HW Chp17.4_17.7

HW Chp17.4_17.7 - Visualizing Concepts 17.6 The product of...

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17.6 The product of the ion concentrations in a saturated solution equals K sp . Use numbers of anions and cations as a measure of concentration to calculate relative ”K sp ” values. Counting cations is not adequate, because excess anions in some of the boxes drive down the cation concentrations. Ion-products must be considered. AgX: (4 Ag + )(4X ) = 16 AgY: (1 Ag + )(9 Y ) = 9 AgZ: (3 Ag + )(6 Y ) = 18 AgY has the smallest K sp . 17.7 Common anions or cations decrease the solubility of salts. Ions that participate in acid-base or complex ion equilibria increase solubility. (a) CO 2 3– from BaCO 3 reacts with H + from HNO 3 , causing solubility of BaCO 3 to increase with increasing HNO 3 concentration. This behavior matches the right diagram. (b) Extra CO 2 3– from Na 2 CO 3 decreases the solubility of BaCO 3 . Solubility of BaCO 3 decreases as [Na 2 CO 3 ] increases. This behavior matches the left diagram. (c) NaNO 3 has no common ions, nor does it enter into acid-base or complex ion equilibria with Ba 2+ or CO 3 2– ; it does not affect the solubility of BaCO 3 . This behavior is shown in the center diagram. 17.8 A metal hydroxide that is soluble at very low and very high pH’s, that is, in strong acid or strong base, is calledamphoteric. Solubility Equilibria and Factors Affecting Solubility 17.46 (a) Solubility is the amount (grams, moles) of solute that will dissolve in a certain volume of solution. Solubility-product constant is an equilibrium constant, the product of the molar concentrations of all the dissolved ions in solution. (b) K sp = [Mn 2+ ][CO 3 2– ]; K sp = [Hg 2+ ][OH ] 2 ; K sp = [Cu 2+ ] 3 [PO 4 3– ] 2 17.47 .Follow the logic in Sample Exercise 17.10. (a) CaF 2 (s) Ca 2+ (aq) + 2F (aq); K sp = [Ca 2+ ][F ] 2 The molar solubility is the moles of CaF 2 that dissolve per liter of solution. Each mole of CaF 2 produces 1 mol Ca 2+ (aq) and 2 mol F (aq). [Ca 2+ ] = 1.24 × 10 –3 M ; [F ] = 2 × 1.24 × 10 –3 M = 2.48 × 10 –3 M K sp = (1.24 × 10 –3 ) (2.48 × 10 –3 ) 2 = 7.63 × 10 –9 (b) SrF 2 (s) Sr 2+ (aq) + 2F (aq); K sp = [Sr 2+ ][F ] 2 Transform the gram solubility to molar solubility. L / SrF mol 10 8 . 8 10 76 . 8 SrF g 6 . 125 SrF mol 1 L 100 . 0 SrF g 10 1 . 1 2 4 4 2 2 2 2 × = × = × × [Sr 2+ ] = 8.76 × 10 –4 M ; [F ] = 2(8.76 × 10 –4 M ) K sp = (8.76 × 10 –4 ) (2(8.76 × 10 –4 )) 2 = 2.7 × 10 –9 (c) Ba(IO 3 ) 2 (s) Ba 2+ (aq) + 2IO 3 (aq); K sp = [Ba 2+ ][IO 3 ] 2 Since 1 mole of dissolved Ba(IO
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This note was uploaded on 04/29/2008 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

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HW Chp17.4_17.7 - Visualizing Concepts 17.6 The product of...

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