Suppose 1.39 g of potassium iodide is dissolved in 200. mL of a 41.0 m M aqueous solution of silver nitrate.Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved init.Be sure your answer has the correct number of significant digits.MExplanationCheck3ol© 2023 McGraw Hit LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibl

Answered: Suppose 1.39 g of potassium iodide is…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

Suppose 1.39 g of potassium iodide is dissolved in 200. mL of a 41.0 m M aqueous solution of silver nitrate.
Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in
it.
Be sure your answer has the correct number of significant digits.
M
Explanation
Check
3
ol
© 2023 McGraw Hit LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibl

Expert Solution

Step 1: Number of moles of potassium iodide

The balanced chemical equation for the reaction between potassium iodide (KI) and silver nitrate (AgNO₃) is

$bold italic 2 bold italic K bold italic I bold italic space bold italic plus bold italic space bold italic A bold italic g bold italic N bold italic O subscript bold italic 3 bold italic space bold italic rightwards arrow bold italic space bold italic 2 bold italic A bold italic g bold italic I bold italic space bold italic plus bold italic space bold italic K bold italic N bold italic O subscript bold italic 3$

First, we need to calculate the number of moles of KI in the solution.

Provided the mass of KI is 1.39 g and the molar mass of KI is 166.0028 g/mol,

by using the formula $bold italic n bold italic o bold italic. bold italic space bold italic o bold italic f bold italic space bold italic m bold italic o bold italic l bold italic e bold italic s bold italic space bold italic K bold italic I bold italic space bold italic equals fraction numerator bold italic G bold italic i bold italic v bold italic e bold italic n bold italic space bold italic m bold italic a bold italic s bold italic s bold italic space bold italic o bold italic f bold italic space bold italic K bold italic I bold italic space over denominator bold italic M bold italic o bold italic l bold italic e bold italic c bold italic u bold italic l bold italic a bold italic r bold italic space bold italic m bold italic a bold italic s bold italic s bold italic space bold italic o bold italic f bold italic space bold italic K bold italic I end fraction$

$rightwards double arrow n o. space o f space m o l e s space K I space equals space fraction numerator 1.39 space g over denominator 166.0028 space g. m o l to the power of negative 1 end exponent space end fraction rightwards double arrow 8.373 cross times 10 to the power of negative 3 end exponent space m o l e s$

Step by step

Solved in 3 steps with 6 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions