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 the lead(II) ions from 150.0 mL of 0.250 M \(Pb(NO_{3})_{2}\) solution, 562.5 mL of 0.100 M \(Na_{3}PO_{4}\) solution is required.

Step-by-step solution

Write the balanced chemical equation for the precipitation reaction

The reaction between lead(II) nitrate (Pb(NO₃)₂) and sodium phosphate (Na₃PO₄) can be written as: \(Pb(NO_{3})_{2} + Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\) However, we need to balance this equation. The balanced chemical equation is: \(2Pb(NO_{3})_{2} + 3Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\)

Calculate moles of lead(II) nitrate

The moles of lead(II) nitrate can be calculated using the formula: Moles = Molarity × Volume in Liters Moles of \(Pb(NO_{3})_{2}\) = 0.250 M × (150.0 mL × (1 L / 1000 mL)) Moles of \(Pb(NO_{3})_{2}\) = 0.0375 mol

Determine moles of sodium phosphate required using stoichiometry

From the balanced equation, we can see that 2 moles of lead(II) nitrate react with 3 moles of sodium phosphate. Thus, the mole ratio of \(Pb(NO_{3})_{2}\) to \(Na_{3}PO_{4}\) is 2:3. Using this ratio, we can calculate the required moles of sodium phosphate to completely react with the moles of lead(II) nitrate that we have: Moles of \(Na_{3}PO_{4}\) = (3 moles \(Na_{3}PO_{4}\) / 2 moles \(Pb(NO_{3})_{2}\)) × 0.0375 moles \(Pb(NO_{3})_{2}\) Moles of \(Na_{3}PO_{4}\) = 0.05625 mol

Calculate the volume of sodium phosphate solution required

We know the molarity of sodium phosphate solution is 0.100 M, and we have found the moles of sodium phosphate required to react completely with the given amount of lead(II) nitrate. Now we can calculate the required volume of sodium phosphate solution using the formula: Volume in Liters = Moles / Molarity Volume of \(Na_{3}PO_{4}\) = 0.05625 mol / 0.100 M Volume of \(Na_{3}PO_{4}\) = 0.5625 L To convert this result to milliliters, multiply by 1000: Volume of \(Na_{3}PO_{4}\) = 0.5625 L × 1000 mL/L Volume of \(Na_{3}PO_{4}\) = 562.5 mL So, 562.5 mL of 0.100 M sodium phosphate solution is required to precipitate all the lead(II) ions from 150.0 mL of 0.250 M lead(II) nitrate solution.

Key concepts

Molarity Calculations

When we talk about molarity, we're referring to the concentration of a solution. It's a way to express the amount of a solute (the substance that's dissolved) in a given volume of solution. Molarity is defined as the number of moles of solute per liter of solution. It is denoted by the letter M. Here's a simple formula to understand and use:

• Molarity (M) = Moles of Solute / Liters of Solution

If you're given the volume of a solution in milliliters, don't forget to convert it to liters since molarity calculations require liters. For example, if you have 150.0 mL of solution, you would convert it to liters by dividing by 1000, because there are 1000 mL in a liter. In problems where you're finding the volume given molarity, you'll rearrange the formula to solve for volume:

• Volume in Liters = Moles of Solute / Molarity

This formula is crucial in solving stoichiometry problems like the one with sodium phosphate and lead(II) nitrate.

Balanced Chemical Equations

Chemical equations need to be balanced to obey the Law of Conservation of Mass. This law states that mass is neither created nor destroyed in a chemical reaction. Therefore, the number of atoms of each element must be the same on both sides of the equation.

To achieve this, we adjust the coefficients (the numbers in front of the compounds) in the reaction. Let's take a look at the precipitation reaction given in the exercise:- **Unbalanced Equation**: \(Pb(NO_{3})_{2} + Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\)- **Balanced Equation**:

• \(2Pb(NO_{3})_{2} + 3Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\)

In balancing, adjust coefficients, ensuring the number of atoms for each element is equal on both sides. Notice how two moles of lead nitrate react with three moles of sodium phosphate? This 2:3 ratio is key in stoichiometry calculations to determine how much of one reactant you need for a given amount of another.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions mix and form an insoluble solid, called a precipitate. In the context of our exercise, when sodium phosphate reacts with lead nitrate, they form lead phosphate as a solid and sodium nitrate remains in solution.
Here's a simple plan to identify a precipitation reaction:1. **Mix two ionic solutions**: Typically involves solutions like our exercise's examples of lead nitrate and sodium phosphate.2. **Look for insoluble products**: Use solubility rules to predict if a product is insoluble in water. In our case, \(Pb_{3}(PO_{4})_{2}\) is insoluble.3. **Form a solid precipitate**: The appearance of a solid indicates a precipitation reaction.Understanding these reactions is crucial in applications involving purification processes or removing waste materials. Knowing which products precipitate can also tell us what ions remain in solution, enhancing our grasp of solution chemistry.