Question

Ignoring protolysis reactions, indicate the concentration of each ion or molecule present in the following solu- tions: (a) \(0.35 \mathrm{MK}_{3} \mathrm{PO}_{4},\) (b) \(5 \times 10^{-4} \mathrm{MCuCl}_{2},(\mathbf{c}) 0.0184\) \(\mathrm{MCH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathbf{d})\) a mixture of \(35.0 \mathrm{~mL}\) of \(0.010 \mathrm{MNa}_{2} \mathrm{CO}_{3}\) and \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{MK}_{2} \mathrm{SO}_{4}\). Assume the volumes are additive.

Short answer

For the given solutions: (a) $0.35 \mathrm{MK}_{3} \mathrm{PO}_{4}$: We have $[K^+] = 1.05 \mathrm{M}$ and $[PO_{4}^{3-}] = 0.35 \mathrm{M}$. (b) \(5 \times 10^{-4} \mathrm{MCuCl}_{2}\): We have $[Cu^{2+}] = 5 \times 10^{-4} \mathrm{M}$ and $[Cl^-] = 1 \times 10^{-3} \mathrm{M}$. (c) $0.0184 \mathrm{MCH}_{3} \mathrm{CH}_{2} \mathrm{OH}$: No ions are formed, and the concentration of CH₃CH₂OH is $0.0184 \mathrm{M}$. (d) Mixture of $35.0 \mathrm{~mL}$ of $0.010 \mathrm{MNa}_{2} \mathrm{CO}_{3}$ and $50.0 \mathrm{~mL}$ of $0.200 \mathrm{MK}_{2} \mathrm{SO}_{4}$: We have $[Na^+] = 0.0082 \mathrm{M}$, $[CO_{3}^{2-}] = 0.0041 \mathrm{M}$, $[K^+] = 0.236 \mathrm{M}$, and $[SO_{4}^{2-}] = 0.118 \mathrm{M}$.

Step-by-step solution

(a) Determine ion concentrations in 0.35 M K₃PO₄ solution

(Write the content here) Given a 0.35 M K₃PO₄ solution, we have the following dissociation reaction: \[K_{3}PO_{4} \rightarrow 3K^{+} + PO_{4}^{3-}\] The stoichiometry shows that for each K₃PO₄, we get 3 K⁺ ions and 1 PO₄³⁻ ion. Therefore, the concentrations are: \[ [K^+] = 3 \times 0.35 \mathrm{M} = 1.05 \mathrm{M} \] \[ [PO_{4}^{3-}] = 0.35 \mathrm{M} \]

(b) Determine ion concentrations in \(5 \times 10^{-4}\) M CuCl₂ solution

(Write the content here) Given a \(5 \times 10^{-4}\) M CuCl₂ solution, we have the following dissociation reaction: \[ CuCl_{2} \rightarrow Cu^{2+} + 2Cl^{-} \] The stoichiometry shows that for each CuCl₂, we get 1 Cu²⁺ ion and 2 Cl⁻ ions. Therefore, the concentrations are: \[ [Cu^{2+}] = 5 \times 10^{-4} \mathrm{M} \] \[ [Cl^{-}] = 2 \times (5 \times 10^{-4}) \mathrm{M} = 1 \times 10^{-3} \mathrm{M} \]

(c) Determine concentration in 0.0184 M CH₃CH₂OH solution

(Write the content here) This is a simple molecular solution. There are no ions formed when ethanol (CH₃CH₂OH) is dissolved in water. So, the concentration of CH₃CH₂OH in the solution is 0.0184 M.

(d) Determine ion concentrations in a mixture of 35.0 mL of 0.010 M Na₂CO₃ and 50.0 mL of 0.200 M K₂SO₄

