Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 87

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{~L}\) of a \(1.0 \mathrm{M} \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}=4.0 \times\) \(10^{-11}\) and \(K_{\mathrm{a}}\) for \(\mathrm{HF}=7.2 \times 10^{-4}\). Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\).

Short Answer

Expert verified
To begin the precipitation of CaF₂ in 1.0 L of 1.0 M HF solution, we first determine the F⁻ ion concentration in the HF solution, which is approximately 0.0268 M. Then, we find the necessary Ca²⁺ concentration to start precipitation using the given Ksp value, which is approximately \(5.6 \times 10^{-6}\) M. Finally, we calculate the mass of Ca(NO₃)₂ needed to achieve that Ca²⁺ concentration, resulting in approximately \(9.2 \times 10^{-4}\) g of Ca(NO₃)₂.

Step by step solution

01

Write the solubility product expression for CaF₂

The solubility product expression for CaF₂ can be written as the concentration of Ca²⁺ ions multiplied by the square of the concentration of F⁻ ions: \[ K_{sp} = [Ca^{2+}][F^-]^2 \] We are given that \(K_{sp} = 4.0 \times 10^{-11}\).
02

Write the dissociation expression for HF

The equilibrium expression for the dissociation of HF is given by: \[ K_a = \frac{[H^+][F^-]}{[HF]} \] We are given that \(K_a = 7.2 \times 10^{-4}\).
03

Determine the F⁻ concentration from the HF dissociation

From the dissociation of HF, we have the equilibrium concentrations: - For F⁻ => \(x\) - For H⁺ => \(x\), since for every F⁻ ion produced, a corresponding H⁺ ion is also produced - For undissociated HF => \(1.0 - x\), since there's initially 1.0 M of HF We can now write the \(K_a\) expression in terms of x: \[ 7.2 \times 10^{-4} = \frac{x^2}{1-x} \] Since the \(K_a\) is relatively small we can assume \(x \ll 1\), so we can write: \[ 7.2 \times 10^{-4} \approx x^2 \] Solving for x: \[ x = \sqrt{7.2 \times 10^{-4}} \approx 0.0268 \,M \] So, the concentration of F⁻ ions in the solution due to the dissociation of HF is approximately 0.0268 M.
04

Determine the concentration of Ca²⁺ needed to start precipitation

From the solubility product expression, we can determine the concentration of Ca²⁺ that is needed to start precipitation: \[ 4.0 \times 10^{-11} = [Ca^{2+}](0.0268)^2 \] Solving for [Ca²⁺]: \[ [Ca^{2+}] \approx \frac{4.0 \times 10^{-11}}{(0.0268)^2} \approx 5.6 \times 10^{-6} \,M \]
05

Calculate the mass of Ca(NO₃)₂ needed

In 1.0 L of solution, we need the concentration of Ca²⁺ to be 5.6 × 10⁻⁶ M. So the amount of Ca²⁺ (in mol) required is: \[ 5.6 \times 10^{-6}\, mol \times 1.0\, L = 5.6 \times 10^{-6}\, mol \] Since the molar mass of Ca(NO₃)₂ = 164 g/mol (40 g/mol for Ca, 14 g/mol each for the two N, and 48 g/mol for the six O), we can calculate the mass of Ca(NO₃)₂ needed: \[ 5.6 \times 10^{-6}\, mol \times 164\, g/mol \approx 9.2 \times 10^{-4}\, g \] Therefore, approximately 9.2 × 10⁻⁴ g of Ca(NO₃)₂ must be added to the 1.0 L of 1.0 M HF solution to begin precipitation of CaF₂.

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

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10 \mathrm{M} \mathrm{KIO}_{3}\) solution is \(2.6 \times\) \(10^{-11} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\text {sp }}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s) .\)

Calculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M \mathrm{H}^{+} .\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HCN}\) is \(6.2 \times 10^{-10}\).)

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}\) is \(1.7 \times 10^{7} . K_{\text {sp }}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10}\). Calculate the molar solubility of AgCl in \(1.0 \mathrm{M} \mathrm{NH}_{3}\).

Approximately \(0.14 \mathrm{~g}\) nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\), dissolves per liter of water at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) b. \(\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}{ }^{3-}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

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