Question

In the laboratory you dissolve 12.2 g of aluminum acetate in a volumetric flask and add water to a total volume of 125 mL.

What is the molarity of the solution?  M.

What is the concentration of the aluminum cation?  M.

What is the concentration of the acetate anion?  M.

Answer 1

i) Molarity of the solution

Number of moles = mass/molar mass

Number of moles of Al(CH3COO)3 = 12.2g/204.13g/mol = 0.05977mol

Molarity = Number of moles of solute per liter of solution

molarity of Al(CH3COO)3 = (0.05977mol/125ml)×1000ml

= 0.4782M

ii) Concentration of aluminium cation

Molecular formula of aluminium acetate = Al(CH3COO)3

Al(CH3COO)3 contain one Al³⁺ ion

Therefore

Concentration of Aluminium ion in molarity = Concentration of Al(CH3COO)3

Therefore,

Concentration of Aluminium ion in molarity= 0.4782M

iii) Concentration of acetate anion

One aluminium acetate contain three CH3COO^- ions

Concentration of acetate ion in molarity = 3× Molarity of Al(CH3COO)3

Therefore

Concentration of acetate ion in molarity = 3 × 0.4782M

= 1.435 M