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

Answer:

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

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 $I^{-}$ = $\frac{0.600}{1.500} L$ = 0.400 L = 400 mL