Concept explainers
A chemist combines
(a)
Interpretation:
The mass of lead
Concept introduction:
The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.
The relation between number of moles and mass of a substance is given by the expression as shown below.
Answer to Problem 152E
The mass of lead
Explanation of Solution
The molarity of lead
The volume of lead
The conversion of volume in
The molarity of potassium iodide solution is
The volume of potassium iodide solution is
The conversion of volume in
The molar mass of lead
The molarity of a solution is given by the expression as shown below.
Where,
•
•
Rearrange the above equation for the value of
Substitute the values of molarity and volume of potassium iodide solution in above expression.
The number of moles of potassium iodide present in solution is
Substitute the values of molarity and volume of lead
The number of moles of lead
The reaction between potassium iodide and lead
Two moles of potassium iodide reacts with one mole of lead
Two moles of potassium iodide produced one mole of lead
Where,
•
•
Substitute the value of
The relation between number of moles and mass of a substance is given by the expression as shown below.
Where,
•
•
Substitute the value of number of moles and molar mass of lead
Therefore, the mass of lead
The mass of lead
(b)
Interpretation:
The final molarity of the potassium ion in solution of
Concept introduction:
The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.
The relation between number of moles and mass of a substance is given by the expression as shown below.
Answer to Problem 152E
The final molarity of the potassium ion in solution of
Explanation of Solution
The number of moles of potassium iodide present in solution is
The dissociation of potassium iodide in aqueous solution is shown below.
One mole of
The total volume of solution of
Where,
•
•
Substitute the values of
The molarity of a solution is given by the expression as shown below.
Where,
•
•
Substitute the values of number of moles of potassium ion and total volume of the solution in the above equation.
Therefore, the final molarity of the potassium ion in solution of
The final molarity of the potassium ion in solution is
(c)
Interpretation:
The final molarity of the ion out of lead
Concept introduction:
The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.
The relation between number of moles and mass of a substance is given by the expression as shown below.
Answer to Problem 152E
The final molarity of the lead
Explanation of Solution
The reaction between potassium iodide and lead
Potassium iodide is the limiting reagent. Therefore, all the iodide ion of potassium iodide has been consumed to precipitated lead
The initial number of moles of lead
The number of moles of lead
One mole of lead
The final number of moles of lead
The dissociation of lead
One mole of lead
The total volume of solution of
Where,
•
•
Substitute the values of
The molarity of a solution is given by the expression as shown below.
Where,
•
•
Substitute the values of number of moles of lead ion and total volume of the solution in the above equation.
Therefore, the final molarity of the lead
The final molarity of the lead
Want to see more full solutions like this?
Chapter 16 Solutions
Introductory Chemistry: An Active Learning Approach
- A 10.0-mL sample of potassium iodide solution was analyzed by adding an excess of silver nitrate solution to produce silver iodide crystals, which were filtered from the solution. KI(aq)+AgNO3(aq)KNO3(aq)+AgI(s) If 2.183 g of silver iodide was obtained, what was the molarity of the original KI solution?arrow_forwardssume a highly magnified view of a solution of HCI that allows you to “see” the HCl. Draw this magnified view. If you dropped in a piece of magnesium, the magnesium would disappear, and hydrogen gas would he released. Represent this change using symbols for the elements, and write the balanced equation.arrow_forward4.51 What is the role of an indicator in a titration?arrow_forward
- Describe in words how you would prepare pure crystalline AgCl and NaNO3 from solid AgNO3 and solid NaCl.arrow_forwardTwenty-five milliliters of a solution (d=1.107g/mL)containing 15.25% by mass of sulfuric acid is added to 50.0 mL of 2.45 M barium chloride. (a) What is the expected precipitate? (b) How many grams of precipitate are obtained? (c) What is the chloride concentration after precipitation is complete?arrow_forwardRelative solubilities of salts in liquid ammonia can differsignificantly from those in water. Thus, silver bromide issoluble in ammonia, but barium bromide is not (thereverse of the situation in water). Write a balanced equation for the reaction of anammonia solution of barium nitrate with an ammoniasolution of silver bromide. Silver nitrate is soluble inliquid ammonia. What volume of a 0.50 M solution of silver bromidewill react completely with 0.215 L of a 0.076 M solutionof barium nitrate in ammonia? What mass of barium bromide will precipitate fromthe reaction in part (b)?arrow_forward
- A 25.0-mL sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution to produce barium sulfate crystals, which were filtered from the solution. Na2SO4(aq)+BaCl2(aq)2NaCl(aq)+BaSO4(s) If 5.719 g of barium sulfate was obtained, what was the molarity of the original Na2SO4 solution?arrow_forwardLaws passed in some states define a drunk driver as one who drives with a blood alcohol level of 0.10% by mass or higher. The level of alcohol can be determined by titrating blood plasma with potassium dichromate according to the following equation 16H+(aq)+Cr2O72(aq)+C2H5OH(aq)4Cr3+(aq)+2CO2(g)+11H2O Assuming that the only substance that reacts with dichromate in blood plasma is alcohol, is a person legally drunk if 38.94 mL of 0.0723 M potassium dichromate is required to titrate a 50.0-g sample of blood plasma?arrow_forwardBone was dissolved in hydrochloric acid, giving 50.0 mL of solution containing calcium chloride, CaCL2. To precipitate the calcium ion from the resulting solution, an excess of potassium oxalate was added. The precipitate of calcium oxalate, CaC2O4, weighed 1.437 g. What was the molarity of CaCl2 in the solution?arrow_forward
- Introductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage LearningIntroductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning