Question

Suppose
1.80g
of lead(II) acetate is dissolved in
350.mL
of a
26.0mM
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.

Answer 1

n(Pb(CH3COO)2) = 1.8 g : 325 g/ mol = 0.00554 mol
n(CH3COO-) = 2n(Pb(CH3COO)2) = 0.011 mol
CM = 0.011 mol / 0.350 L = 0.031 M
Answer: 0.031 M