Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 47

What is the concentration of fluoride ion in a water solution saturated with \(\mathrm{BaF}_{2}, K_{s \mathrm{p}}=1.8 \times 10^{-7}\) ?

Short Answer

Expert verified
Answer: The concentration of fluoride ions in the saturated water solution is approximately \(5.88 \times 10^{-3}\,\text{M}\).

Step by step solution

01

Write the dissolution equilibrium equation

Write the balanced chemical equation for the dissolution of BaF2 in water: \[\text{BaF}_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{F}^-(aq)\]
02

Write the solubility product expression

Write the expression for Ksp for BaF2: \[K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2\]
03

Define variables for ion concentrations

Let x be the concentration of Ba2+ ions at equilibrium. Then, the concentration of F- ions at equilibrium will be 2x because there are two F- ions for each Ba ion: \[[\text{Ba}^{2+}] = x\] \[[\text{F}^-] = 2x\]
04

Substitute concentrations into the Ksp expression and solve for x

Insert the expressions for the ion concentrations into the Ksp expression, and then solve for x: \[K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 = (x)(2x)^2\] \[K_\text{sp} = 4x^3\] \[x = \sqrt[3]{\frac{K_\text{sp}}{4}}\]
05

Calculate the value of x

Substitute the given value for Ksp into the equation and solve for x: \[x = \sqrt[3]{\frac{1.8 \times 10^{-7}}{4}}\] \[x ≈ 2.94 \times 10^{-3}\,\text{M}\]
06

Calculate the concentration of fluoride ions

Now that we have the concentration of Ba2+ ions at equilibrium (x), we can find the concentration of F- ions by multiplying x by 2: \[[\text{F}^-] = 2x = 2 \times 2.94 \times 10^{-3}\,\text{M} \approx 5.88 \times 10^{-3}\,\text{M}\] In conclusion, the concentration of fluoride ions in the saturated water solution is approximately \(5.88 \times 10^{-3}\,\text{M}\).

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

State the oxidation number of \(\mathrm{N}\) in (a) \(\mathrm{NO}_{2}^{-}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{HNO}_{3}\) (d) \(\mathrm{NH}_{4}{ }^{+}\)

The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine; \(\mathrm{ClO}^{-}\) is reduced to \(\mathrm{Cl}^{-}\). The amount of iodine produced by the redox reaction is determined by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ; \mathrm{I}_{2}\) is reduced to \(\mathrm{I}^{-}\). The sodium thiosulfate is oxidized to sodium tetrathionate, \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\). In this analysis, potassium iodide was added in excess to \(5.00 \mathrm{~mL}\) of bleach \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\). If \(25.00 \mathrm{~mL}\) of \(0.0700 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was required to reduce all the iodine produced by the bleach back to iodide, what is the mass percent of \(\mathrm{NaClO}\) in the bleach?

Give the Lewis structure of (a) \(\mathrm{NO}_{2}\) (b) \(\mathrm{NO}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{SO}_{3}\)

Name the following compounds. (a) \(\mathrm{HBrO}_{3}\) (b) KIO (c) \(\mathrm{NaClO}_{2}\) (d) \(\mathrm{NaBrO}_{4}\)

37\. Iodine can be prepared by allowing an aqueous solution of hydrogen iodide to react with manganese dioxide, \(\mathrm{MnO}_{2}\). The reaction is $$2 \mathrm{I}^{-}(a q)+4 \mathrm{H}^{+}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s)$$ If an excess of hydrogen iodide is added to \(0.200 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\), how many grams of iodine are obtained, assuming \(100 \%\) yield?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

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