Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 54

A solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{BaCl}_{2}\) and \(125 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\). What are the concentrations of barium and sulfate ions in this solution? Assume only \(\mathrm{SO}_{4}^{2-}\) ions (no \(\left.\mathrm{HSO}_{4}^{-}\right)\) are present.

Short Answer

Expert verified
The final concentrations of barium and sulfate ions in the solution are \(0.0075\,M\) and \(0.025\,M\), respectively.

Step by step solution

01

Find the moles of each ion in the initial solutions

First, we need to calculate the moles of Ba²⁺ and SO₄²⁻ in the initial solutions. For the BaCl₂ solution: \[moles\,of\,Ba^{2+} = volume \times concentration = 75.0\,mL \times 0.020\,M = 1.5\,mmol\] For the K₂SO₄ solution: \[moles\,of\,SO_{4}^{2-} = volume \times concentration = 125\,mL \times 0.040\,M = 5\,mmol\]
02

Calculate the total volume of the final solution

Next, we'll determine the total volume of the final solution, which is the sum of the volumes of the two initial solutions: \[total\,volume\,of\,final\,solution = 75.0\,mL + 125\,mL = 200\,mL\]
03

Divide the moles of each ion by the total volume to find their concentrations

Finally, we will divide the moles of each ion by the total volume to find their concentrations in the final solution: For Ba²⁺ ion: \[concentration\,of\,Ba^{2+} = \frac{moles\,of\,Ba^{2+}}{total\,volume} = \frac{1.5\,mmol}{200\,mL} = 0.0075\,M\] For SO₄²⁻ ion: \[concentration\,of\,SO_{4}^{2-} = \frac{moles\,of\,SO_{4}^{2-}}{total\,volume} = \frac{5\,mmol}{200\,mL} = 0.025\,M\] Therefore, the final concentrations of barium and sulfate ions in the solution are 0.0075 M and 0.025 M, respectively.

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

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{CoF}_{6}^{3-}\) b. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

What mass of \(\operatorname{ZnS}\left(K_{\text {sp }}=2.5 \times 10^{-22}\right.\) ) will dissolve in \(300.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\).

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\)

A \(50.0-\mathrm{mL}\) sample of \(0.00200 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{NaIO}_{3} .\) What is the equilibrium concentration of \(\mathrm{Ag}^{+}\) in solution? \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgIO}_{3}\) is \(\left.3.0 \times 10^{-8} .\right)\)

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{CaF}_{2}(s), K_{\mathrm{sp}}=4.0 \times 10^{-11}\), or \(\mathrm{BaF}_{2}(s), K_{\mathrm{sp}}=2.4 \times 10^{-5}\) b. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s), K_{\mathrm{sp}}=1.3 \times 10^{-32}\), or \(\mathrm{FePO}_{4}(s)\) \(K_{\text {? }}=1.0 \times 10^{-22}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

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