Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 47

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 \(\mathrm{S}^{2-}\).

Short Answer

Expert verified
Approximately \(1.46 \times 10^{-8}\) g of ZnS will dissolve in 300.0 mL of 0.050 M Zn(NO₃)₂ solution.

Step by step solution

01

Write the balanced dissolution equation for ZnS

\(\mathrm{ZnS_{(s)}} \leftrightarrows \mathrm{Zn^{2+}_{(aq)}} + \mathrm{S^{2-}_{(aq)}}\)
02

Find equilibrium concentration of S²⁻ ions using Ksp

The Ksp expression is given by: \[K_{sp} = [\mathrm{Zn^{2+}}][\mathrm{S^{2-}}]\] Plugging in the given value for Ksp and the concentrations at equilibrium, we get: \[2.5 \times 10^{-22} = (x)(x)\] \[x^2 = 2.5 \times 10^{-22}\] \[x = \sqrt{2.5 \times 10^{-22}} = 5 \times 10^{-12} \mathrm{M}\] So, the equilibrium concentration of S²⁻ ions is \(5 \times 10^{-12}\) M.
03

Calculate the total concentration of Zn²⁺ ions

The total concentration of Zn²⁺ ions comes from the dissolution of both \(\mathrm{ZnS}\) and \(\mathrm{Zn(NO_3)_2}\). The concentration of Zn²⁺ ions from the dissolution of \(\mathrm{ZnS}\) is equal to the concentration of S²⁻ ions. Therefore, the total concentration of Zn²⁺ ions is: \[\mathrm{Total\, Zn^{2+}\, concentration} = [\mathrm{Zn^{2+}}]_{\mathrm{from\, ZnS}} + [\mathrm{Zn^{2+}}]_{\mathrm{from\, Zn(NO_3)_2}\] \[= 5 \times 10^{-12} \mathrm{M} + 0.050 \mathrm{M}\]
04

Determine the concentration of dissolved ZnS

The concentrations of Zn²⁺ ions and S²⁻ ions at equilibrium are the same. The concentration of dissolved ZnS is equal to the concentration of S²⁻ ions, which we calculated earlier: \[[\mathrm{ZnS}]= 5 \times 10^{-12} \mathrm{M}\]
05

Convert the concentration of ZnS to mass

Now we will find the mass of ZnS that will dissolve in 300.0 mL of solution: Molar mass of ZnS = 65.38 g/mol (Zn) + 32.07 g/mol (S) = 97.45 g/mol Mass of ZnS = concentration x volume x molar mass \[m = (5 \times 10^{-12}\, \mathrm{M})(300 \times 10^{-3}\, \mathrm{L})(97.45\, \mathrm{g/mol})\] \[m = 1.46 \times 10^{-8}\, \mathrm{g}\] So, approximately \(1.46 \times 10^{-8}\) g of ZnS will dissolve in 300.0 mL of 0.050 M Zn(NO₃)₂ solution.

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 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? \(\left(K_{6}\right.\) for \(\mathrm{CaF}_{2}=4.0 \times 10^{-11}\) )

The concentration of \(\mathrm{Mg}^{2+}\) in seawater is \(0.052 \mathrm{M}\). At what \(\mathrm{pH}\) will \(99 \%\) of the \(\mathrm{Mg}^{2+}\) be precipitated as the hydroxide salt? \(\left[K_{\text {sp }}\right.\) for \(\left.\mathrm{Mg}(\mathrm{OH})_{2}=8.9 \times 10^{-12} .\right]\)

Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{~S}\), and \(\mathrm{Na}_{3} \mathrm{PO}_{4}\). What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\text {spp }}\) values are \(K_{\text {sp }}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29}\), and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-34}\)

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 \(8.0 \mathrm{M} \mathrm{NaF}\). $$ \begin{aligned} \mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}^{+}(a q) & & K_{1}=7.9 \times 10^{4} \\ \mathrm{BeF}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{2}(a q) & & K_{2}=5.8 \times 10^{3} \\ \mathrm{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{3}-(a q) & & K_{3}=6.1 \times 10^{2} \\ \mathrm{BeF}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{4}{ }^{2-}(a q) & & K_{4}=2.7 \times 10^{1} \end{aligned} $$ 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.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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