/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 What volume of \(0.100 M \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Expert verified
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

01

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: \(2Pb(NO_{3})_{2} + 3Na_{3}PO_{4} \rightarrow Pb_{3}(PO_{4})_{2} + 6NaNO_{3}\)
02

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
03

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
04

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A student titrates an unknown amount of potassium hydrogen phthalate \(\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\right.\), often abbreviated \(\mathrm{KHP}\) ) with \(20.46 \mathrm{~mL}\) of a \(0.1000 M\) NaOH solution. KHP (molar mass \(=204.22\) \(\mathrm{g} / \mathrm{mol}\) ) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

Complete and balance each acid-base reaction. a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains three acidic hydrogens b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) Contains two acidic hydrogens c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) Contains two acidic hydrogens d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains two acidic hydrogens.

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of \(0.5894 \mathrm{~g}\) were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is \(53.66 \% \mathrm{C}\) and \(4.09 \% \mathrm{H}\) by mass. A titration required \(18.02 \mathrm{~mL}\) of \(0.0406 \mathrm{M} \mathrm{NaOH}\) to neutralize \(0.3602 \mathrm{~g}\) carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?

Some of the substances commonly used in stomach antacids are \(\mathrm{MgO}, \mathrm{Mg}(\mathrm{OH})_{2}\), and \(\mathrm{Al}(\mathrm{OH})_{3}\) a. Write a balanced equation for the neutralization of hydrochloric acid by each of these substances. b. Which of these substances will neutralize the greatest amount of \(0.10 \mathrm{M} \mathrm{HCl}\) per gram?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.