Question

What volume of \(0.100 M \mathrm{Na}_{3} \mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from \(150.0 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short answer

To precipitate all lead(II) ions from \(150.0 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), \(562.5 \mathrm{mL}\) of \(0.100 M \mathrm{Na}_{3}\mathrm{PO}_{4}\) solution is required.

Step-by-step solution

Write down the balanced chemical equation

First, we need to write down the balanced chemical equation for the reaction between sodium phosphate (Na3PO4) and lead(II) nitrate (Pb(NO3)2). The equation is given by: \[2\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} + 3\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 6\mathrm{NaNO}_{3}+ \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\]

Calculate the moles of Pb(NO3)2

Now, we will calculate the number of moles of Pb(NO3)2 using the given volume and molarity. Moles of Pb(NO3)2 = Molarity × Volume Moles of Pb(NO3)2 = \(0.250 M × 0.150 L = 0.0375 mol\)

Calculate the moles of Na3PO4 required

Next, we will use the stoichiometry of the balanced chemical equation to determine the number of moles of Na3PO4 needed to react with the moles of Pb(NO3)2 we calculated in step 2. According to the balanced chemical equation, 3 moles of Na3PO4 react with 2 moles of Pb(NO3)2. Moles of Na3PO4 required = \(\frac{3}{2} \times \text{moles of Pb(NO} _3)_2\) Moles of Na3PO4 required = \(\frac{3}{2} \times 0.0375 mol = 0.05625 mol\)

Calculate the volume of Na3PO4 solution needed

Finally, we will use the molarity of the given Na3PO4 solution and the number of moles calculated in step 3 to find the volume of Na3PO4 solution required to precipitate all the lead(II) ions. Volume = Moles ÷ Molarity Volume = \(\frac{0.05625 mol}{0.100 M} = 0.5625 L\) Since the question asked for the volume in milliliters, we multiply by 1000: Volume = \(0.5625 L \times 1000 = 562.5 \mathrm{mL}\) Hence, \(562.5 \mathrm{mL}\) of \(0.100 M \mathrm{Na}_{3}\mathrm{PO}_{4}\) solution is required to precipitate all the lead(II) ions from \(150.0 \mathrm{mL}\) of \(0.250 \mathrm{M}\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).

Key concepts

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants and products involved, along with their respective quantities. Consider the equation given in the exercise:

• \[2\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} + 3\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 6\mathrm{NaNO}_{3}+ \mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\]

This equation tells us that two moles of lead(II) nitrate react with three moles of sodium phosphate to produce lead(II) phosphate and sodium nitrate.
Each compound is represented by a formula where the different elements and their numbers are indicated. The numbers in front of the compounds (called coefficients) show the ratio in which the substances react or are produced.
Beneath the surface representation, chemical equations also respect the Law of Conservation of Mass. This principle means that the total mass of reactants equals the total mass of products in any chemical reaction, ensuring that atoms are balanced on both sides of the equation.

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula is:

• \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\]

It tells us how much of a substance is dissolved in a specific volume of liquid.
In the context of the exercise, we have the molarity of lead(II) nitrate given as 0.250 M and we need to use it to find out how much sodium phosphate is necessary for the reaction.
By using molarity, we can easily convert volumes of solutions into moles of substance, which is essential for stoichiometry calculations.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions react and form an insoluble solid, known as the precipitate. In this exercise, lead(II) ions (\(\mathrm{Pb}^{2+}\)) react with phosphate ions (\(\mathrm{PO}_{4}^{3-}\)) to form an insoluble solid, lead(II) phosphate (\(\mathrm{Pb}_{3}(\mathrm{PO}_{4})_{2}\)).
This kind of reaction is crucial because it allows for the separation of solid substances from a solution. Not all ionic compounds are soluble in water, and knowing which compounds precipitate under certain conditions is key.

• The resulting lead(II) phosphate is what defines the success of this reaction, as the objective is to remove lead from the solution.
• The formation of a precipitate confirms that the reaction has occurred and the ions have been effectively combined and eliminated from the liquid phase.

Calculating Moles

Calculating moles is the foundation of stoichiometry, allowing chemists to quantitatively predict the outcomes of chemical reactions. The number of moles measures the amount of a substance, linking the macroscopic scale (grams) with the microscopic scale (atoms, molecules).

• To find moles, we multiply the molarity by the volume in liters (or convert milliliters to liters first).
• For instance, in the exercise, we determined the moles of \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) by multiplying its molarity (0.250 M) by its volume (0.150 L), resulting in 0.0375 moles.

These calculated moles are then used in the stoichiometric ratios derived from the balanced equation to determine how much of another substance is needed or produced in the reaction.
Such conversions are crucial for calculating the actual amount of reactants or products involved, as seen when determining the volume of the \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) solution necessary for the precipitation reaction.