Chapter-9[1] chem

Chapter-9[1] chem - Chapter 9: Dissolution and...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 9: Dissolution and Precipitation Equilibria Pure water solvent unsaturated solution, A( aq ) add solute, A dissolution reaction add more solute, A precipitation reaction saturated solution, A( aq ) A( s ) A( aq ) Dissolution-precipitation equilibrium, K= [A( aq )] Equilibrium constant of dissolution-precipitation reaction tells us about the maximum amount of a substance that can dissolve in a given amount of the solvent . If [A( aq )] > K then that solution is said to be supersaturated solution (a). If a dissolution reaction is endothermic then increasing temperature will increase the solubility. This is the most common situation. Effect of Temperature : (b). If a dissolution reaction is exothermic then increasing temperature will decrease solubility. Interactions : In a dissolution reaction, solute-solute interactions are replaced by solute-solvent interactions. In a precipitation reaction, solute-solvent interactions are replaced by solute-solute interactions
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Solubility of salts : For highly soluble salts, the concentration of ions can be very large, so much so that ions tend to associate. Such solutions are said to be non-ideal solutions and require complex equilibrium equations. For this reason, let us restrict consideration to sparingly soluble or insoluble salts for which concentrations in saturated solutions are < 0.1 M. For such solutions, equilibrium can be represented by just one equation. Solubility Product : (a). AgCl( s ) Ag + ( aq ) + Cl - ( aq ) K = [Ag + ( aq )][Cl - ( aq ] = K sp =1.6 x 10 -10 (b). PbI 2 ( s ) Pb 2+ ( aq ) + 2 I - ( aq ) K sp = [Pb 2+ ( aq )][I - ( aq )] 2 c). Ag 2 CrO 4 ( s ) 2Ag + ( aq ) + CrO 4 2- ( aq ) K sp = [Ag + ( aq )] 2 [CrO 4 2- (aq)] (d). Al(OH) 3 ( s ) Al 3+ ( aq ) + 3 OH - ( aq ) K sp = [Al 3+ ( aq )][[OH - ( aq )] 3 Solubility and K sp : sp for AgCl is 1.6x10 -10 at 25 o C. How many grams of AgCl dissolves in water at 5 o C? AgCl( s ) Ag + ( aq ) + Cl - ( aq ) K sp = [Ag + ( aq )][Cl - ( aq ] =1.6 x 10 -10 y x y = y 2 = 1.6 x 10 -10 y = [Ag + ( aq )] = [Cl - ( aq ] = (1.6x10 -10 ) 1/2 = 1.26 x 10 -5 M i.e. (1.26 x 10 -5 moles x 143.3 g/mol) = 1.8 x 10 -3 g of AgCl dissolves per 1 L solution
Background image of page 2
sp for CaF 2 is 3.9x10 -11 at 25 o C. What is the solubility of CaF 2 in water at 25 o C? CaF 2 ( s ) Ca 2+ ( aq ) + 2F - ( aq ) K sp = [Ca 2+ ( aq )][F - ( aq ] 2 =3.9 x 10 -11 y x (2y) 2 = 4y 3 = 3.9 x 10 -11 [F - (aq)]= 2 x 2.14 x 10 -4 = 4.28 x 10 -4 Solubility of CaF 2 = (2.14x10 -4 moles x 78.1 ) = 0.017 g __ mol g __ L ____ L y = [Ca 2+ ( aq )] = 3.9x10 -11 1/3 = 2.14 x 10 -4 M _______ 4 Determine the mass of lead(II) iodate present in 2.50 L saturated aqueous solution. Pb(IO 3 ) 2 ( s ) Pb 2+ ( aq ) + 2 IO 3 - ( aq ) K sp = 2.6 x 10 -13 K sp = 2.6 x 10 -13 = [Pb 2+ ( aq ) ][IO 3 - ( aq )] 2 = y x (2y) 2 = 4 y 3 y = [Pb 2+ ( aq )] = 2.6 x 10 -13 ________ 4 = 4.0 x 10 -5 1/3 M 4.0 x 10 -5 moles of Pb(IO 3 ) 2 are present in 1 L of saturated solution ____ mol 557 g grams of Pb(IO 3 ) 2 present in 1 L of saturated solution = 4.0 x10
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/13/2008 for the course CHEM 102a taught by Professor Hanusa during the Fall '06 term at Vanderbilt.

Page1 / 17

Chapter-9[1] chem - Chapter 9: Dissolution and...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online