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Pb(IO(3))(2) is a sparingly soluble salt...

`Pb(IO_(3))_(2)` is a sparingly soluble salt `(K_(sp) = 2.6 xx 10^(-13))`. To `35mL` of `0.15M Pb(NO_(3))_(2)` solution, `15mL` of `0.8M KIO_(3)` solution is added, and a precipiatte of `Pb(IO_(3))_(2)` is formed.
Which is the limiting reactant of teh reaction that takes place in the solution?

A

`Pb(IO_(3))_(2)`

B

`Pb(NO_(3))_(2)`

C

`KIO_(3)`

D

Both (b) and (c).

Text Solution

Verified by Experts

The correct Answer is:
B
`{:(2KIO_(3)+,Pb(NO_(3))_(2)rarr,Pb(IO_(3))_(2)+,2KNO_(3),),(0.8xx15,35xx0.15,,,),(=12,=5.25,,,):}`
1 mol of `Pb(NO_(3))_(2)` reacts with 2 mol of `KIO_(3)`
`5.25` mmol of `Pb(NO_(3))_(2)` reacts `= 2 xx 5.25 = 10.5` mmol
mmol of `KIO_(3)` left `= 12 - 10.5 = 1.5`
Hence `Pb(NO_(3))_(2)` is the limiting reactant.
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