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In the laboratory you dissolve 13.2 g of iron(III) acetate in a volumetric flask and add...

In the laboratory you dissolve 13.2 g of iron(III) acetate in a volumetric flask and add water to a total volume of 500 mL.

What is the molarity of the solution?  M.

What is the concentration of the iron(III) cation?  M.

What is the concentration of the acetate anion?  M.

Expert Solution

Solution :-

Iron (lll) acetate = Fe(C2H3O2)3 = 232.97 g per mol

Lets first calculate the moles of the iron(III) acetate

Moles = mass / molar mass

Moles of iron(III) acetate = 13.2 g / 232.97 g per mol

= 0.0567 mol

Now lets calculate its molarity

Molarity = moles / volume in liter

Molarity = 0.0567 mol / 0.500 L

= 0.113 M iron(III) acetate

Now lets calculate the concentration of the iron (lll) cation

Mole ratio is 1 : 1

Therefore

0.113 M Fe(C2H3O2)3 * 1 M Fe^3+ / 1 M Fe(C2H3O2)3 = 0.113 M Fe^3+

Threfore the concentration of the Iron (lll) cation is 0.113 M

Now lets calculate the concentration of the acetate ion

Mole ratio is 1 : 3

0.113 M Fe(C2H3O2)3 * 3 M C2H3O2^- / 1 M Fe(C2H3O2)3 = 0.339 M C2H3O2^-

Threfore the concentration of the acetate ion is 0.339 M

queen_honey_blossom answered 11 months ago

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