Chap17-Part4

Chap17-Part4 - Titration of Polyprotic Acids H3PO3(aq) +...

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Titration of Polyprotic Acids H 3 PO 3 ( aq ) + OH - ( aq ) H 2 PO 3 - ( aq ) + H 2 O( l ) H 2 PO 3 - ( aq ) + OH - ( aq ) HPO 3 2- ( aq ) + H 2 O( l ) 1
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Match each panel with the following descriptions: (a) strong acid added to strong base; (b) strong base added to weak acid; 2
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Match each panel with the following descriptions: (a) strong acid added to strong base; (b) strong base added to weak acid; (a)
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Match each panel with the following descriptions: (a) strong acid added to strong base; (b) strong base added to weak acid; (a) (b)
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Solubility Equilibria The dissolution or precipitation of solids from liquid solutions can be treated as an equilibrium problem. The equilibrium is heterogeneous . The solubility is described quantitatively by an equilibrium reaction; e.g. BaSO 4 ( s ) Ba 2+ ( aq ) + SO 4 2- ( aq ) K sp = [Ba 2+ ( aq )][SO 4 2- ( aq )] K sp is called the solubility product and describes how soluble the solid is in the liquid. There must be some solid present for this equation to hold, although the factor [BaSO 4 ( s )] does not appear in the expression for K sp .
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Write the solubility constants K sp for (a) MnCO 3 ( s ) (b) Hg(OH) 2 ( s ) (c) Cu 3 (PO 4 ) 2 ( s )
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Write the solubility constants K sp for (a) MnCO 3 ( s ) MnCO 3 ( s ) Mn 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Mn 2+ ][CO 3 2- ] (b) Hg(OH) 2 ( s ) (c) Cu 3 (PO 4 ) 2 ( s )
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Write the solubility constants K sp for (a) MnCO 3 ( s ) MnCO 3 ( s ) Mn 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Mn 2+ ][CO 3 2- ] (b) Hg(OH) 2 ( s ) Hg(OH) 2 ( s ) Hg 2+ ( aq ) + 2 OH - ( aq ) K sp = [Hg 2+ ][OH - ] 2 (c) Cu 3 (PO 4 ) 2 ( s )
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Write the solubility constants K sp for (a) MnCO 3 ( s ) MnCO 3 ( s ) Mn 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Mn 2+ ][CO 3 2- ] (b) Hg(OH) 2 ( s ) Hg(OH) 2 ( s ) Hg 2+ ( aq ) + 2 OH - ( aq ) K sp = [Hg 2+ ][OH - ] 2 (c) Cu 3 (PO 4 ) 2 ( s ) Cu 3 (PO 4 ) 2 ( s ) 3 Cu 2+ + 2 PO 4 3- K sp = [Cu 2+ ] 3 [PO 4 3- ] 2
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The solubility of a substance a) is not the same as its solubility product b) is the quantity of a substance that dissolves to form a saturated solution; and c) may be expressed in terms of mols per volume or mass per volume.
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The solubility of PbBr 2 at 25 o C is 0.010 M. Calculate K sp .
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The solubility of PbBr 2 at 25 o C is 0.010 M. Calculate K sp . PbBr 2 Pb 2+ + 2 Br - K sp = [Pb 2+ ][Br - ] 2 [Pb 2+ ] = 0.010 M [Br - ] = 0.020 M K sp = (0.010)(0.020) 2 = 4.0 x 10 -6
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If 0.0490 gm of AgIO 3 dissolves per liter of solution, calculate the solubility product. The equilibrium is AgIO 3 Ag + + IO 3 -
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If 0.0490 gm of AgIO 3 dissolves per liter of solution, calculate the solubility product. The equilibrium is AgIO 3 Ag + + IO 3 - mol AgIO 3 = 0.0490 gm / [283 gm / mol] = 1.73 x 10 -4 mol K sp = [Ag + ][IO 3 - ] [Ag + ] = 1.73 x 10 -4 M & [IO 3 - ] = 1.73 x 10 -4 M K sp = (1.73 x 10 -4 ) 2 = 3.00 x 10 -8
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1) temperature
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Chap17-Part4 - Titration of Polyprotic Acids H3PO3(aq) +...

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