A 50.0 mL sample of 0.436 M NH4NO3 is diluted with water to a total volume of 250.0 mL. What is the nitrate ion concentration in the resulting solution?
A 50.0 mL sample of 0.436 M NH4NO3 is diluted with water to a total volume of 250.0 mL. What is the nitrate ion concentration in the resulting solution?
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...
Related questions
Question
![**Problem Statement**
A 50.0 mL sample of 0.436 M NH₄NO₃ is diluted with water to a total volume of 250.0 mL. What is the nitrate ion concentration in the resulting solution?
**Solution**
To find the nitrate ion concentration in the resulting diluted solution, you need to use the concept of dilution.
First, calculate the moles of NH₄NO₃ in the 50.0 mL sample:
\[ \text{Moles of NH₄NO₃} = \text{Concentration (M)} \times \text{Volume (L)} \]
\[ \text{Moles of NH₄NO₃} = 0.436 \, \text{M} \times \frac{50.0 \, \text{mL}}{1000 \, \text{mL/L}} \]
\[ \text{Moles of NH₄NO₃} = 0.0218 \, \text{moles} \]
Since NH₄NO₃ dissociates completely in water:
\[ \text{NH₄NO₃} \rightarrow \text{NH₄}^+ + \text{NO₃}^- \]
Each mole of NH₄NO₃ produces one mole of NO₃⁻ (nitrate ion).
Therefore, there are 0.0218 moles of NO₃⁻ in the diluted solution.
Next, calculate the concentration of NO₃⁻ in the total volume of the diluted solution (250.0 mL):
\[ \text{Concentration (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} \]
\[ \text{Concentration (M)} = \frac{0.0218 \, \text{moles}}{0.250 \, \text{L}} \]
\[ \text{Concentration (M)} = 0.0872 \, \text{M} \]
Thus, the nitrate ion concentration in the resulting solution is 0.0872 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd321eb40-4702-45a2-842a-571d0c67a0bc%2Ff7151316-9832-43d2-bce7-b7fbea37b5ae%2Flfjc0fi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
A 50.0 mL sample of 0.436 M NH₄NO₃ is diluted with water to a total volume of 250.0 mL. What is the nitrate ion concentration in the resulting solution?
**Solution**
To find the nitrate ion concentration in the resulting diluted solution, you need to use the concept of dilution.
First, calculate the moles of NH₄NO₃ in the 50.0 mL sample:
\[ \text{Moles of NH₄NO₃} = \text{Concentration (M)} \times \text{Volume (L)} \]
\[ \text{Moles of NH₄NO₃} = 0.436 \, \text{M} \times \frac{50.0 \, \text{mL}}{1000 \, \text{mL/L}} \]
\[ \text{Moles of NH₄NO₃} = 0.0218 \, \text{moles} \]
Since NH₄NO₃ dissociates completely in water:
\[ \text{NH₄NO₃} \rightarrow \text{NH₄}^+ + \text{NO₃}^- \]
Each mole of NH₄NO₃ produces one mole of NO₃⁻ (nitrate ion).
Therefore, there are 0.0218 moles of NO₃⁻ in the diluted solution.
Next, calculate the concentration of NO₃⁻ in the total volume of the diluted solution (250.0 mL):
\[ \text{Concentration (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} \]
\[ \text{Concentration (M)} = \frac{0.0218 \, \text{moles}}{0.250 \, \text{L}} \]
\[ \text{Concentration (M)} = 0.0872 \, \text{M} \]
Thus, the nitrate ion concentration in the resulting solution is 0.0872 M.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
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.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY