Question

Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 1.500 M HI(aq) must be added to a solution containing 0.300 mol of to completely precipitate the lead?

Answer 1

400 mL of 1.500 M HI(aq.) must be added for complete precipitation of lead

Step-by-step explanation:

Balanced reaction for formation of lead iodide-

`Pb^(2+)+2I^(-)\rightarrow PbI_(2)`

Hence, according to balanced equation, for precipitation of 1 mol of
`Pb^(2+)`, 2 moles of
`I^(-)` are required.

Therefore, to precipitate 0.300 moles of
`Pb^(2+)`, 0.600 moles of
`I^(-)` are required.

1.500 m HI(aq.) solution means 1 L of HI (or
`I^(-)`) solution contains 1.500 moles of HI (or
`I^(-)`).

So, volume of HI solution containing 0.600 moles
`I^(-)` =
`(0.600)/(1.500) L` = 0.400 L = 400 mL