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.

(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: \[ [K^+] = 3 \times 0.35 \mathrm{M} = 1.05 \mathrm{M} \] \[ [PO_{4}^{3-}] = 0.35 \mathrm{M} \]

(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: \[ [Cu^{2+}] = 5 \times 10^{-4} \mathrm{M} \] \[ [Cl^{-}] = 2 \times (5 \times 10^{-4}) \mathrm{M} = 1 \times 10^{-3} \mathrm{M} \]

(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.

(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: \[ [K^+] = 2 \times 0.118 \mathrm{M} = 0.236 \mathrm{M} \] \[ [SO_{4}^{2-}] = 0.118 \mathrm{M} \]