a) What volume of a 1.0 M KI solution must be added to 470.0 mL of a solution that is 0.11 M in Pb2 ion to precipitate all the lead ion

0.168mL of the 1.0M KI solution must be added

Explanation:

The PbI₂ is in equilibrium with water as follows:

PbI₂ ⇄ Pb²⁺ + 2I⁻

Where Ksp is:

Ksp = 1.4x10⁻⁸ = [Pb²⁺] / [I⁻]²

As molarity of Pb²⁺ is 0.11M:

1.4x10⁻⁸ = [0.11M] / [I⁻]²

1.27x10⁻⁷ = [I⁻]²

3.57x10⁻⁴M = [I⁻]

Thus, to precipitate all Pb²⁺ you need to add 3,57x10⁻⁴M of I⁻. As volume of the solution is 470.0mL = 0.47L, you need to add:

0.47L * (3,57x10⁻⁴moles / L) = 1.68x10⁻⁴ moles of I⁻ = Moles of KI.

That comes from the 1.0M KI. You need to add:

1.68x10⁻⁴ moles of KI * (L / 1.0 mol) = 1.68x10⁻⁴L =

bcde is the incorrect answer it is a

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol