/*! 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} Q13 E Assuming that no equilibria othe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Q13 E (page 872)

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

\(\begin{array}{l}(a)KH{C_4}{H_4}{O_6}\\(b)Pb{I_2}\\(c)A{g_4}\left[ {Fe{{(CN)}_6}} \right],a salt containing the Fe{(CN)_4}^ - ion\\(d)H{g_2}{I_2}\end{array}\)

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.

a)\({\rm{x}}\;{\rm{ = }}\;{\rm{2 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\)

b) \({\rm{x}}\;{\rm{ = }}\;{\rm{1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\)

c)\({\rm{x}}\;{\rm{ = }}\;{\rm{2}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{M}}\)

d)\({\rm{x}}\;{\rm{ = }}\;{\rm{2}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 10}}}}{\rm{M}}\)

Step by step solution

01

Step 1: To calculate the molar solubility of each of the following from its solubility product:

Solubility product of the salt \({\rm{KH}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{\rm{4}}}{{\rm{O}}_{\rm{6}}}{\rm{ = }}\;{\rm{6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\) from Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{KH}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{\rm{4}}}{{\rm{O}}_{\rm{6}}}} \right){\rm{ = 6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{{\rm{K}}^{\rm{ + }}}} \right]\left[ {{\rm{H}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{\rm{4}}}{\rm{O}}_{\rm{6}}^{\rm{ - }}} \right]\\{\rm{ = }}\;{\rm{x}} \times \;{\rm{x}}\;{\rm{ = }}\;{{\rm{x}}^{\rm{2}}}\\{\rm{x}}\;{\rm{ = }}\sqrt {{{\rm{K}}_{{\rm{sp}}}}} {\rm{ = }}\sqrt {{\rm{6}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}} \\\;{\rm{x}}\;{\rm{ = }}\;{\rm{2 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{M}}\end{array}\)

02

Step 2: the following salts is the most soluble, in terms of moles per liter, in pure water:  

Solubility product of\({\rm{Pb}}{{\rm{I}}_{\rm{2}}}\;{\rm{ = }}\;1.4\; \times \;{10^{ - 8}}\)from Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{Pb}}{{\rm{I}}_{\rm{2}}}} \right){\rm{ = 1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{P}}{{\rm{b}}^{{\rm{2 + }}}}} \right]{\left[ {{{\rm{I}}^{\rm{ - }}}} \right]^{\rm{2}}}\\{\rm{ = }}\;{\rm{x}}\; \times {\rm{4}}{{\rm{x}}^{\rm{2}}}\;{\rm{ = }}\;{\rm{4}}{{\rm{x}}^{\rm{3}}}\\{\rm{ = }}\,\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div 4}}}}\\{\rm{ = }}\,\sqrt[{\rm{3}}]{{{\rm{1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{ \div 4}}}}\\\;{\rm{x}}\;{\rm{ = }}\;{\rm{1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{M}}\end{array}\)

03

the following  salts is the most soluble, in terms of moles per liter, in pure water:  

Solubility product of \({\rm{A}}{{\rm{g}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]\; = \;{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 41}}}}\)from Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{A}}{{\rm{g}}_{\rm{4}}}\left[ {{\rm{Fe(CN}}{{\rm{)}}_{\rm{6}}}} \right]} \right){\rm{ = 1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 41}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}{\left[ {{\rm{A}}{{\rm{g}}^{\rm{ + }}}} \right]^{\rm{4}}}\left[ {{\rm{Fe(CN)}}_{\rm{4}}^{\rm{ - }}} \right]\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{{\rm{(4x)}}^{\rm{4}}}{\rm{ \times x}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\;{\rm{256}}\;{{\rm{x}}^{\rm{5}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{5}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div 256}}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{5}}]{{{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 41}}}}{\rm{ \div 256}}}}\quad \\{\rm{x}}\;{\rm{ = }}\;{\rm{2}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{M}}\end{array}\)

04

 Step 4: the following salts is the most soluble, in terms of moles per liter, in pure water:  

Solubility product of \({{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{H}}{{\rm{g}}_{\rm{2}}}{{\rm{I}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{4}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 29}}}}\)from Ksp table

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}}\left( {{\rm{H}}{{\rm{g}}_{\rm{2}}}{{\rm{I}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{4}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 29}}}}\\{{\rm{K}}_{{\rm{sp}}}}{\rm{ = }}\left[ {{\rm{H}}{{\rm{g}}^{{\rm{2 + }}}}} \right]{\left[ {{{\rm{I}}^{\rm{ - }}}} \right]^{\rm{2}}}\\{\rm{ = }}\;{\rm{x}}\; \times {\rm{4}}{{\rm{x}}^{\rm{2}}}{\rm{ = }}\;{\rm{4}}{{\rm{x}}^{\rm{3}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{3}}]{{{{\rm{K}}_{{\rm{sp}}}}{\rm{ \div 4}}}}\\{\rm{x}}\;{\rm{ = }}\;\sqrt[{\rm{3}}]{{{\rm{4}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{ - 29}}}}{\rm{ \div 4}}}}\\{\rm{x}}\;{\rm{ = }}\;{\rm{2}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 10}}}}{\rm{M}}\end{array}\)

Thus, the solubility product of the salts are sorted out.

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

To a \({\bf{0}}.{\bf{10M}}\) solution of \({\bf{Pb}}{\left( {{\bf{N}}{{\bf{O}}_{\bf{3}}}} \right)_{\bf{2}}}\) is added enough \({\bf{HF}}\left( {\bf{g}} \right)\) to make\(\left[ {{\bf{HF}}} \right]{\rm{ }} = {\rm{ }}{\bf{0}}.{\bf{10}}{\rm{ }}{\bf{M}}\).

(a) Does \({\bf{Pb}}{{\bf{F}}_{\bf{2}}}\) precipitate from this solution? Show the calculations that support your conclusion.

(b) What is the minimum pH at which \({\bf{Pb}}{{\bf{F}}_{\bf{2}}}\) precipitates?

Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):

(a) \(AgI\)

(b) \(A{g_2}S{O_4}\)

(c) \(Mn{(\;OH\;)_2}\)

(d) \(Sr{(\;OH\;)_2} \times 8{H_2}O\)

(e) The mineral brucite, \(Mg{(\;OH\;)_2}\)

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 Fe3+ equilibrium concentration when 0.0888 mole of K3[Fe (CN)6] is added to a solution with 0.0.00010 M CN鈥�.

Question: 28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate \({K_{sp}}\)for each of the slightly soluble solids indicated:

(a) \(AgBr:\left( {A{g^ + }} \right) = 5.7 \times 1{0^{ - 7}}M,\left( {B{r^ - }} \right) = 5.7 \times 1{0^{ - 7}}M\)

(b) \(CaC{O_3}:\left( {C{a^{2 + }}} \right) = 5.3 \times 1{0^{ - 3}}M,\left( {C{O_3}^{2 - }} \right) = 9.0 \times 1{0^{ - 7}}M\)

(c) \(Pb{F_2}:\left( {P{b^{2 + }}} \right) = 2.1 \times 1{0^{ - 3}}M,\left( {{F^ - }} \right) = 4.2 \times 1{0^{ - 3}}M\)

(d) \(A{g_2}Cr{O_4}:\left( {A{g^ + }} \right) = 5.3 \times 1{0^{ - 5}}M,3.2 \times 1{0^{ - 3}}M\)

(e) \(In{F_3}:\left( {I{n^{3 + }}} \right) = 2.3 \times 1{0^{ - 3}}M,\left( {{F^ - }} \right) = 7.0 \times 1{0^{ - 3}}M\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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.