Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 118

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the F- ion concentration is adjusted to about 1 ppm. Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is 8 ppm. Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Short Answer

Expert verified
A precipitate of \(\mathrm{CaF}_{2}\) will not form under the given conditions, as the ion product (Q) is less than the solubility product constant (Ksp).

Step by step solution

01

Convert concentration from ppm to mol/L

First, we need to convert the given concentrations from ppm (parts per million) to mol/L (moles per liter). To do this conversion, we can use the following relation for each ion: given concentration (ppm) × (1 mol / atomic or molecular weight (g)) × (1 g / 1,000,000 mL) × (1,000 L / 1,000,000 mL) For F⁻ ion concentration (considering atomic weight of F = 19 g/mol): 1 ppm × (1 mol/ 19 g) × (1 g / 1,000,000 mL) × (1,000 L / 1,000,000 mL)= \( \frac{1}{19,000} \) mol/L For Ca²⁺ ion concentration (considering atomic weight of Ca = 40 g/mol): 8 ppm × (1 mol / 40 g) × (1 g / 1,000,000 mL) × (1,000 L / 1,000,000 mL) = \( \frac{1}{5,000} \) mol/L
02

Calculate ion product (Q) under given conditions

The ion product (Q) for \(\mathrm{CaF}_{2}\) can be calculated by multiplying the concentrations of the ions in the solution: Q = [Ca²⁺] × [F⁻]² = (\(\frac{1}{5,000}\)) × (\(\frac{1}{19,000}\))²
03

Compare Q with Ksp

Now, we'll compare the ion product (Q) with the solubility product constant (Ksp) for \(\mathrm{CaF}_{2}\), which is \(3.9\times10^{-11}\). If Q > Ksp, a precipitate will form, otherwise, it won't. \( Q = \frac{1}{5,000} \times \left(\frac{1}{19,000}\right)^2 = 2.77 \times 10^{-12} \) As \( 2.77 \times 10^{-12} < 3.9 \times 10^{-11} \), which means Q < Ksp.
04

Conclusion

A precipitate of \(\mathrm{CaF}_{2}\) will not form under the given conditions, as the ion product (Q) is less than the solubility product constant (Ksp).

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

Which of the following solutions is a buffer? (a) A solution made by mixing 100 \(\mathrm{mL}\) of 0.100 \(\mathrm{MCH}_{3} \mathrm{COOH}\) and 50 mL of \(0.100 M \mathrm{NaOH},(\mathbf{b})\) a solution made by mixing 100 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and 500 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{NaOH}\) , (c) A solution made by mixing 100 \(\mathrm{mL}\) of 0.100 \(\mathrm{MCH}_{3} \mathrm{COOH}\) and 50 \(\mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HCl},\) \((\mathbf{d} ) \) A solution made by mixing of 0.100 \(\mathrm{MCH}_{3} \mathrm{COOK}\) and 50 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{KCl}\) .

A sample of 7.5 \(\mathrm{L}\) of \(\mathrm{NH}_{3}\) gas at \(22^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 -L solution of 0.40 \(\mathrm{M}\) HCl. Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14} .\) (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?\) \((\mathbf{b} ) \)The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

Rainwater is acidic because \(\mathrm{CO}_{2}(\mathrm{g})\) dissolves in the water, creating carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}\) . If the rainwater is too acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, CaCO_ \(_{3} ) .\) Calculate the concentrations of carbonic acid, bicarbonate ion \(\left(\mathrm{HCO}_{3}^{-}\right)\) and carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) that are in a raindrop that has a pH of 5.60 , assuming that the sum of all three species in the raindrop is \(1.0 \times 10^{-5} M .\)

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The \(\mathrm{pH}\) is adjusted to about \(1,\) and \(\mathrm{H}_{2} \mathrm{S}\) is bubbled through. Again, no precipitate forms. The \(\mathrm{pH}\) of the solution is then adjusted to about \(8 .\) Again, \(\mathrm{H}_{2} \mathrm{S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4} .\) No precipitate forms. Which of these metal cations are either possibly present or definitely absent: \(\mathrm{Al}^{3+}, \mathrm{Na}^{+}, \mathrm{Ag}^{+}, \mathrm{Mg}^{2+} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

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