Question

Suppose 13.2 g of lead (II) acetate is dissolved in 200. mL of a 0.50 M aqueous solution of ammonium sulfate Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the lead (II) acetate is dissolved in it. Round your answer to 3 significant digits.

Answer 1

Answer

0.406M

Explanation

Number of moles = mass /molar mass

number of moles of Pb(CH3COO)2 = 13.2g /325.3g/mol = 0.04058mol

Number of moles of CH3COO^- = 2×0.04058mol = 0.08116mol

molarity = number of moles of solute per liter of solution

Final molarity of acetate anion in the solution = (0.08116mol/200ml)×1000ml = 0.406M