Answer:
Molarity of Pb (II) ion in original solution is 0.0226 M
Explanation:
[tex]Pb(NO_{3})_{2}[/tex] is precipitated as [tex]PbI_{2}[/tex] when NaI is added to solution of [tex]Pb(NO_{3})_{2}[/tex]
Balanced reaction: [tex]Pb(NO_{3})_{2}+2NaI\rightarrow PbI_{2}+2NaNO_{3}[/tex]
Molar mass of [tex]PbI_{2}[/tex] = 461.01 g/mol
So, 0.636 g of [tex]PbI_{2}[/tex] = [tex]\frac{0.636}{461.01}[/tex] mol of
= 0.00138 mol of [tex]PbI_{2}[/tex]
= 0.00138 mol of Pb (II) ion
So, 61.1 mL of [tex]Pb(NO_{3})_{2}[/tex] solution contains 0.00138 mol of Pb (II) ion
Hence, molarity of Pb (II) ion in original solution = [tex]\frac{0.00138}{61.1}\times 1000M[/tex] = 0.0226 M
