Question

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?

Answer 1

Answer:

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

Explanation:

$Pb(NO_{3})_{2}$ is precipitated as $PbI_{2}$ when NaI is added to solution of $Pb(NO_{3})_{2}$

Balanced reaction: $Pb(NO_{3})_{2}+2NaI\rightarrow PbI_{2}+2NaNO_{3}$

Molar mass of $PbI_{2}$ = 461.01 g/mol

So, 0.636 g of $PbI_{2}$ = $\frac{0.636}{461.01}$ mol of

                              = 0.00138 mol of $PbI_{2}$

                              = 0.00138 mol of Pb (II) ion

So, 61.1 mL of $Pb(NO_{3})_{2}$ solution contains 0.00138 mol of Pb (II) ion

Hence, molarity of Pb (II) ion in original solution = $\frac{0.00138}{61.1}\times 1000M$ = 0.0226 M