Question

If 52.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.248 g of precipitate, what is the molarity of lead(II) ion in the original solution

Answer 1

Explanation:

The volume of given lead nitrate solution is:

52.5 mL.

The amount of lead iodide formed is ---0.248 g.

To get the molarity of lead (II) ion follow the below-shown procedure:

The number of moles of lead iodide formed is:

$number of moles of lead iodide(n)=mass of lead iodide/its molecular mass\\n=0.248 g/461.01g/mol\\n=0.000537mol$

0.000537 mole of lead iodide contains --- 0.000537 moles of lead (II) ion.

Thus, the number of moles are there, volume is there, and to get the molarity of lead (II) ion use the formula:

$Molarity=\frac{number of moles}{volume in L.} \\M=0.000537 mol / 0.0525 L.\\M=0.0102mol/L$

Molarity of lead iodide is --- 0.0102 M.