Question

Suppose

2.10g

of lead(II) acetate is dissolved in

350.mL

of a

65.0mM

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.

Be sure your answer has the correct number of significant digits.

Answer 1

number of mole of Pb(CH3COO)2 = (given mass)/(molar mass of)
molar mass of Pb(CH3COO)2 = 325.3 g/mol
number of mole of Pb(CH3COO)2 = 2.10/325.3
= (6.5*10^-3) mole

1 mole of Pb(CH3COO)2 give 2 mole of CH3COO-
(6.5*10^-3) mole of Pb(CH3COO)2 give (2*6.5*10^-3) mole of (CH3COO
number of mole of CH3COO- = (2*6.5*10^-3) mole
= 0.013 mole

volume of solution = 350.0 mL
= 0.350 L

molarity of CH3COO- = (number of mole)/(volume)
= 0.013/0.350
= 0.037 M

Answer : 0.037 M