/*! 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} Q11 E The Handbook of Chemistry and Ph... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Q11 E (page 870)

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

\(\begin{array}{l}(a)BaSe{O_4},0.0118\;g/100\;mL\\(b)Ba{\left( {Br{O_3}} \right)_2} \times {H_2}O,0.30\;g/100\;mL\\(c)N{H_4}MgAs{O_4} \times 6{H_2}O,0.038\;g/100\;mL\\(d)L{a_2}{\left( {Mo{O_4}} \right)_3},0.00179\;g/100\;mL\end{array}\)

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.

Step by step solution

01

Step 1: To calculate the solubility product for each of the following.

(a) \(BaSe{O_4},0.0118\;g/100\;mL\)

First, convert the given value to\(mol/L ,or M.\)

\(\frac{{0.118\;g/L}}{{280.28\;g/mol}} = 4.21 \times 1{0^{ - 4}}M\)

\(Then calculate the solubility product :\)

\(\begin{array}{l}{K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {SeO_4^{2 - }} \right]\\{K_{sp}} = \left( {4.21 \times 1{0^{ - 4}}} \right)\left( {4.21 \times 1{0^{ - 4}}} \right)\\{K_{sp}} = 1.77 \times 1{0^{ - 7}}\end{array}\)

02

Step 2: To calculate the solubility product for each of the following.

(b) \(Ba{\left( {Br{O_3}} \right)_2} \times {H_2}O,0.30\;g/100\;mL\)

First, convert the given value to\(mol/L ,or M.\)

\(\frac{{3\;g/L}}{{411.147\;g/mol}} = 7.3 \times 1{0^{ - 3}}M\)

Then calculate the solubility product:

\(\begin{array}{l}{K_{sp}} = \left[ {B{a^{2 + }}} \right]{\left[ {BrO_3^ - } \right]^2}\\{K_{sp}} = \left( {7.3 \times 1{0^{ - 3}}} \right){\left( {2 \times 7.3 \times 1{0^{ - 3}}} \right)^2}\\{K_{sp}}\; = 1.6 \times 1{0^{ - 6}}\end{array}\)

03

 Step 3: To calculate the solubility product for each of the following .

(c) \(N{H_4}MgAs{O_4} \times 6{H_2}O,0.038\;g/100\;mL\)

First, convert the given value to\(mol/L ,or M.\)

\(\frac{{0.38g/L}}{{289.3544g/mol}} = 1.3 \times 1{0^{ - 3}}M\)

Then calculate the solubility product:

\(\begin{array}{l}{K_{sp}} = \left[ {NH_4^ + } \right]\left[ {M{g^{2 + }}} \right]\left[ {AsO_4^{3 - }} \right]\\{K_{sp}}\; = \;\left[ {\rm{x}} \right]\left[ {\rm{x}} \right]\left[ {\rm{x}} \right]\\{\rm{x}}\; = \;1.3 \times 1{0^{ - 3}}\\{K_{sp}} = {\left( {1.3 \times 1{0^{ - 3}}} \right)^3}\\{K_{sp}} = 2.2 \times 1{0^{ - 9}}\end{array}\)

04

Step 4: To calculate the solubility product for each of the following .

(d) \(L{a_2}{\left( {Mo{O_4}} \right)_3},0.00179\;g/100\;mL\)

First, convert the given value to\(mol/L ,or M.\)

\(\frac{{0.0179\;g/L}}{{757.62\;g/mol}} = 2.36 \times 1{0^{ - 5}}M\)

Then calculate the solubility product:

\(\begin{array}{l}{K_{sp}} = {\left[ {2 \times L{a^{3 + }}} \right]^2}{\left[ {3 \times MoO_4^{2 - }} \right]^3}\\{K_{sp}} = {\left( {2 \times 2.36 \times 1{0^{ - 5}}} \right)^2}{\left( {3 \times 2.36 \times 1{0^{ - 5}}} \right)^3}\\{K_{sp}} = 7.91 \times 1{0^{ - 22}}\end{array}\)

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

What is the effect on the amount of\(CaHP{O_4}\) that dissolves and the concentrations of \(C{a^{2 + }}\;and\;HPO_4^ - \)when each of the following are added to a mixture of solid \(CaHP{O_4}\)and water at equilibrium?

\(\;(a)\;CaC{l_2}\)

\(\;(b)\;HCl\)

\(\;(c)\;KCl{O_4}\)

\(\;(d)\;NaOH\)

\(\;(e)\;CaHP{O_4}\)

Question: A roll of \(35 - mm\) black and white photographic film contains about \(0.27g\) of unexposed \(AgBr\) before developing. What mass of \(N{a_2}{S_2}{O_3} \cdot 5{H_2}O\) (sodium thiosulfate penta hydrate or hypo) in \(1.0L\)of developer is required to dissolve the \(AgBr\)as \(Ag\left( {{S_2}{O_3}} \right)_2^{3 - }\left( {{K_f} = 4.7 \times 1{0^{13}}} \right)?\)

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

\(\begin{array}{l}(a)BaSi{F_6},0.026\;g/100\;mL(contains Si{F_6}^2 - ions)\\(b)Ce{\left( {I{O_3}} \right)_4},1.5 \times 1{0^{ - 2}}\;g/100\;mL\\(c)G{d_2}{\left( {S{O_4}} \right)_3},3.98\;g/100\;mL\\(d){\left( {N{H_4}} \right)_2}PtB{r_6},0.59\;g/100\;mL(contains PtB{r_6}^{2 - } ions)\end{array}\)

Question: Calculate the cadmium ion concentration, \(\left( {C{d^{2 + }}} \right)\)in a solution prepared by mixing 0.100 L of 0.0100 M \(Cd{\left( {N{O_3}} \right)_2}\)with 1.150 L of 0.100 \(N{H_3}(aq)\).

Question: Using the dissociation constant, \({K_d} = 3.4 \times 1{0^{ - 15}}\), calculate the equilibrium concentrations of \(Z{n^{2 + }}\;and\;O{H^ - }in{\rm{\;}}\)\({\rm{\;}}a\;0.0465 - M\)solution of \(Zn(OH)_4^{2 - }\).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

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.