/*! 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 33 Calculate the molar solubility o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 33

Calculate the molar solubility of \(\mathrm{Al}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2 \times 10^{-32}\).

Short Answer

Expert verified
The molar solubility of Al(OH)3 is approximately \(6.34 \times 10^{-9} \: M\).

Step by step solution

01

Write the balanced chemical equation for the dissolution of Al(OH)3 in water

We start by writing the balanced chemical equation for the dissolution of Al(OH)3: \[ Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3OH^-(aq) \]
02

Define the equilibrium concentrations of the ions in terms of molar solubility

Let's denote the molar solubility of Al(OH)3 as \(x\). Since one mole of Al(OH)3 produces one mole of Al3+ and three moles of OH-, at equilibrium we have: \[ [Al^{3+}] = x \] \[ [OH^-] = 3x \]
03

Write the expression for the solubility product constant, Ksp

The expression for the solubility product constant is given by the product of the concentrations of the ions raised to their stoichiometric coefficients: \[ K_{sp} = [Al^{3+}] [OH^-]^3 \]
04

Substitute the equilibrium concentrations into the Ksp expression

Now we substitute the expressions for the equilibrium concentrations from Step 2 into the Ksp expression: \[ K_{sp} = (x)(3x)^3 \]
05

Solve for the molar solubility, x

Now we have an equation with only one variable, x. Plug in the value of \( K_{sp} = 2 \times 10^{-32} \), and solve for x: \[ 2 \times 10^{-32} = x (27x^3) \] \[ x = \sqrt[4]{\frac{2 \times 10^{-32}}{27}} \] \[ x = \sqrt[4]{7.41 \times 10^{-34}} \] Evaluating this expression, we get: \[ x \approx 6.34 \times 10^{-9} \: M \]
06

State the molar solubility

The molar solubility of Al(OH)3 is approximately \(6.34 \times 10^{-9} \: M\).

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

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

A solution is prepared by mixing \(50.0 \mathrm{mL}\) of \(0.10\) \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) KCl. Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\mathrm{sp}} \text { for } \mathrm{PbCl}_{2}(s) \text { is } 1.6 \times 10^{-5} .\right]\)

A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) NaX with \(50.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}^{+}\) forms complex ions with \(X^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \quad K_{2}=1.0 \times 10^{4}$$ $$\mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ 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. \(\mathrm{Cu}^{+}\)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr},\) and \(\mathrm{AgI} ?\)

Will a precipitate form when \(100.0 \mathrm{mL}\) of \(4.0 \times 10^{-4} \mathrm{M}\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{mL}\) of \(2.0 \times 10^{-4}\) \(M\) \(\mathrm{NaOH} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

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.