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Zina [86]
3 years ago
7

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

the following is false?
Select one:

A. The final concentration of Pb2+ ions is 0.174 M.

B. You form 111 g of lead(II) iodide.

C. The final concentration of K+ is 0.821 M.

D. The final concentration of NO3– is 0.821 M.

E. All are true.
Chemistry
1 answer:
SIZIF [17.4K]3 years ago
4 0

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

Moles=1.60 \times {300.0\times 10^{-3}}\ moles

<u>Moles of potassium iodide = 0.48 moles </u>

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=1.20\times {285\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.342 moles </u>

According to the given reaction:

2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

<u>Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.</u>

Also, consumed lead(II) nitrate = 0.24 moles  (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.102/0.585 M = 1.174 M</u>

<u>Statement A is correct.</u>

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

<u>Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g </u>

<u>Statement B is correct.</u>

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.48/0.585 M = 0.821 M</u>

<u>Statement C is correct.</u>

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate  produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate  produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>

<u>Statement D is incorrect.</u>

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