Question

Suppose 2.27g of lead(II) nitrate is dissolved in 300.mL of a 52.0mM aqueous solution of ammonium sulfate. Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) nitrate is dissolved in it.

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

Answer 1

Given weight of lead(II) nitrate=2.27 g and molar mass =331.2 g/mol,

Moles of lead(II) nitrate Pb(NO3)2 =2.27 g/331.2 g/mol=6.8538x10-3 moles.

1 mole of  Pb(NO3)2 gives 1 mole of Pb⁺² and 2 moles of NO₃^-.

So moles of NO₃^- =2x moles of  Pb(NO3)2=2x6.8538x10^-3 moles=0.0137 moles.

0.0137 moles of nitrate anion is dissolved in 300 mL (0.3L) of 52.0 mM aqueous solution of ammonium sulfate.

So the moles of nitrate anion will not be changed after mixing with ammonium sulfate, because the total moles of nitrate anion is same after making solution with ammonium sulfate.

So molarity of nitrate anion=moles of nitrate/volume=0.0137 moles/0.3 L=0.04569 M.