(i)
Interpretation: The mathematical relation between solubility product,
Concept introduction: At equilibrium, the measure of maximum amount of solute that is to be dissolved in a solvent is known as solubility. Solubility product is defined as the product of concentration of ions in a saturated solution where each ion is raised to the power of their coefficients.
(ii)
Interpretation: The mathematical relation between solubility product,
Concept introduction: At equilibrium, the measure of maximum amount of solute that is to be dissolved in a solvent is known as solubility. Solubility product is defined as the product of concentration of ions in a saturated solution where each ion is raised to the power of their coefficients.
(iii)
Interpretation: The mathematical relation between solubility product,
Concept introduction: At equilibrium, the measure of maximum amount of solute that is to be dissolved in a solvent is known as solubility. Solubility product is defined as the product of concentration of ions in a saturated solution where each ion is raised to the power of their coefficients.
(iv)
Interpretation: The mathematical relation between solubility product,
Concept introduction: At equilibrium, the measure of maximum amount of solute that is to be dissolved in a solvent is known as solubility. Solubility product is defined as the product of concentration of ions in a saturated solution where each ion is raised to the power of their coefficients.
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Chemistry: An Atoms First Approach
- 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. (a) BaSeO4, 0.0118 g/100 mL. (b) Ba(BrO3)2H2O, 0.30 g/100 mL. (c) NH4MgAsO46H2O, 0.033 g/100 mL. (d) La2(MoO4)3, 0.00179 g/100 mLarrow_forwardConsider the following four titrations. i. 100.0 mL of 0.10 M HCl titrated by 0.10 M NaOH ii. 100.0 mL of 0.10 M NaOH titrated by 0.10 M HCl iii. 100.0 mL of 0.10 M CH3NH2 titrated by 0.10 M HCl iv. 100.0 mL of 0.10 M HF titrated by 0.10 M NaOH Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing pH initially before any titrant has been added. c. increasing pH at the halfway point in equivalence. d. increasing pH at the equivalence point. How would the rankings change if C5H5N replaced CH3NH2 and if HOC6H5 replaced HF?arrow_forwardFigure 15-3 outlines the classic scheme for separating a mixture of insoluble chloride salts from one another. Explain the chemistry involved in the various steps of the figure.arrow_forward
- PbCl2; molar solubility = 1.43 x 10-² M Ksp = 1.17 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Note that Ksp is the solubility-product constant, the equilibrium constant for a chemical equation representing the dissolution of an ionic compound: Ksp = [Pb²+][C1-1² ■√ ΑΣΦ 10-8 Part C Prepare an ICE table (where I means initial, C means change, and E means equilibrium), where the equilibrium concentrations of the formed ions are expressed in terms of S (the number of moles per liter of PbCl2 dissolved) and substitute them into the expression for Ksp. Ksp CaF2; molar solubility = 3.32 × 10-4 M = Submit ΑΣΦ Request Answerarrow_forwardThe molar solubility of Mg(CN)2 is 1.4 × 10-5 M at a certain temperature. Determine the value of Ksp for Mg(CN). 1 2 NEXT > Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Mg(CN):(s) Mg²-(aq) 2 CN-(aq) Initial (M) Change (M) Equilibrium (M) 5 RESET 1.4 x 10-5 -1.4 x 105 2.8 x 10-5 -2.8 x 105 +x +2x -2x 1.4 x 10-5 + x 1.4 x 10-5 + 2x 1.4 x 10-5 - x 1.4 x 10-5 - 2x 2.8 x 105 + x 2.8 x 10-5 + 2x -X 2.8 x 10-5. 2.8 x 10-5 - 2x 1Larrow_forwardWhen 200.0 mL of 5.15 x 10-4 M Ba(NO3)2 is added to 150.0 mL of 8.25 x 10-4 M Na2SO4. Ksp(BaSO4) = 1.08 x 10-10. A precipitate choose your answer.... Ksp because Qsp choose your answer... choose your answer... SIGMA = formarrow_forward
- The Ksp of zinc hydroxide, ZN(OH)2 is 3.00 x 10-17. Calculate the solubility of this compound in grams per liter.arrow_forwardDetermine the molar solubility of MgCO3 in pure water. Ksp (MgCO3) =6.82•10-6arrow_forwardThe molar solubility of bismuth sulfide Bi,S, at 25°C is 1.0 x 10-15 M. Calculate Ksp- O 1x 10 15 O 1x 10 75 O 1x 10 73 O 6 x 10 30arrow_forward
- Solubility Product Constants. Analyze and solve the given problems. Show your completesolution. 1. Determine the Ksp of mercury(I) bromide (Hg2Br2), given that its molar solubility i 2.52 x 10¯8 mole per liter. 2. The value of Ksp of AgCl is 1.8 x 10-10. What would be the molar concentration of Ag+ and Clin AgCl in pure water placed in contact with solid AgCl(s)? 3. Calculate the molar solubility of CaF2 at 25oC in 0.010M Ca(NO3)2 solution. (Ksp = 3.1x 10-5 M)arrow_forwardDetermine the molar solubility for CaF2 by constructing an ICE table, writing the solubility constant expression, and solving for molar solubility. The value of Ksp for CaF2 is 3.5 × 1011. Complete Parts 1-3 before submitting your answer. Using the values from the ICE table (Part 1), construct the expression for the solubility constant, Ksp. Each reaction participant must be represented by one tile. Do not combine terms. Ksp || = 3.5 × 10-11 RESET [0] [0.00300] [0.00600] [0.00300]² [0.00600]² [x] [x] 2 [2x] [2x]² [0.00300 + x] [0.00300 - x] [0.00300 + 2x]² [0.00300 - 2x]² [0.00600 + x] [0.00600 - x] [0.00600 + 2x]² [0.00600 - 2x]²arrow_forwardThe molar solubility of SrF2 is 2.5 x 10-9 M. Calculate the Ksp of SrF2.arrow_forward
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