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

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.