/*! 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} Q21 E Refer to Appendix \(J\) for solu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Q21 E (page 872)

Refer to Appendix \(J\) for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.

Short Answer

Expert verified

The solubility product constant represented as Ksp can be defined as state in which a solid and its respective ions in given solution are in equilibrium. Its value indicates the extent to which a compound can undergo dissociation in water.

Step by step solution

01

To determine the x value for \(Ca{(\;OH)_2}\)

\(\begin{array}{l}{\rm{Ca}}{({\rm{OH}})_2} \to {\rm{C}}{{\rm{a}}^{2 + }} + 2O{{\rm{H}}^ - }\\{{\rm{K}}_{{\rm{sp}}}} = {\rm{x}}{(2{\rm{x}})^2} = 4{{\rm{x}}^3} = 7.9 \times {10^{ - 6}}\\{\rm{x}} = \sqrt[3]{{\frac{{7.9 \times {{10}^{ - 6}}}}{4}}} = 0.013\;{\rm{M}}\end{array}\)

02

To determine the x value for \(CaC{O_3}\)

\(\begin{array}{l}{\rm{CaC}}{{\rm{O}}_3} \to {\rm{C}}{{\rm{a}}^{2 + }} + {\rm{CO}}_3^{2 - }\\{{\rm{K}}_{s{\rm{p}}}} = {{\rm{x}}^2} = 4.8 \times {10^{ - 9}}\\{\rm{x}} = 6.9 \times {10^{ - 5}}\;{\rm{M}}\end{array}\)

03

To determine the x value for \(CaS{O_4} \times {H_2}O\)

\(\begin{array}{l}{\rm{CaS}}{{\rm{O}}_4} \cdot {{\rm{H}}_2}{\rm{O}} \to {\rm{C}}{{\rm{a}}^{2 + }} + {\rm{SO}}_4^{(2 - )}.2{{\rm{H}}_{\rm{2}}}{\rm{O}}\\{{\rm{K}}_{{\rm{sp}}}} = {\rm{xx}}{(2x)^2} = 4{{\rm{x}}^4} = 2.4 \times {10^{ - 5}}\\{\rm{x}} = \sqrt[4]{{\frac{{2.4 \times {{10}^{ - 5}}}}{4}}} = 0.049\;{\rm{M}}\end{array}\)\(\)

04

To determine the x value for \(Ca{C_2}{O_4} \times {H_2}O\)

\(\begin{array}{l}{\rm{Ca}}{{\rm{C}}_2}{{\rm{O}}_4} \cdot {{\rm{H}}_2}{\rm{O}} \to {\rm{C}}{{\rm{a}}^{2 + }} + {{\rm{C}}_2}{\rm{O}}_4^{2 - }.{{\rm{H}}_2}{\rm{O}}\\{{\rm{K}}_{{\rm{sp}}}} = {\rm{xxx}} = {{\rm{x}}^3} = 2.27 \times {10^{ - 9}}\\{\rm{x}} = \sqrt[3]{{2.27 \times {{10}^{ - 9}}}} = 1.3 \times {10^{ - 3}}\;{{\rm{M}}_{_{}}}\end{array}\)

05

To determine the x value for \(C{a_3}{\left( {P{O_4}} \right)_2}\)

\(\begin{array}{l}{\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} \to 3{\rm{C}}{{\rm{a}}^{2 + }} + 2{\rm{PO}}_4^{(3 - )}\\{K_{{\rm{sp}}}} = {{\rm{x}}^3}{\left( {\frac{2}{3}{\rm{x}}} \right)^2} = 1 \times {10^{ - 25}}\\{\rm{x}} = \sqrt[5]{{\frac{{1 \times {{10}^{ - 25}} \times 9}}{4}}} = 1 \times {10^{ - 5}}{\rm{M}}\end{array}\)

In this case, solubility equals to one third of\(\left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\), that is\(4 \times {10^{ - 6}}\).

06

To determine the x value for \(CaHP{O_4}\)

\(\begin{array}{l}{\rm{CaHP}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{a}}^{2 + }} + {\rm{HPO}}_4^{(2 - )}\\{{\rm{K}}_{{\rm{sp}}}} = {{\rm{x}}^2} = 7 \times {10^{ - 7}}\\{\rm{x}} = 8.4 \times {10^{ - 4}}\;{\rm{M}}\end{array}\)

07

To determine the x value for \(Ca{F_2}\)

\(\begin{array}{l}{\rm{Ca}}{{\rm{F}}_2} \to {\rm{C}}{{\rm{a}}^{2 + }} + {{\rm{F}}^{(2 - )}}\\{{\rm{K}}_{{\rm{sp}}}} = {{\rm{x}}^2} = 4 \times {10^{ - 11}}\\{\rm{x}} = 6.3 \times {10^{ - 6}}\;{\rm{M}}\end{array}\)

Finally we get,

a) \(0.013\;\;{\rm{M}}\)

b) \(6.9 \times {10^{ - 5}}\;{\rm{M}}\)

c) \(0.049\;\;{\rm{M}}\)

d) \(1.3 \times {10^{ - 3}}\;{\rm{M}}\)

e) \(4 \times {10^{ - 6}}\;{\rm{M}}\)

f) \(8.4 \times {10^{ - 4}}\;{\rm{M}}\)

g) \(6.3 \times {10^{ - 6}}\;{\rm{M}}\)

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

Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)

Iron concentrations greater than 5.4脳10-6M in water used for laundrypurposes can cause staining. What [OH-] is required to reduce [Fe2+] to this level by precipitation of Fe(OH)2?

The solubility product of \(CaS{O_4}.2{H_2}O\) is\(2.4 \times 1{0^{ - 5}}\). What mass of this salt will dissolve in \(1.0\;L\) of \(0.010M\) \(SO{4^{2 - }}\)?

Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is theof a saturated solution of Ca(OH)2?

Question: Calculate the equilibrium concentration of Zn2+ in a 3.0 M solution of\(Zn(CN)_4^{2 - }\) .
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

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.