Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 84

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 $\mathrm{mg} \mathrm{F}^{-}$ per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? $\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)$

Short Answer

Expert verified
The maximum molarity of calcium ions in hard water, with the fluoride concentration at the USPHS recommended level, is \(1.45 \times 10^{-2}\) M.

Step by step solution

01

Write the balanced chemical equation for the dissolution of CaF₂

First, we need to write the balanced chemical equation for the dissolution of calcium fluoride (CaF₂) in water: \[ \mathrm{CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2 F^{-} (aq)}\]
02

Write the expression for Ksp

Next, we need to write the expression for the solubility product constant (Ksp) of CaF₂. From the balanced chemical equation, we can write the Ksp expression as: \[ K_\mathrm{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2\] We are given the value of Ksp for CaF₂: \[ K_\mathrm{sp} = 4.0\times10^{-11}\]
03

Use the recommended fluoride concentration

We know that the recommended concentration of fluoride ions is 1 mg F⁻ per liter. First, we need to convert this concentration into moles per liter (molarity). The molar mass of F⁻ is 19 g/mol. Since 1 mg = 0.001 g, we can write the molarity of fluoride ions as: \[ [\mathrm{F^{-}}] = \frac{0.001\ \text{g}}{19\ \text{g/mol}} = 5.26\times10^{-5}\ \text{M}\]
04

Find the maximum molarity of calcium ions

Now, substitute the given Ksp and the molarity of fluoride ions into the Ksp expression to find the maximum molarity of calcium ions: \[ K_\mathrm{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2\] \[ 4.0\times10^{-11} = [\mathrm{Ca^{2+}}](5.26\times10^{-5})^2\] Now, solve for the molarity of calcium ions: \[ [\mathrm{Ca^{2+}}] = \frac{4.0\times10^{-11}}{(5.26\times10^{-5})^2} = 1.45\times10^{-2}\ \text{M}\] So, the maximum molarity of calcium ions in hard water, with the fluoride concentration at the USPHS recommended level, is 1.45 × 10⁻² 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 solution is prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of $8.0 M\( \)\mathrm{NaF}.$ $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\operatorname{Be} \mathrm{F}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\operatorname{Be} \mathrm{F}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \qquad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of $\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+},\( \)\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},$ and \(\mathrm{BeF}_{4}^{2-}\) in this solution.

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is $1.6 \times 10^{-7} \mathrm{mol} / \mathrm{L}\( at \)25^{\circ} \mathrm{C},$ calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

In the presence of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) forms the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) If the equilibrium concentrations of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) are $1.8 \times 10^{-17} \mathrm{M}\( and \)1.0 \times 10^{-3} \mathrm{M},\( respectively, in a \)1.5-M \mathrm{NH}_{3}$ solution, calculate the value for the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}.\) $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K_{\mathrm{overall}}=?$$

Solutions of sodium thiosulfate are used to dissolve unexposed \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for black-and-white film. What mass of \(\mathrm{AgBr}\) can dissolve in \(1.00 \mathrm{L}\) of \(0.500 M\) $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\( reacts with \)\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ to form a complex ion: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \\ K &=2.9 \times 10^{13} \end{aligned}$$

A solution contains $1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\( and \)5.0 \times 10^{-7} M$ \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.$
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Organic 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