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 50.00-mL sample of 0.0250 M silver nitrate is mixed with 0.0400 M chromium(III) chloride. (a) What is the minimum volume of chromium(III) chloride required to completely precipitate silver chloride? (b) How many grams of silver chloride are produced from (a)?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_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 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_forwardA 25-mL sample of 0.50 M NaOH is combined with a 75-mL sample of 0.50 M NaOH. What is the concentration of the resulting NaOH solution?arrow_forwardCarminic 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% C and 4.09% H by mass. A titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 g carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?arrow_forward
- 4.51 What is the role of an indicator in a titration?arrow_forwardA 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_forward
- Introductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher: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
- Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning