(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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Chapter 16 Solutions
Chemistry
- 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
- When 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_forwardThe solubility-product constants, Ksp, at 25 °C for two compounds [iron(II) carbonate, FeCO3, and cadmium(II) carbonate, CdCO3] are given by the table Substance Ksp FeCO3 2.10 x 10-11 CdCO3 1.80 × 10-14 Part A A solution of Na2CO3 is added dropwise to a solution that contains 1.02x10-2 MFe²+ and 1.48x10-2 M Cd²+. What concentration of CO3²- is need to initiate precipitation? Neglect any volume changes during the addition. Express your answer with the appropriate units. ► View Available Hint(s) [CO3²- ] = Submit μA Value Part B Complete previous part(s) Part C Complete previous part(s) Units ?arrow_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_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_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_forwardDetermine the molar solubility of MgCO3 in pure water. Ksp (MgCO3) =6.82•10-6arrow_forward
- The Ksp of CaF, and CacO3 are 4.9 x 10-11 and 4.7 x 10° respectively. Calculate the molar solubility of each salt. O Molar solubility of CaF2 = 2.31x 104. Molar solubility of CaCO3 = 6.86 x 105 O Molar solubility of CaF2 = 6.86x 10-5; Molar solubility of CaCO3 = 6.86 x 10-5 Molar solubility of CaF2 = 2.31x 104; Molar solubility of CaCO3 = 4.7x 109 Molar solubility of CaF2 = 4.9x 10-11; Molar solubility of CaCO3 = 4.7x 10-9arrow_forwardSolubility 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_forward
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