Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 108

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is $3.0 \times 10^{-16}$. The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

Short Answer

Expert verified
To dissolve 0.015 mol of Zn(OH)₂ in a liter of solution, the concentration of OH⁻ required can be calculated using the given solubility product and formation constant, as well as the equilibrium equations for the dissolution of Zn(OH)₂ and the formation of Zn(OH)₄²⁻ complex. After solving the system of non-linear equations for the amount of Zn(OH)₄²⁻ formed (x), the concentration of OH⁻ required can be determined using the relation: \(OH^{-} = 0.030 - 2x\).

Step by step solution

01

Write the balanced equation for the dissolution of Zn(OH)₂

For the dissolution of Zn(OH)₂, the balanced equation is: \[Zn(OH)_{2} \rightleftharpoons Zn^{2+} + 2OH^{-}\]
02

Write the balanced equation for the formation of Zn(OH)₄²⁻ complex

For the formation of the Zn(OH)₄²⁻ complex, the balanced equation is: \[Zn^{2+} + 4OH^{-} \rightleftharpoons Zn(OH)_{4}^{2-}\]
03

Write the expressions for the solubility product and the formation constant

The solubility product (Ksp) expression for Zn(OH)₂ is: \[K_{sp} = [Zn^{2+}][OH^{-}]^{2}\] And the formation constant (Kf) expression for the Zn(OH)₄²⁻ complex is: \[K_{f} = \frac{[Zn(OH)_{4}^{2-}]}{[Zn^{2+}][OH^{-}]^{4}}\]
04

Use the given data to set the equilibrium equations

The dissolution process will produce 0.015 mol of Zn(OH)₂, 1 mol of which produces 2 mol of OH⁻. The formation of Zn(OH)₄²⁻ will consume some OH⁻ ions. We can use x as the amount of Zn(OH)₄²⁻ formed and set up the relation between the concentrations of the species involved: \[Zn^{2+} = 0.015 - x\] \[OH^{-} = 0.030 - 2x\] \[Zn(OH)_{4}^{2-} = x\]
05

Substitute the given values into the equations and solve for x

Substitute the values of Ksp, Kf, and concentrations into the equilibrium expressions: \(K_{sp} = (0.015 - x)(0.030 - 2x)^{2} = 3.0 \times 10^{-16}\) \(K_{f} = \frac{x}{(0.015 - x)(0.030 - 2x)^{4}} = 4.6 \times 10^{17}\) Now, we can solve these equations simultaneously for x, which is the amount of Zn(OH)₄²⁻ formed. Your best option is to use a graphing calculator or online software to solve this system of non-linear equations to find the value of x.
06

Determine the concentration of OH⁻ required

After solving for x, the concentration of OH⁻ required can be calculated using the relation: \[OH^{-} = 0.030 - 2x\] After substituting the value of x, you will get the concentration of OH⁻ required to dissolve 0.015 mol of Zn(OH)₂ in a liter of 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

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

A buffer contains 0.20 mol of acetic acid and 0.25 mol of sodium acetate in \(2.50 \mathrm{~L}\). (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.05 mol of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of \(0.05 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH},\) and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the Chemistry and Life box on page 790 , fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), whose $K_{s p}=1.0 \times 10^{-60}$ (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid $\mathrm{H}_{2} \mathrm{~A}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)$ or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a 1.0 \(M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) 6.50? (Ignore any volume change.)

A 10.0-mL sample of \(0.250 \mathrm{M}\) acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\( is titrated with \)0.100 \mathrm{M}$ KOH solution. Calculate the pH after the following volumes of base have been added: (a) \(0 \mathrm{~mL},\) (b) \(12.5 \mathrm{~mL}\) (c) \(24.5 \mathrm{~mL}\) (d) \(25.0 \mathrm{~mL}\) (e) \(25.5 \mathrm{~mL}\) (f) \(30.0 \mathrm{~mL}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

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