The reason as to why inspite of the low solubility of Ca ( OH ) 2 it becomes soluble in acidic solution is to be stated in terms of Le Chatelier’s Principle. Concept introduction: The equilibrium is a state of the chemical reaction in which the rate of the forward reaction and the rate of the backward reaction become the same. There is no net change in the concentration of reactants and products at the equilibrium condition.

Question

Write the chemical equations for the dissociation of each of the following substances and determine their molar solubilities using the Solubility Product Constant Table. Note: Include states-of-matter under the given conditions in your answer. Please use the double-sided equilibrium arrows. (a) AgI chemPad XX→ (b) CuS chemPad X.X° →→← Aluminum Al(OH)3 Barium Ba(OH)2 BaCO3 BaSO4 Substance Ksp at 25 °C Cd(OH)2 CdCO3 CdS M M Calcium Help Greek Help Greek 5.0 x 10-3 8.1 x 10-9 1.1 x 10-10 Ba3(PO4)2 3.4 x 10-23 Cadmium 1.9 x 10-33 2.5 x 10-14 5.2 x 10-12 8.0 x 10-27 Ca(OH)2 5.5 x 10-6 CaCO3 4.8 x 10-⁹ Ca3(PO4)2 1.0 × 10-26 CaF2 3.9 x 10-11 Copper Cu(OH)2 Cu₂S CUCO3 CuS Iron Fe(OH)2 FeS Fe(OH)3 FePO4 Lead Pb(OH)2 PbF2 PbCl₂ PbBr2 Pbl₂ DOO 2.2 x 10-20 1.6 x 10-48 1.4 x 10-10 6.3 x 10-36 8.0 x 10-16 6.3 x 10-18 2.5 x 10-39 9.9 x 10-29 2.8 x 10-16 3.7 x 10-8 1.7 x 10-5 6.3 x 10-6 6.5 x 10-9 Manganese Mn(OH)2 MnCO3 MnS Nickel Ni(OH)2 Silver AgCl AgBr Agl AgCN Ag₂S Ag₂ CrO4 Ag3PO4 Tin Sn(OH)2 1.9…

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Question

Chapter 18, Problem 39E

Interpretation Introduction

Interpretation:

The reason as to why inspite of the low solubility of Ca(OH)2 it becomes soluble in acidic solution is to be stated in terms of Le Chatelier’s Principle.

Concept introduction:

The equilibrium is a state of the chemical reaction in which the rate of the forward reaction and the rate of the backward reaction become the same. There is no net change in the concentration of reactants and products at the equilibrium condition.

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Write the chemical equations for the dissociation of each of the following substances and determine their molar solubilities using the Solubility Product Constant Table. Note: Include states-of-matter under the given conditions in your answer. Please use the double-sided equilibrium arrows. (a) AgI chemPad XX→ (b) CuS chemPad X.X° →→← Aluminum Al(OH)3 Barium Ba(OH)2 BaCO3 BaSO4 Substance Ksp at 25 °C Cd(OH)2 CdCO3 CdS M M Calcium Help Greek Help Greek 5.0 x 10-3 8.1 x 10-9 1.1 x 10-10 Ba3(PO4)2 3.4 x 10-23 Cadmium 1.9 x 10-33 2.5 x 10-14 5.2 x 10-12 8.0 x 10-27 Ca(OH)2 5.5 x 10-6 CaCO3 4.8 x 10-⁹ Ca3(PO4)2 1.0 × 10-26 CaF2 3.9 x 10-11 Copper Cu(OH)2 Cu₂S CUCO3 CuS Iron Fe(OH)2 FeS Fe(OH)3 FePO4 Lead Pb(OH)2 PbF2 PbCl₂ PbBr2 Pbl₂ DOO 2.2 x 10-20 1.6 x 10-48 1.4 x 10-10 6.3 x 10-36 8.0 x 10-16 6.3 x 10-18 2.5 x 10-39 9.9 x 10-29 2.8 x 10-16 3.7 x 10-8 1.7 x 10-5 6.3 x 10-6 6.5 x 10-9 Manganese Mn(OH)2 MnCO3 MnS Nickel Ni(OH)2 Silver AgCl AgBr Agl AgCN Ag₂S Ag₂ CrO4 Ag3PO4 Tin Sn(OH)2 1.9…

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