Question

Suppose 27.7g of lead(II) acetate is dissolved in 200.mL of a 0.70M aqueous solution of ammonium sulfate.

Calculate the final molarity of lead(II) cation in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

M(Pb(CH3COO)2)=325g/mol

n(Pb(CH3COO)2)=n/M=27.7g/325g/mol=0.085 mol

Pb(CH3COO)2=Pb^2+ + 2CH3COO^-

n(Pb^2+)= 0.085 mol

C(Pb^2+)=n/V=0.085mol/0.2L= 0.425 M