1) Answer is: the concentration of potassium cations (K⁺) is 0.3 M.
Balanced chemical reaction (dissociation) of potassium sulfide in the water:
K₂S(aq) → 2K⁺(aq) + S²⁻(aq).
c(K₂S) = 0.15 M.
From balanced reaction: n(K₂S) : n(K⁺) = 1 : 2, because volume of the solution is constant: c(K⁺) = 2 · c(K₂S).
c(K⁺) = 0.3 M; concentration of potassium cations.
2) Answer is: its concentration is half that of the Cl⁻ ion.
Balanced chemical reaction (dissociation) of calcium chloride (CaCl₂) in the water:
CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq).
From balanced reaction: n(Ca²⁺) : n(Cl⁻) = 1 : 2, because volume of the solution is constant: c(Ca²⁺) = c(Cl⁻) ÷ 2.
n is amount of the substance.
3) Answer is: 46.176 grams of Na₃PO₄ will be needed.
Balanced chemical reaction (dissociation) of sodium phosphate (Na₃PO₄) in the water:
Na₃PO₄(aq) → 3Na⁺(aq) + PO₄³⁻(aq).
V = 650 mL ÷ 1000 mL.
V = 0.650 L; volume of the solution.
c(Na⁺) = 1.3 M; concentratium of sodium cations.
From balanced reaction: n(Na₃PO₄) : n(Na⁺) = 1 : 3, because volume of the solution is constant: c(Na₃PO₄) = c(Na⁺) ÷ 3.
c(Na₃PO₄) = 1.3 M ÷ 3.
c(Na₃PO₄) = 0.43 M.
n(Na₃PO₄) = c(Na₃PO₄) · V(solution).
n(Na₃PO₄) = 0.43 M · 0.65 L.
n(Na₃PO₄) = 0.281 mol.
m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).
m(Na₃PO₄) = 0.281 mol · 163.94 g/mol.
m(Na₃PO₄) = 46.176 g.