁻), 6 ions total.Na2CO3: 3 M produces 3 ions (2Na⁺ and 1 CO3²⁻), 9 ions total.Total for Mixture 1: 15 ions(2) NaCl: 2 M produces 2 ions (Na⁺ and Cl⁻), 4 ions total.KCl: 2 M produces 2 ions (K⁺ and Cl⁻), 4 ions total.Total for Mixture 2: 8 ions(3) MnSO4: 2 M produces 2 ions (Mn²⁺ and SO4²⁻), 4 ions total.K2SO4: 2 M produces 3 ions (2K⁺ and 1 SO4²⁻), 6 ions total.Total for Mixture 3: 10 ions(4) K2SO4: 2 M produces 3 ions (2K⁺ and 1 SO4²⁻), 6 ions total.(NH4)2SO4: 2 M produces 3 ions (2NH4⁺ and SO4²⁻), 6 ions total.Total for Mixture 4: 12 ions

- Compare the total number of ions for each mixture

Compare the total number of ions calculated for each mixture:Mixture 1: 15 ionsMixture 2: 8 ionsMixture 3: 10 ionsMixture 4: 12 ionsMixture 2 contains the least number of ions (8 ions).

Key concepts

Ionic Compounds

Ionic compounds are a type of chemical compound composed of cations (positively charged ions) and anions (negatively charged ions). These ions are held together by ionic bonds, which are strong electrostatic forces of attraction between oppositely charged ions. When ionic compounds dissolve in water, they dissociate into their constituent ions. For example, when sodium chloride (NaCl) dissolves, it dissociates into sodium ions (Na⁺) and chloride ions (Cl⁻).
Sodium chloride is a simple ionic compound, but there are many others with more complex structures.

• Common ionic compounds include salts, such as potassium sulfate (K₂SO₄) and ammonium sulfate ((NH₄)₂SO₄).
• The structure of ionic compounds usually forms a crystalline lattice which minimizes the potential energy of the system.

Understanding the nature of ionic compounds is fundamental to grasping how they behave in different conditions, such as during dissociation in water.

Dissociation Equations

Dissociation equations describe the process in which an ionic compound separates into its constituent ions when it dissolves in water. This is a physical change, not a chemical change, because the compound itself is not altered; rather, its ions are simply dispersed in solution. For example, the dissociation of sodium chloride can be written as:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
Each dissociation equation shows what ions are produced from one formula unit of the ionic compound in water.

• For potassium sulfate (K₂SO₄): \[ \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \]
• For ammonium sulfate ((NH₄)₂SO₄): \[ (\text{NH}_4)_2\text{SO}_4 \rightarrow 2\text{NH}_4^+ + \text{SO}_4^{2-} \]

Writing dissociation equations helps in understanding how many ions a compound will produce when dissolved, which is essential for calculating the total number of ions in a solution and is crucial in problems related to molarity and ion counting.

Molarity

Molarity (M) is a way to express the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The formula for molarity is:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
For example, a 2 M solution of potassium sulfate (K₂SO₄) means there are 2 moles of K₂SO₄ dissolved in 1 liter of solution. Molarity is important because it helps quantify the concentration of ions in a solution after dissociation. When calculating how many ions are in a solution, you need to know the molarity of the original solution.
In a problem that involves comparing ion concentrations, knowing the molarity of each component helps determine which mixture has more or fewer ions. For instance, if you have 2 M K₂SO₄ and 3 M Na₂CO₃, the molarity tells you how many ions form when each dissolves.

• 2 M K₂SO₄ produces 2 K⁺ and 1 SO₄²⁻, hence producing 6 ions for the 2 moles total.
• 3 M Na₂CO₃ produces 2 Na⁺ and 1 CO₃²⁻ for each formula unit, leading to a total of 9 ions from 3 moles.

Understanding molarity allows you to accurately compare different solutions’ ion content.

Ion Counting

Ion counting involves determining the total number of ions present in a solution. This process is essential in understanding chemical reactions and solution properties. To count ions, follow these steps:
First, write the dissociation equation for each compound in the solution. Then, use the molarity to determine how many moles of each ion are produced. Finally, sum up the ions.
Here's an example:

• For 2 M K₂SO₄:\[ \text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-} \]Each mole of K₂SO₄ produces 3 ions (2 K⁺ and 1 SO₄²⁻). So, 2 M K₂SO₄ gives 6 ions total.
• For 3 M Na₂CO₃:\[ \text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-} \] Each mole of Na₂CO₃ produces 3 ions, giving us 9 ions total.

Adding up these ions tells you the total ion count. Comparing mixtures then becomes easy—just see which has fewer total ions. For example, if one mixture totals 15 ions and another 8, the latter has fewer ions. This method helps simplify complex problems, making them manageable and ensuring accurate results in determining ion concentrations.