Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 77

A \(35.2-\mathrm{mL}, 1.66 M \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

Short Answer

Expert verified
The final concentration of the \( \mathrm{KMnO}_{4} \) solution when 35.2 mL of a 1.66 M solution is mixed with 16.7 mL of a 0.892 M solution is 1.414 M.

Step by step solution

01

Understanding Terms

Molarity (M) is defined as the moles of solute divided by the volume of the solution in liters. Thus, if the molarity and volume of a solution are known, the moles of the solute can be determined using the equation \( moles = Molarity \times Volume \). In this case, we have two \( \mathrm{KMnO}_{4} \) solutions with known molarities and volumes which we can use to find the total moles of \( \mathrm{KMnO}_{4} \).
02

Calculate the Moles of \( \mathrm{KMnO}_{4} \) in Each Solution

The moles of \( \mathrm{KMnO}_{4} \) in the 35.2-mL, 1.66 M solution and in the 16.7 mL, 0.892 M solution can be found with the formula moles = Molarity × Volume. Remember, Volume should be in liters. \n\nFor the 1.66 M solution, \nmoles of \( \mathrm{KMnO}_{4} \) = 1.66 M × 35.2 mL = 1.66 M × 0.0352 L = 0.058432 moles. \n\nFor the 0.892 M solution, \nmoles of \( \mathrm{KMnO}_{4} \) = 0.892 M × 16.7 mL = 0.892 M × 0.0167 L = 0.0149054 moles.
03

Calculate the Total Moles of \( \mathrm{KMnO}_{4} \)

The total moles of \( \mathrm{KMnO}_{4} \) when the two solutions mix can be found by adding the moles from each solution. Thus, \n Total moles of \( \mathrm{KMnO}_{4} \) = moles in 1.66 M solution + moles in 0.892 M solution = 0.058432 moles + 0.0149054 moles = 0.0733374 moles.
04

Calculate the Total Volume

The total volume of the solution when the two are mixed is simply the sum of the individual volumes. Thus, Total volume = Volume of 1.66 M solution + Volume of 0.892 M solution = 35.2 mL + 16.7 mL = 51.9 mL or 0.0519 L.
05

Calculating the Final Concentration

The final concentration of the \( \mathrm{KMnO}_{4} \) solution can now be determined by the equation Molarity = moles/Vol (in Liters). Thus, \n Final Concentration of \( \mathrm{KMnO}_{4} \) = Total moles / Total volume = 0.0733374 moles / 0.0519 L = 1.414 M

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A quantity of \(18.68 \mathrm{~mL}\) of a KOH solution is needed to neutralize \(0.4218 \mathrm{~g}\) of \(\mathrm{KHP}\). What is the concentration (in molarity) of the KOH solution?

How many moles of \(\mathrm{MgCl}_{2}\) are present in \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{MgCl}_{2}\) solution?

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a 0.1022 -g sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

A 22.02-mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64-mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH}\). Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

Explain how you would prepare potassium iodide (KI) by means of (a) an acid- base reaction and (b) a reaction between an acid and a carbonate compound.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free