Suppose of lead(II) acetate is dissolved in of a 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 significant digits.

0.294 M

Explanation:

The computation of the final molarity of acetate anion is shown below:-

Lead acetate = Pb(OAc)2

Lead acetate involves two acetate ion.

14.3 gm lead acetate = Mass ÷ Molar mass

= 14.3 g ÷ 325.29 g/mol

= 0.044 mole

Volume of solution = 300 ml.

then

Molarity of lead is

= 0.044 × 1,000 ÷ 300

= 0.147 M

Therefore the molarity of acetate anion is

= 2 × 0.147

= 0.294 M

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