Question

20mL of 0.200 M aluminum acetate solution is added to a solution of magnesium nitrate.

How many moles of the acetate ion are present in the aluminum acetate?

The answer is 7.2x10²¹ C₂H₃O₂^- ions but I don't know how to get that.

Answer 1

20mL = 20mL x (1L / 1000mL) = 0.02L

0.02 L of 0.2 M aluminum acetate = (0.02 L x 0.2 M) = 0.004 moles of  aluminum acetate

The molecular formula of aluminium acetate = Al(CH3COOH)₃ which dissociates  as,

Al(CH3COO)₃  $\rightleftharpoons$ Al³⁺ + 3 CH3COO^-

thus, 1 mole of  Al(CH3COO)₃ contains 3 moles of acetate ion

or, 0.004 mole of  Al(CH3COO)₃ contains (3 x 0.004)= 0.012 moles of acetate ions are present

1 mole of acetate ion = 6.023 x 10²³ number of acetate ion

or, 0.012 moles of acetate ion = 6.023 x 10²³ x 0.012 = 7.23 x 10²¹ number of acetate ion