Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide:
$\mathrm { Pb } ^ { 2 + }$ (aq) + $2 \mathrm { I } ^ { - }$ (aq) ? $\mathrm { PbI } _ { 2 }$ (s) Lead iodide is virtually insoluble in water so that the reaction appears to go to completion.How many millilitres of 3.550 mol L^-1 HI(aq) must be added to a solution containing 0.700 mol of $\mathrm { Pb } \left( \mathrm { NO } _ { 3 } \right) _ { 2 } ^ { ( a q ) }$ to completely precipitate the lead?

Question

Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide:

\mathrm { Pb } ^ { 2 + }

(aq) +

2 \mathrm { I } ^ { - }

(aq) ?

\mathrm { PbI } _ { 2 }

(s) Lead iodide is virtually insoluble in water so that the reaction appears to go to completion.How many millilitres of 3.550 mol L^-1 HI(aq) must be added to a solution containing 0.700 mol of

\mathrm { Pb } \left( \mathrm { NO } _ { 3 } \right) _ { 2 } ^ { ( a q ) }

to completely precipitate the lead?

Quizplus · Accepted Answer

According to the balanced chemical equation, 1 mole of $\mathrm{Pb(NO_3)_2}$ reacts with 2 moles of HI to produce 1 mole of $\mathrm{PbI_2}$. Therefore, if we have 0.700 mol of $\mathrm{Pb(NO_3)_2}$, we need 2 x 0.700 = 1.400 mol of HI to completely precipitate the lead.

To calculate the volume of 3.550 mol/L HI solution required to provide 1.400 mol of HI, we use the following equation:

Molarity = moles / volume

Rearranging,

Volume = moles / molarity

Volume = 1.400 mol / 3.550 mol/L = 0.394 L = 394 mL

Therefore, the answer is C, 394 mL.

Lead Ions Can Be Precipitated from Aqueous Solutions by the Addition