molar concentration of Sn²⁺ ions = 0.273 M
molar concentration of Al³⁺ ions = 0.415 M
molar concentration of Cl⁻ ions = 1.692 M
Explanation:
First we calculate the number of moles of tin (II) chloride (SnCl₂) in the first solution:
number of moles = molar concentration × volume (L)
number of moles of SnCl₂ = 0.55 × 0.45 = 0.248 moles
And the number of moles of aluminum chloride (AlCl₃) in the second solution:
number of moles = molar concentration × volume (L)
number of moles of AlCl₃ = 0.7 × 0.65 = 0.455 moles
0.248 moles of SnCl₂ contains 0.248 moles of Sn²⁺ ions and 2 × 0.248 = 0.496 moles of Cl⁻ ions
0.455 moles of AlCl₃ contains 0.455 moles of Al³⁺ ions and 3 × 0.455 = 1.365 moles of Cl⁻ ions
Volume of the final solution = 450 mL + 650 mL = 1100 mL = 1.1 L
molar concentration = number of moles / volume (L)
molar concentration of Sn²⁺ ions = 0.248 / 1.1 = 0.273 M
molar concentration of Al³⁺ ions = 0.455 / 1.1 = 0.415 M
molar concentration of Cl⁻ ions = (0.496 + 1.365) / 1.1 = 1.692 M
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