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?

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

Explanation:

Balanced reaction for formation of lead iodide-

Hence, according to balanced equation, for precipitation of 1 mol of , 2 moles of are required.

Therefore, to precipitate 0.300 moles of , 0.600 moles of are required.

1.500 m HI(aq.) solution means 1 L of HI (or ) solution contains 1.500 moles of HI (or ).

So, volume of HI solution containing 0.600 moles = = 0.400 L = 400 mL

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

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