Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 55

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{~mL}\) of a \(0.100 M\) solution of \(\mathrm{AgNO}_{3}\) ?

Short Answer

Expert verified
To precipitate all the silver ions from a \(75.0\,\mathrm{mL}\) solution of \(0.100\, \mathrm{M}\, \mathrm{AgNO}_{3}\), approximately \(0.607\, \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required.

Step by step solution

01

Write the balanced chemical equation

The reaction that occurs when sodium chromate (Na₂CrO₄) reacts with silver nitrate (AgNO₃) to precipitate silver ions (Ag⁺) can be described by the following balanced chemical equation: \[2 AgNO_3(aq) + Na_2CrO_4(aq) \rightarrow Ag_2CrO_4(s) + 2 NaNO_3(aq)\]
02

Find the moles of AgNO₃ in the given solution

We are given the volume (\(75.0\,\text{mL}\)) and concentration (\(0.100\,\text{M}\)) of the AgNO₃ solution. We can use these values to calculate the moles of AgNO₃ in the solution. Moles = Molarity × Volume The volume needs to be converted to liters, so \(\text{Volume} = 75.0\,\text{mL} \times \frac{1 \,\text{L}}{1000\, \text{mL}} = 0.075\, \text{L}\). Then, Moles of AgNO₃ = \(0.100\, \text{M} \times 0.075\, \text{L} = 0.0075\, \text{mol}\)
03

Find the moles of Na₂CrO₄ required to precipitate all the Ag⁺ ions

From the balanced chemical equation, we can see that 2 moles of AgNO₃ react with 1 mole of Na₂CrO₄ to form 2 moles of NaNO₃ and 1 mole of Ag₂CrO₄. So, the mole ratio of AgNO₃ to Na₂CrO₄ is 2:1. To find the moles of Na₂CrO₄ required to precipitate all the silver ions, we can use this mole ratio: Moles of Na₂CrO₄ = \(\text{Moles of AgNO₃} \, \times \, \frac{1\, \text{mol Na₂CrO₄}}{2\, \text{mol AgNO₃}} = 0.0075\, \text{mol} \, \times \, \frac{1}{2} = 0.00375\, \text{mol}\)
04

Calculate the mass of Na₂CrO₄ needed

Now that we know the moles of Na₂CrO₄ required, we can calculate the mass of Na₂CrO₄ using its molar mass. The molar mass of Na₂CrO₄ is approximately \(2(22.99) + 51.996 + 4(16.00) = 161.984\, \text{g/mol}\). Then, Mass of Na₂CrO₄ = \(\text{Moles of Na₂CrO₄} \, \times \, \text{Molar mass of Na₂CrO₄} = 0.00375\, \text{mol} \, \times \, 161.984\, \text{g/mol} = 0.607\, \text{g}\) So, the mass of Na₂CrO₄ required to precipitate all the silver ions from a \(75.0\,\mathrm{mL}\) solution of \(0.100\, \mathrm{M}\, \mathrm{AgNO}_{3}\) is approximately \(0.607\, \mathrm{g}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Many plants are poisonous because their stems and leaves contain oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) or sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\); when ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in aqueous solution.

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2}\), or \(200.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{NaCl} ?\)

A standard solution is prepared for the analysis of fluoxymesterone \(\left(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{FO}_{3}\right)\), an anabolic steroid. A stock solution is first prepared by dissolving \(10.0 \mathrm{mg}\) of fluoxymesterone in enough water to give a total volume of \(500.0 \mathrm{~mL}\). A \(100.0-\mu \mathrm{L}\) aliquot (portion) of this solution is diluted to a final volume of \(100.0 \mathrm{~mL}\). Calculate the concentration of the final solution in terms of molarity.

A stock solution containing \(\mathrm{Mn}^{2+}\) ions was prepared by dissolving \(1.584 \mathrm{~g}\) pure manganese metal in nitric acid and diluting to a final volume of \(1.000 \mathrm{~L}\). The following solutions were then prepared by dilution: For solution A. \(50.00 \mathrm{~mL}\) of stock solution was diluted to \(1000.0 \mathrm{~mL}\) For solution \(B, 10.00 \mathrm{~mL}\) of solution \(A\) was diluted to \(250.0 \mathrm{~mL}\). For solution \(C, 10.00 \mathrm{~mL}\) of solution \(B\) was diluted to \(500.0 \mathrm{~mL}\). Calculate the concentrations of the stock solution and solutions \(A, B\), and \(C\).

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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.

Sign-up for free