emilyahope39
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Be sure to answer all parts.

One way to remove lead ions from water is to add a source of iodide ions so that lead iodide will precipitate out of solution:

[tex]\[
Pb^{2+}(aq) + 2I^{-}(aq) \rightarrow PbI_2(s)
\][/tex]

(a) What volume of a [tex]$1.0 \, M \, KI$[/tex] solution must be added to [tex]$360.0 \, mL$[/tex] of a solution that is [tex]$0.20 \, M$[/tex] in [tex]$Pb^{2+}$[/tex] ion to precipitate all the lead ions?

[tex]\[\square \, mL \][/tex]

(b) What mass of [tex]$PbI_2$[/tex] should precipitate?

[tex]\[\square \, g \, PbI_2\][/tex]



Answer :

GinnyAnswer
Sure! Let's work through this problem step by step.

### Part (a)

Objective: Determine the volume of a 1.0 M KI solution required to precipitate all the [tex]\( \text{Pb}^{2+} \)[/tex] ions in a 360.0 mL solution that is 0.20 M in [tex]\( \text{Pb}^{2+} \)[/tex] ions.

1. Calculate the moles of [tex]\( \text{Pb}^{2+} \)[/tex]:
[tex]\[ \text{Moles of } \text{Pb}^{2+} = \text{Molarity } \times \text{ Volume in liters} \][/tex]
Since we are given 360.0 mL, we need to convert it to liters:
[tex]\[ \text{Volume in liters} = \frac{360.0 \text{ mL}}{1000} = 0.360 \text{ L} \][/tex]
[tex]\[ \text{Moles of } \text{Pb}^{2+} = 0.20 \text{ M} \times 0.360 \text{ L} = 0.072 \text{ moles} \][/tex]

2. Determine the moles of [tex]\( \text{I}^{-} \)[/tex] needed:
From the reaction [tex]\( \text{Pb}^{2+}(aq) + 2\text{ I}^{-}(aq) \rightarrow \text{PbI}_2(s) \)[/tex], 1 mole of [tex]\( \text{Pb}^{2+} \)[/tex] reacts with 2 moles of [tex]\( \text{I}^{-} \)[/tex]. Therefore:
[tex]\[ \text{Moles of } \text{I}^{-} \text{ required} = 2 \times \text{Moles of } \text{Pb}^{2+} = 2 \times 0.072 = 0.144 \text{ moles} \][/tex]

3. Calculate the volume of [tex]\( 1.0 \text{ M KI} \)[/tex] solution needed:
[tex]\[ \text{Molarity (M)} = \frac{\text{Moles}}{\text{Volume in liters}} \][/tex]
Rearranging to solve for volume:
[tex]\[ \text{Volume in liters} = \frac{\text{Moles of } \text{I}^{-}}{\text{Molarity of KI}} = \frac{0.144}{1.0} = 0.144 \text{ L} \][/tex]
Convert this volume to mL:
[tex]\[ \text{Volume in mL} = 0.144 \text{ L} \times 1000 = 144.0 \text{ mL} \][/tex]

So, the volume of a 1.0 M KI solution needed is [tex]\( 144.0 \)[/tex] mL.

### Part (b)

Objective: Determine the mass of [tex]\( \text{PbI}_2 \)[/tex] that should precipitate.

1. Calculate the moles of [tex]\( \text{PbI}_2 \)[/tex] formed:
From the chemical reaction, the moles of [tex]\( \text{PbI}_2 \)[/tex] formed will be equal to the moles of [tex]\( \text{Pb}^{2+} \)[/tex] that reacted, which is 0.072 moles.

2. Determine the molar mass of [tex]\( \text{PbI}_2 \)[/tex]:
[tex]\[ \text{Molar mass of } \text{PbI}_2 = \text{Molar mass of } \text{Pb} + 2 \times \text{Molar mass of } \text{I} \][/tex]
Given (assuming the principles of atomic masses):
- [tex]\( \text{Pb} \approx 207.2 \text{ g/mol} \)[/tex]
- [tex]\( \text{I} \approx 126.9 \text{ g/mol} \)[/tex]

[tex]\[ \text{Molar mass of } \text{PbI}_2 = 207.2 + 2 \times 126.9 = 461.0 \text{ g/mol} \][/tex]

3. Calculate the mass of [tex]\( \text{PbI}_2 \)[/tex] precipitated:
[tex]\[ \text{Mass of } \text{PbI}_2 = \text{Moles of } \text{PbI}_2 \times \text{Molar mass of } \text{PbI}_2 \][/tex]
[tex]\[ \text{Mass of } \text{PbI}_2 = 0.072 \text{ moles} \times 461.0 \text{ g/mol} = 33.192 \text{ g} \][/tex]

So, the mass of [tex]\( \text{PbI}_2 \)[/tex] that should precipitate is [tex]\( 33.192 \)[/tex] grams.

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