If 61.1 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.636 g of precipitate, what is the molarity of lead(II) ion in the original solution?

Molarity of Pb (II) ion in original solution is 0.0226 M

Explanation:

is precipitated as when NaI is added to solution of

Balanced reaction:

Molar mass of = 461.01 g/mol

So, 0.636 g of = mol of

                              = 0.00138 mol of

                              = 0.00138 mol of Pb (II) ion

So, 61.1 mL of solution contains 0.00138 mol of Pb (II) ion

Hence, molarity of Pb (II) ion in original solution = = 0.0226 M

b is the correct