/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Calculate the molar solubility o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 30

Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-11}\).

Short Answer

Expert verified
The molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) is approximately \(1.63 \times 10^{-4} \,M\).

Step by step solution

01

Write the balanced chemical equation for the dissolution of Cd(OH)2

The balanced chemical equation for the dissolution of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be written as: \[ \mathrm{Cd}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Cd}^{2+} + 2\mathrm{OH}^-\] As the solid dissolves in water, it creates Cd2+ and OH- ions.
02

Define the solubility product expression and molar solubility

For the dissolution of \(\mathrm{Cd}(\mathrm{OH})_{2}\), let the molar solubility be represented by the variable s. Then, the concentration of the ions in the saturated solution can be expressed as follows: \[\mathrm{[Cd}^{2+}] = s\] \[\mathrm{[OH}^\mathrm{-}] = 2s\] The solubility product expression is given by the product of the concentrations of the dissolved ions: \[K_{\mathrm{sp}} = \mathrm{[Cd}^{2+}] \times \mathrm{[OH}^\mathrm{-}]^{2}\]
03

Substitute the solubility product constant and molar solubility into the equation

Substitute the given values of \(K_{\mathrm{sp}}\) and the expressions for the ion concentrations in terms of molar solubility into the equation from Step 2: \[5.9 \times 10^{-11} = (s) \times (2s)^2\]
04

Solve the equation for molar solubility (s)

First, simplify the equation obtained in Step 3: \[5.9 \times 10^{-11} = 4s^3\] Now, divide by 4: \[s^3 = \frac{5.9 \times 10^{-11}}{4}\] Next, take the cube root of both sides to get: \[s = \sqrt[3]{\frac{5.9 \times 10^{-11}}{4}}\] Calculate the value of s: \[s \approx 1.63 \times 10^{-4}\] So the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) is approximately \(1.63 \times 10^{-4} \,M\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Explain the following phenomenon: You have a test tube with about \(20 \mathrm{~mL}\) silver nitrate solution. Upon adding a few drops of sodium chromate solution, you notice a red solid forming in a relatively clear solution. Upon adding a few drops of a sodium chloride solution to the same test tube, you notice \(a\) white solid and a pale yellow solution. Use the \(K_{s p}\) values in the book to support your explanation, and include the balanced equations.

A solution is formed by mixing \(50.0 \mathrm{~mL}\) of \(10.0 \mathrm{M} \mathrm{NaX}\) with \(50.0 \mathrm{~mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}(\mathrm{I})\) forms com- plex ions with \(\mathrm{X}^{-}\) as follows: $$\begin{aligned}\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) & K_{1}=1.0 \times 10^{2} \\ \mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) & K_{2}=1.0 \times 10^{4} \\ \mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) & K_{3}=1.0 \times 10^{3} \end{aligned}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}\) c. \(\mathrm{Cu}^{+}\)

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+}\right.\) is the cation in solution.)

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of "nano machines." One common building block of nanostructured architectures is manganese oxide nanoparticles. The particles can be formed from manganese oxalate nanorods, the formation of which can be described as follows: Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}:\) \(K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}{ }^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}\)

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}\). EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\), are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00 .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

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.