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Chapter 4: Problem 47

If \(38.5 \mathrm{~mL}\) of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield \(0.628 \mathrm{~g}\) of precipitate, what is the molarity of lead(II) ion in the original solution?

Short Answer

Expert verified
The molarity of the lead(II) ion in the original solution is 0.0353 M.

Step by step solution

01

Write the balanced chemical equation

The reaction between lead(II) nitrate \((Pb(NO_3)_2)\) and sodium iodide \((NaI)\) forms lead(II) iodide \((PbI_2)\) and sodium nitrate \((NaNO_3)\). The balanced equation is: \[ Pb(NO_3)_2 + 2 NaI \rightarrow PbI_2 + 2 NaNO_3 \]
02

Identify the precipitate and calculate its moles

Lead(II) iodide (PbI_2) is the precipitate. First, calculate its molar mass: \[ \text{Molar mass of } PbI_2 = 207.2 + 2 \times 126.9 = 461 \text{ g/mol} \] Then, use the mass of the precipitate to find moles: \[ \text{Moles of } PbI_2 = \frac{0.628 \text{ g}}{461 \text{ g/mol}} = 0.00136 \text{ moles} \]
03

Relate moles of precipitate to moles of lead(II) nitrate

According to the balanced equation, 1 mole of Pb(NO_3)_2 produces 1 mole of PbI_2. Therefore, moles of PbI_2 = moles of Pb(NO_3)_2 = 0.00136.
04

Calculate the molarity of the lead(II) ion solution

Molarity is defined as moles of solute per liter of solution. Given the volume of lead(II) nitrate solution is 38.5 mL, convert this to liters: \[ 38.5 \text{ mL} = 0.0385 \text{ L} \] Then, calculate the molarity: \[ \text{Molarity} = \frac{0.00136 \text{ moles}}{0.0385 \text{ L}} = 0.0353 \text{ M} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lead(II) nitrate
Lead(II) nitrate has the chemical formula \text{Pb(NO\(_3\))\(_2\)}. It's composed of one lead ion (Pb\(^{2+}\)) and two nitrate ions (NO\(_3^-\)).
This compound is commonly used in chemistry labs and can act as a source of lead ions in reactions.
In the exercise, lead(II) nitrate reacts with sodium iodide to form a precipitate.
Lead compounds are toxic, so always handle lead(II) nitrate with care and follow safety guidelines.
Precipitation reaction
A precipitation reaction occurs when two solutions are mixed and form an insoluble solid, called the precipitate.
In the provided exercise, the reaction between lead(II) nitrate and sodium iodide forms lead(II) iodide, which is the precipitate.
The chemical equation for this reaction is:
$$ \text{Pb(NO\(_3\))\(_2\) + 2 NaI 鈫� PbI\(_2\) + 2 NaNO\(_3\)} $$
Lead(II) iodide (PbI\(_2\)) is an insoluble yellow solid that forms and settles out from the solution.
Stoichiometry
Stoichiometry helps us relate quantities of substances in a chemical reaction.
It involves using the balanced chemical equation to determine the ratios of reactants and products.
For the reaction in the exercise, the balanced equation tells us that 1 mole of lead(II) nitrate produces 1 mole of lead(II) iodide.
If we know the moles of one substance, we can find the moles of another substance using this ratio.
Solution concentration
Solution concentration is a measure of the amount of solute dissolved in a given volume of solvent.
Molarity (M) is a common way to express concentration, defined as moles of solute per liter of solution.
In the problem, we calculated the molarity of the lead(II) nitrate solution by dividing the moles of lead(II) nitrate by the volume of the solution in liters:
$$ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} $$
Using this formula, we found the molarity of the lead(II) ion in the original solution.

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