{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter-9

# Chapter-9 - Chapter 9 Dissolution and Precipitation...

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

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

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

View Full Document
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 staurated 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
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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

Chapter-9 - Chapter 9 Dissolution and Precipitation...

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

View Full Document
Ask a homework question - tutors are online