Calculate the molar solubility of calcium hydroxide in a solution buffered at each of the following pH's:?
Calcium hydroxide dissolves and dissociates according to: Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq) So the ionic molarities in a saturated solution satisfy the following equilibrium equation Ksp = [Ca²⁺] ∙ [OH⁻]² The solubility constant at 25°C is [1] Ksp = 5.02×10^–6 Let s be the molar solubility of Ca(OH)₂. When you dissolve s moles per liter, all the salt dissolves and dissociates, forming one calcium ion per salt molecule. Hence the calcium ion molarities is: [Ca²⁺] = s The hydroxide ion concentration does not change due to buffering. It is fixed to a level, which is specified by the given pH: [OH⁻] = 10^(pH - 14) Therefore Ksp = s ∙ (10^(pH - 14))² = s ∙ 10^(2∙pH - 28) => s = Ksp / 10^(2∙pH - 28) = Ksp ∙ 10^(28 - 2∙pH) at pH=5 s = 5.02×10^–6 ∙ 10^(28 - 2∙5) = 5.02×10^–6 ∙ 10^(18) = 5.02×10^12 mol/L at pH=7 s = 5.02×10^–6 ∙ 10^(28 - 2∙7) = 5.02×10^–6 ∙ 10^(14) = 5.02×10^8 mol/L at pH=8 s = 5.02×10^–6 ∙ 10^(28 - 2∙7) = 5.02×10^–6 ∙ 10^(12) = 5.02×10^6 mol/L
Calculate the solubility product of calcium hydroxide if the solubility of Ca(OH)2(s) in water at 25?
you havent mentioned the value of solubility. is 25 the temperature?
Solubility Product of Calcium Hydroxide?
_____Ksp PbI2 <-------> Pb+2 + 2I- _______Ksp = [Pb+2][I-]^2 Ca(IO3)2 <-----> Ca+2 + 2 IO3- _______Ksp = similar Sat soln of Ca(OH)2 Ca(OH)2 <----> Ca+2 + 2OH- _______Ksp = similar [Ca+] = [OH-]/2 [OH-] moles/liter = 0.130 moles HCl/liter * (1 mole OH-/ mole H+) * [22.7 ml HCl/25 ml Ca(OH)2] = ?? Plug and SOLVE Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.
Determining Solubility and Ksp by Titration?
You add an excess of Ca(OH)2 to water maintained at a particular temperature, stir until the solution is saturated, filter, then determine the [OH−] in the solution by titration with acid. Titration of 15.0 mL of the calcium hydroxide solution to the endpoint requires 6.64 mL of 0.055 M HCl solution. I found the molar quantity of OH− in the 15.0 mL of solution to be 0.365 mmol . How do I go about finding concentrations of the anions/cations? and solubility? I can find the Ksp part. What are the concentrations of Ca2+ and OH−? What is the solubility of Ca(OH)2 under these conditions? What is the Ksp for Ca(OH)2 under these conditions? Thank you so much in advance. Cheers
Determine the solubility product of calcium hydroxide?
1) The amount of hydrochloric acid used equals concentration times volume used in titration: ∆n_HCl = [HCl]∙∆V_HCl = 0.05 mol/dm³ ∙ 21.10×10⁻³ dm³ = 1.055×10⁻³ mol 2) The reaction occurring in the titration process is: OH⁻ + HCl → H₂O + Cl⁻ According to this reaction equation you need on mole of Hcl to neutralize one mole of hydroxide ions. So the solution contained: n_OH⁻ = ∆n_HCl = 1.055×10⁻³ mol 3) The concentration of hydroxide equals the amount from above divided by the volume of saturated solution, which is titrated: [OH⁻] = n_OH⁻/V_Ca(OH)₂ = 1.055×10⁻³ mol / 25.0×10⁻³ dm³ = 0.0422 mol/dm³ 4) Calcium hydroxide dissolves according to the reaction equation: Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2 OH⁻(aq) That means the solution contains twice as much hydroxide ions as calcium ions. In other words the calcium ion concentration is one half of the hydroxide ion concentration: [Ca²⁺] = (1/2)∙[OH⁻] = (1/2)∙ 0.0422 mol/dm³ = 0.0211 mol/dm³ 5) According to dissolution reaction equation the solubility product for calcium hydroxide is given by: Ksp = [Ca²⁺] ∙ [OH⁻]² Commonly Ksp is computed with ionic concentrations in M = mol/dm³ So Ksp = 0.0211 ∙ (0.0422)² = 3.76×10⁻⁵
Solubility of calcium iodate?
The net reaction between iodate and thiosulfate: IO3- (aq) + 6S2O3 2- (aq) + 6H3O+(aq) --------> I- (aq) + 3S4O6 2- (aq) + 9H2O(l) moles S2O3 2- = M S2O3 2- x L S2O3 2- = (0.011)(0.02162) = 0.000238 moles S2O3 2- The balanced equation tells us that it takes 1 mole of IO3- to react with 6 moles of S2O3 2-. 0.000238 moles S2O3 2- x (1 mole IO3- / 6 moles S2O3 2-) = 0.0000397 moles IO3- In the formula Ca(IO3)2, there are 2 moles of IO3- for every 1 mole of Ca2+. 0.0000397 moles IO3- x (1 mole Ca2+ / 2 moles IO3-) = 0.0000198 moles Ca2+ M IO3- = moles IO3- / L solution = 0.0000397 / 0.010 = 0.00397 M IO3- M Ca2+ = moles Ca2+ / L solution = 0.0000198 / 0.010 = 0.00198 M Ksp Ca(IO3)2 = [Ca2+][IO3-]^2 = (0.00198)(0.00397)^2 = 3.1 x 10^-8
Determining Solubility and Ksp by Titration?
You add an excess of Ca(OH)2 to water maintained at a particular temperature, stir until the solution is saturated, filter, then determine the [OH−] in the solution by titration with acid. Titration of 40.0 mL of the calcium hydroxide solution to the endpoint requires 20.78 mL of 0.065 M HCl solution. -What is the molar quantity of OH− in the 40.0 mL of solution? -What are the concentrations of Ca2+ and OH−? (seperately) -What is the solubility of Ca(OH)2 under these conditions? (Express g/100 mL as g/dL).? -What is the Ksp for Ca(OH)2 under these conditions?
Help with finding Ksp, molar solubility, and equilibrium?
Can someone help me and show me how to find questions 8 through 11? How to I find OH- and Ca2+ in equlibrium, Molar Solubility, and Ksp of the following? A. Molar Solubility and Solubility Product of Calcium Hydroxide 1. Conc of standardized HCl solution (mol/L) = 0.05 mol/L 2. Buret reading, initial (mL) = 0.1 mL 3. Buret reading, final (mL) = 22.9 mL 4. Volume of HCl added (mL) = 22.8 5. Amount og HCl added (mol) = 1.14E-3 mol 6. Amount of OH- in satd solution (mol) = 25.2 mol 7. Volume of satd Ca(OH)2 solution (mL) = 25.0 mL 8. [OH-], equilibrium (mol/L) = ? 9. [Ca2+], equilibrium (mol/L) = ? 10. Molar solubility of Ca(OH)2 = ? 11. Ksp of Ca(OH)2 = ?