Log out Go to App
Go to App

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.

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

Calculate the solubility of \(\mathrm{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\) in a buffered solution with a pH of \(11.00\).

Calculate the molar solubility of \(\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}\).

Silver chloride dissolves readily in \(2 \mathrm{M} \mathrm{NH}_{3}\), but is quite insoluble in \(2 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3}\). Explain.

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

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.)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

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