/*! 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} Q25 E The solubility product of \(CaS{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Q25 E (page 872)

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 \(0.010M\) \(SO{4^{2 - }}\)?

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.

\({\rm{ Mass of dissolved salt }} = 0.34\;{\rm{g}}/{\rm{L}}\)

Step by step solution

01

To find the mass of the dissolved salt

\(CaS{O_4}.2{H_2}O\)

Because of the common ion, the concentration of\({\rm{SO}}_4^{2 - }\)present in the solution is a limiting factor to the dissolving of \({\rm{CaS}}{{\rm{O}}_4} \times 2{{\rm{H}}_2}{\rm{O}}\).

\({\rm{CaS}}{{\rm{O}}_4}({\rm{s}}) \to {\rm{C}}{{\rm{a}}^{2 + }}({\rm{aq}}) + {\rm{SO}}_4^{2 - }({\rm{aq}})\)

\(\begin{array}{l}{{\rm{K}}_{{\rm{sp}}}} = \left[ {{\rm{C}}{{\rm{a}}^{{\rm{2 + }}}}} \right]\left[ {{\rm{SO}}_{\rm{4}}^{{\rm{2 - }}}} \right] = {\rm{x}} \times ({\rm{x}} + 0.01) = 2.4 \times {10^{ - 5}}\\\end{array}\)

\({{\rm{x}}^2} + 0.01x - 2.4 \times {10^{ - 5}} = 0\)

\({\rm{x}} = \frac{{ - 0.01 \pm \sqrt {1 \times {{10}^{ - 4}} + 9.6 \times {{10}^{ - 5}}} }}{2} = 2 \times {10^{ - 3}}\)

This value corresponds to the concentration of \({\rm{CaS}}{{\rm{O}}_4} \times 2{{\rm{H}}_2}{\rm{O}}\) that has dissolved.

02

Next step is to find the mass of the salt in 1L of the solution

To find the Mass of the salt in \(1\;{\rm{L}}\)of the solution calculations are done in the following way

\({\rm{m}} = 2 \times {10^{ - 3}}\;{\rm{mol}}/{\rm{L}} \times 172.16\;{\rm{g}}/{\rm{mol}} = 0.34\;{\rm{g}}/{\rm{L}}\)

Finally, we get,

\({\rm{ Mass of dissolved salt }} = 0.34\;{\rm{g}}/{\rm{L}}\)

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

Question: The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, 鈥揘H2, and a carboxylic acid group, 鈥揅O2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3.

(a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in 1 M KOH.

(b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the 鈥揘H2 and 鈭扖O2鈭� groups.)

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: Use the simulation (http://openstaxcollege.org/l/16solublesalts) from the earlier Link to Learning to complete the following exercise: Using 0.01 g\(Ca{F_2},\;\)give the \({K_{sp}}\)values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.

What is the molar solubility of \({\bf{Tl}}{\left( {{\bf{OH}}} \right)_{\bf{3}}}\) in a \({\bf{0}}.{\bf{10}}{\rm{ }}{\bf{M}}\) solution of \({\bf{N}}{{\bf{H}}_{\bf{3}}}\) ?

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

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

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.