K sp Mg 2 CO 3 2 a MgCO 3 s Mg 2 aq CO 3 2 aq K sp Fe 2 OH 2 b FeOH 2 s Fe 2 aq

K sp mg 2 co 3 2 a mgco 3 s mg 2 aq co 3 2 aq k sp fe

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K sp = [Mg 2+ ][CO 3 2- ] (a) MgCO 3 ( s ) Mg 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Fe 2+ ][OH - ] 2 (b) Fe(OH) 2 ( s ) Fe 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 (c) Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3- ( aq ) (d) Ag 2 S( s ) 2Ag + ( aq ) + S 2- ( aq ) S 2- ( aq ) + H 2 O( l ) HS - ( aq ) + OH - ( aq ) Ag 2 S( s ) + H 2 O( l ) 2Ag + ( aq ) + HS - ( aq ) + OH - ( aq ) K sp = [Ag + ] 2 [HS - ][OH - ]
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Equilibria of Slightly Soluble Ionic Compounds Determining K sp from Solubility PROBLEM: (a) Lead ( II ) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10 -3 g/100mL solution. What is the K sp of PbSO 4 ? (b) When lead ( II ) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64g/L. Calculate the K sp of PbF 2 . PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. K sp = [Pb 2+ ][SO 4 2- ] = 1.40x10 -4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 = SOLUTION: PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq ) (a) 1000mL L 4.25x10 -3 g 100mL soln 303.3g PbSO 4 mol PbSO 4 1.96x10 -8
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Equilibria of Slightly Soluble Ionic Compounds Determining K sp from Solubility continued (b) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 = 2.6x10 -3 M K sp = (2.6x10 -3 )(5.2x10 -3 ) 2 = 0.64g L soln 245.2g PbF 2 mol PbF 2 7.0x10 -8
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Equilibria of Slightly Soluble Ionic Compounds Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, Formula K sp Aluminum hydroxide, Al(OH) 3 Cobalt ( II ) carbonate, CoCO 3 Iron ( II ) hydroxide, Fe(OH) 2 Lead ( II ) fluoride, PbF 2 Lead ( II ) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn( I O 3 ) 2 3 x 10 -34 1.0 x 10 -10 4.1 x 10 -15 3.6 x 10 -8 1.6 x 10 -8 4.7 x 10 -29 8 x 10 -48 Mercury ( I ) iodide, Hg 2 I 2 3.9 x 10 -6
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Equilibria of Slightly Soluble Ionic Compounds Determining Solubility from K sp PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5x10 -6 . PLAN: Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. SOLUTION: Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ][OH - ] 2 - Initial Change Equilibrium - - 0 0 +S + 2S S 2S K sp = (S)(2S) 2 S = 6.5x10 6 4 3 = 1.2x10x -2 M Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) Concentration (M)
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Equilibria of Slightly Soluble Ionic Compounds Relationship Between K sp and Solubility at 25 0 C No. of Ions Formula Cation:Anion K sp Solubility (M) 2 MgCO 3 1:1 3.5 x 10 -8 1.9 x 10 -4 2 PbSO 4 1:1 1.6 x 10 -8 1.3 x 10 -4 2 BaCrO 4 1:1 2.1 x 10 -10 1.4 x 10 -5 3 Ca(OH) 2 1:2 5.5 x 10 -6 1.2 x 10 -2 3 BaF 2 1:2 1.5 x 10 -6 7.2 x 10 -3 3 CaF 2 1:2 3.2 x 10 -11 2.0 x 10 -4 3 Ag 2 CrO 4 2:1 2.6 x 10 -12 8.7 x 10 -5
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Equilibria of Slightly Soluble Ionic Compounds The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2+ ( aq ) + CrO 4 2- ( aq ) PbCrO 4 ( s ) Pb 2+ ( aq ) + CrO 4 2- ( aq ) CrO 4 2- added
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Equilibria of Slightly Soluble Ionic Compounds Calculating the Effect of a Common Ion on Solubility PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5x10 -6 . PLAN: Set up a reaction equation and table for the dissolution of Ca(OH) 2 .
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