Homework Help is Here – Start Your Trial Now!
Science
Chemistry
A 45.0-mL sample of a 0.837 M HNO, solution is titrated with a 0.338 M KOH solution. What will be the final volume of solution when the HNO, has been exactly neutralized by the KOH?

A 45.0-mL sample of a 0.837 M HNO, solution is titrated with a 0.338 M KOH solution. What will be the final volume of solution when the HNO, has been exactly neutralized by the KOH?

Chemistry
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
icon
Related questions
icon
Concept explainers
Question
**Titration Calculation Example** **Problem Statement:** A 45.0 mL sample of a 0.837 M HNO₃ solution is titrated with a 0.338 M KOH solution. What will be the final volume of the solution when the HNO₃ has been exactly neutralized by the KOH? **Explanation:** In this titration problem, we have a sample of nitric acid (HNO₃) being neutralized by potassium hydroxide (KOH). The goal is to find out the final volume of the solution once the neutralization is complete. To solve this problem, you will need to use the concepts of molarity and the neutralization reaction between an acid and a base. Here, we can use the neutralization equation and the given molarities to determine the volume of KOH required to neutralize the HNO₃. Let's start by writing the balanced equation for the neutralization reaction: \[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \] 1. **Calculate the moles of HNO₃:** \[ \text{moles of HNO}_3 = \text{Molarity of HNO}_3 \times \text{Volume of HNO}_3 \] \[ \text{moles of HNO}_3 = 0.837 \, \text{M} \times 45.0 \, \text{mL} \] \[ \text{moles of HNO}_3 = 0.837 \, \text{M} \times 0.0450 \, \text{L} \] \[ \text{moles of HNO}_3 = 0.037665 \, \text{moles} \] 2. **Calculate the volume of KOH needed to neutralize the HNO₃:** Since the reaction ratio between HNO₃ and KOH is 1:1 (from the balanced equation), the moles of KOH needed are equal to the moles of HNO₃. \[ \text{moles of KOH} = 0.037665 \, \text{moles} \] Now, using the molarity of the KOH solution, we calculate the volume of KOH required: \[ \
Transcribed Image Text:**Titration Calculation Example** **Problem Statement:** A 45.0 mL sample of a 0.837 M HNO₃ solution is titrated with a 0.338 M KOH solution. What will be the final volume of the solution when the HNO₃ has been exactly neutralized by the KOH? **Explanation:** In this titration problem, we have a sample of nitric acid (HNO₃) being neutralized by potassium hydroxide (KOH). The goal is to find out the final volume of the solution once the neutralization is complete. To solve this problem, you will need to use the concepts of molarity and the neutralization reaction between an acid and a base. Here, we can use the neutralization equation and the given molarities to determine the volume of KOH required to neutralize the HNO₃. Let's start by writing the balanced equation for the neutralization reaction: \[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \] 1. **Calculate the moles of HNO₃:** \[ \text{moles of HNO}_3 = \text{Molarity of HNO}_3 \times \text{Volume of HNO}_3 \] \[ \text{moles of HNO}_3 = 0.837 \, \text{M} \times 45.0 \, \text{mL} \] \[ \text{moles of HNO}_3 = 0.837 \, \text{M} \times 0.0450 \, \text{L} \] \[ \text{moles of HNO}_3 = 0.037665 \, \text{moles} \] 2. **Calculate the volume of KOH needed to neutralize the HNO₃:** Since the reaction ratio between HNO₃ and KOH is 1:1 (from the balanced equation), the moles of KOH needed are equal to the moles of HNO₃. \[ \text{moles of KOH} = 0.037665 \, \text{moles} \] Now, using the molarity of the KOH solution, we calculate the volume of KOH required: \[ \
Expert Solution
steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Ionic Equilibrium
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
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS