How Do You Calculate Molar Solubility In Grams100mL Class 11 ...

CoursesCourses for KidsFree study materialOffline CentresMoreStore IconStore

Talk to our experts

1800-120-456-456

Sign In
  • Question Answer
  • Class 11
  • Chemistry
  • How do you calculate Molar sol...
seo-qnaheader left imagearrow-right Answerdown arrowQuestion Answers for Class 12down arrowClass 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11down arrowClass 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10down arrowClass 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9down arrowClass 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8down arrowClass 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7down arrowClass 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6down arrowClass 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5down arrowClass 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4down arrowClass 4 ScienceClass 4 EnglishClass 4 MathsSearchIconbannerHow do you calculate Molar solubility in grams/$100$mL of calcium iodate in water at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$? Ksp = $7.1 \times {10^{ - 7}}$AnswerVerifiedVerified547.5k+ viewsHint: ${{\text{K}}_{{\text{sp}}}}$ represents the solubility product constant at equilibrium. When a solid substance dissolves into water it dissociates into ion. The product of solubilities of the ions with the number of each ion in power is known as the solubility product. First we will determine the molarity and then the equilibrium constant.Formula used:${{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\text{x}}{{\text{S}}^{\text{x}}}\,{{ \times }}\,{\text{y}}{{\text{S}}^{\text{y}}}$Complete step-by-step solution:The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.An ionic compound dissociates into the water as follows:${{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}$ Where,${\text{AB}}$ is an ionic compound.The general representation for the solubility product of an ionic compound is as follows:\[{{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\left( {{\text{x}}{{\text{A}}^{{\text{ + y}}}}\,} \right)^{\text{x}}}\, \times \,\,{\left( {{\text{y}}{{\text{A}}^{ - x}}\,} \right)^{\text{y}}}\]${{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\text{x}}{{\text{S}}^{\text{x}}}\,{{ \times }}\,{\text{y}}{{\text{S}}^{\text{y}}}$Where,${{\text{K}}_{{\text{sp}}}}$is the solubility product constant.${\text{S}}$is the solubility of each ion.$x$ and $y$ represents the number of ions.The compound calcium iodate dissociates into the water as follows:${\text{Ca(I}}{{\text{O}}_3}{)_2} \rightleftarrows \,\,\,{\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{ + }}\,\,2\,{\text{IO}}_3^ - $ The solubility product for calcium iodate is represented as follows:${{\text{K}}_{{\text{sp}}}}\, = \,\,\left[ {{\text{C}}{{\text{a}}^{{\text{2 + }}}}} \right]{\left[ {{\text{2IO}}_4^ - } \right]^2}$$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\,\left[ {\,{\text{S}}} \right]{\left[ {{\text{2S}}} \right]^2}$ $\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\,{\text{4}}{{\text{S}}^3}$$\Rightarrow {\text{S}}\,\,{\text{ = }}\,\,\,\sqrt[3]{{\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{4}}}\,$On substituting $7.1 \times {10^{ - 7}}$ for Ksp in above equation,${\text{S}}\,\,{\text{ = }}\,\,\,\sqrt[3]{{\dfrac{{7.1 \times {{10}^{ - 7}}}}{4}}}\,$$\Rightarrow {\text{S}}\,\,{\text{ = }}\,\,5.62 \times {10^{ - 3}}\,$Solubility tells the concentration of solute compounds in solution in molarity. Molarity tells the mole of solute dissolve in per litre of the solution.Here, solubility is $5.62 \times {10^{ - 3}}\,$, it means that $5.62 \times {10^{ - 3}}\,$ moles of calcium iodate is dissolved in one litre of water.We know, ${\text{1}}\,{\text{L}}\,{\text{ = }}\,{\text{1000}}\,{\text{mL}}$So, we can also say $5.62 \times {10^{ - 3}}$moles of calcium iodate is dissolved in ${\text{1000}}$ mL of water.We have to determine the molar solubility in g/$100$mL so, we will convert the mole of calcium iodate into gram as followsThe formula of mole is as follows:${\text{mole = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$ Molar mass of calcium iodate is $389.88$ g/mol.On substituting $5.62 \times {10^{ - 3}}\,$mol for moles of calcium iodate and $389.88$ g/mol for molar mass.$5.62 \times {10^{ - 3}}\,{\text{mol}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{389}}{\text{.88}}\,{\text{g/mol}}}}$$\Rightarrow {\text{mass}}\,{\text{ = }}\,\,5.62 \times {10^{ - 3}}\,{\text{mol}}\, \times \,{\text{389}}{\text{.88}}\,{\text{g/mol}}$$\Rightarrow {\text{mass}}\,{\text{ = }}\,\,2.19\,{\text{g}}$So, the mass of calcium iodate is $2.19$gram.So, the molar solubility of calcium iodate is $2.19$gram/$1000$ mL. So, the molar solubility in $100$mL is,${\text{1000}}\,{\text{mL}}$water = $2.19$gram calcium iodate$100$mL water = $0.219$gram calcium iodateSo, the molar solubility of calcium iodate is $0.219$gram/ $100$mL.Therefore, the Molar solubility in grams/$100$mL of calcium iodate in water at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$is $0.219$gram/ $100$mL.Note: By the addition of atomic mass the molar mass is determined. Solubility tells the concentration of solid compounds dissolved in water. Concentration in the determination of solubility products is taken in terms of molarity. The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. Molar solubility represents the number of ions dissolved per liter of solution. Here, solubility represents the number of ions dissolved in a given amount of solvent.Recently Updated PagesWhy are manures considered better than fertilizers class 11 biology CBSEarrow-rightFind the coordinates of the midpoint of the line segment class 11 maths CBSEarrow-rightDistinguish between static friction limiting friction class 11 physics CBSEarrow-rightThe Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSEarrow-rightThe first National Commission on Labour NCL submitted class 11 social science CBSEarrow-rightNumber of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSEarrow-rightWhy are manures considered better than fertilizers class 11 biology CBSEarrow-rightFind the coordinates of the midpoint of the line segment class 11 maths CBSEarrow-rightDistinguish between static friction limiting friction class 11 physics CBSEarrow-rightThe Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSEarrow-rightThe first National Commission on Labour NCL submitted class 11 social science CBSEarrow-rightNumber of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSEarrow-right
  • 1
  • 2
Trending doubtsDifferentiate between an exothermic and an endothermic class 11 chemistry CBSEarrow-right10 examples of friction in our daily lifearrow-rightOne Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSEarrow-rightDifference Between Prokaryotic Cells and Eukaryotic Cellsarrow-rightState the laws of reflection of lightarrow-rightExplain zero factorial class 11 maths CBSEarrow-rightDifferentiate between an exothermic and an endothermic class 11 chemistry CBSEarrow-right10 examples of friction in our daily lifearrow-rightOne Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSEarrow-rightDifference Between Prokaryotic Cells and Eukaryotic Cellsarrow-rightState the laws of reflection of lightarrow-rightExplain zero factorial class 11 maths CBSEarrow-right
  • 1
  • 2

Repeaters Course for NEET 2022 - 23

NEET Repeater 2023 - Aakrosh 1 Year Course

Tag » How To Calculate Molar Solubility