(Write the content here) First, let's find the moles and then the new concentration when the solutions of Na₂CO₃ and K₂SO₄ are mixed together. As the volumes are additive, we will consider the total volume for calculating new concentrations. Initial moles of Na₂CO₃: \[ n_{Na_2CO_3} = C_{1}V_{1} = 0.010 \mathrm{M} \times 0.035 \mathrm{L} = 3.5 \times 10^{-4} \text{moles} \] Initial moles of K₂SO₄: \[ n_{K_2SO_4} = C_{2}V_{2} = 0.200 \mathrm{M} \times 0.050 \mathrm{L} = 0.01 \text{moles} \] The total volume after mixing: \[ V_{\text{total}} = V_{1} + V_{2} = 35.0 \mathrm{~mL} + 50.0 \mathrm{~mL} = 85.0 \mathrm{~mL} = 0.085 \mathrm{L} \] Now let's find the new concentrations: Concentration of Na₂CO₃: \[ C'_{Na_2CO_3} = \frac{n_{Na_2CO_3}}{V_{total}} = \frac{3.5 \times 10^{-4} \text{moles}}{0.085 \text{L}} = 0.0041 \mathrm{M} \] Concentration of K₂SO₄: \[ C'_{K_2SO_4} = \frac{n_{K_2SO_4}}{V_{total}} = \frac{0.010 \text{moles}}{0.085 \text{L}} = 0.118 \mathrm{M} \] Now we have the new concentrations of both salts in the mixed solution. For the Na₂CO₃ dissociation reaction: \[ Na_{2}CO_{3} \rightarrow 2Na^{+} + CO_{3}^{2-} \] We now calculate the ion concentrations: \[ [Na^+] = 2 \times 0.0041 \mathrm{M} = 0.0082 \mathrm{M} \] \[ [CO_{3}^{2-}] = 0.0041 \mathrm{M} \] For the K₂SO₄ dissociation reaction: \[ K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-} \] We now calculate the ion concentrations: \[ [K^+] = 2 \times 0.118 \mathrm{M} = 0.236 \mathrm{M} \] \[ [SO_{4}^{2-}] = 0.118 \mathrm{M} \]

Key concepts

Dissociation Reactions

Dissociation reactions are fundamental in understanding how ionic compounds separate in a solution. When an ionic solid, like potassium phosphate ( K₃PO₄ ), dissolves in water, it splits into its respective ions. This process is essential for calculating the resulting ion concentrations. Let's take the dissociation of K₃PO₄ as an example:

• The chemical equation for dissociation is K₃PO₄ → 3K⁺ + PO₄³⁻
• This means one formula unit of K₃PO₄ yields three potassium ions ( K⁺ ) and one phosphate ion ( PO₄³⁻ ).

Understanding stoichiometry in dissociation provides us with the means to predict and calculate the ion concentrations in a solution. Each compound has a characteristic dissociation equation displaying the resulting ions and their ratios, which is crucial for subsequent concentration and stoichiometric calculations.

Solution Concentration Calculations

Calculating the concentration of ions in a solution requires understanding the molarity concept. Molarity ( M ) is the concentration of a solution expressed as moles of solute per liter of solution. For an ionic compound like CuCl₂ , which dissociates as shown:

• CuCl₂ → Cu²⁺ + 2Cl⁻

This equation tells us that one mole of CuCl₂ will produce one mole of copper ions ( Cu²⁺ ) and two moles of chloride ions ( Cl⁻ ). If we have a 5 × 10^{-4} M solution of CuCl₂ :

• The concentration of Cu²⁺ ions will be 5 × 10^{-4} M.
• The concentration of Cl⁻ ions will be 2 × (5 × 10^{-4}) M = 1 × 10^{-3} M. Each reflects how molarity is used to convert between the overall concentration of the compound and the resulting ions.

These calculations are at the heart of solution stoichiometry and help predict how solutions are prepared and behave.

Stoichiometry in Solutions

Stoichiometry in aqueous solutions involves using the principles of chemical reactions in water. It's about balancing the quantities of reactants and products when solutions are prepared or mixed. Consider a mixture of solutions where Na₂CO₃ and K₂SO₄ are mixed:

• First, calculate the moles of each compound using molarity (M) and volume (L).
• Combine the molarities by adding to find a new total volume and calculate the resulting concentrations.

For instance, if we mix two solutions: 35.0 mL of 0.010 M Na₂CO₃ and 50.0 mL of 0.200 M K₂SO₄ :

• Calculate the new molarity for each compound using the moles of solute and the total volume (85.0 mL).
• Apply dissociation equations: Na₂CO₃ → 2Na⁺ + CO₃²⁻ and K₂SO₄ → 2K⁺ + SO₄²⁻ .
• Calculate individual ion concentrations by multiplying by their respective coefficients from the dissociation equation.

By doing these stoichiometric calculations, we predict how the ions interact in a mixture to give insights into chemical behavior and reaction progress in solution.