Successfully reported this slideshow.
Upcoming SlideShare
×

# Chapter 18.4 : Solubility Equilibrium

9,454 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• I'd advise you to use this service: ⇒ www.HelpWriting.net ⇐ The price of your order will depend on the deadline and type of paper (e.g. bachelor, undergraduate etc). The more time you have before the deadline - the less price of the order you will have. Thus, this service offers high-quality essays at the optimal price.

Are you sure you want to  Yes  No
• Dating for everyone is here: ♥♥♥ http://bit.ly/2ZDZFYj ♥♥♥

Are you sure you want to  Yes  No

Are you sure you want to  Yes  No

### Chapter 18.4 : Solubility Equilibrium

1. 1. Chapter 18.4
2. 2. 1. Explain what is meant by solubility product constants, and calculate their values. 2. Calculate solubilities using solubility product constants. 3. Carry out calculations to predict whether precipitates will form when solutions are combined.
3. 3. • A saturated solution contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance. • A saturated solution is not necessarily a concentrated solution. • The equilibrium principles developed in this chapter apply to all saturated solutions of sparingly soluble salts.
4. 4. • The heterogeneous equilibrium system in a saturated solution of silver chloride containing an excess of the solid salt is represented by  - AgCl( ) Ag ( Cl ( )s aq) + aq • The solubility product constant, Ksp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation. • The equation for the solubility equilibrium expression for the dissolution reaction of AgCl is spK - [Ag ][Cl ]  • The equilibrium expression is written without including the solid species. • The numerical value of Ksp can be determined from solubility data.
5. 5. • For a saturated solution of CaF2, the equilibrium equation is spK 2 – 2 [Ca ][F ]    2 – 2CaF ( ) Ca ( ) 2F ( )s aq aq • The expression for the solubility product constant is • The solubility of CaF2 is is 8.6  10−3/100 g of water at 25°C. Expressed in moles per liter this concentration becomes 1.1  10−3 mol/L.
6. 6. • CaF2 dissociates to yield twice as many F− ions as Ca2+ ions. [Ca2+] = 1.1  10−3 mol/L [F− ] = 2.2  10−3 mol/L   spK 2 2 [Ca ][F ] spK –3 3 2 (1.1 10 )(2.2 10 )    Ksp = 5.3  10-9 • Calculations of Ksp ordinarily should be limited to two significant figures.
7. 7. • The solubility product constant is an equilibrium constant representing the product of the molar concentrations of its ions in a saturated solution. • The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent. • It has only one value for a given solid at a given temperature. • It has an infinite number of possible values at a given temperature and is dependent on other conditions, such as the presence of a common ion.
8. 8. • Calculate the solubility product constant, Ksp ,for copper(I) chloride, CuCl, given that the solubility of this compound at 25°C is 1.08  10–2 g/100. g H2O. 2 2 2 1 g H Og CuCl 1000 mL 1 molCuCl solubility in mol/L 100. g H O 1 mL H O 1 L g CuCl       – CuCl( ) Cu ( ) Cl ( )s aq aq Ksp=[Cu+][Cl–] Unknown: Ksp Given: solubility of CuCl = 1.08  10−2 g CuCl/100. g H2O Solution: [Cu+] = [Cl–] = solubility in mol/L
9. 9. –2 2 2 2 solubility in mol/L 1 g H O1.08 10 g CuCl 1000 mL 1 molCuCl 100. g H O 1 mL H O 1 L 99.0 g CuCl       1.09  10-3 mol/L CuCl Ksp = (1.09  10-3)(1.09  10-3) = 1.19  10-6 [Cu+] = [Cl–]=1.09  10-3 mol/L
10. 10. • The solubility product constant can be used to determine the solubility of a sparingly soluble salt.   2 2– 3 3BaCO ( ) Ba ( ) CO ( )s aq aq spK 2 2– –9 3[Ba ][CO ] 5.1 10    x2 2– 3[Ba ] [CO ]   2 spK x x x2 2– –9 3[Ba ][CO ] ( )( ) 5.1 10      x –9 –5 5.1 10 7.1 10    • The molar solubility of BaCO3 is 7.1  10−5 mol/L. • How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C?
11. 11. 1. Calculate the solubility of silver bromide, AgBr, in mol/L, using the Ksp value for this compound. Given: Ksp = 5.0 10−13 Unknown: solubility of AgBr Solution: [Ag+] = [Br−], so let [Ag+] = x and [Br−] = x  s aq) + aqAgBr( ) Ag ( Br ( ) spK – [Ag ][Br ]  spK x x x– 2 [Ag ][Br ] ( )( )    x2 –13 5.0 10  x –13 5.0 10  7 –13 solubility 7 of AgBr 5. .1 1 0 10 0 mol/L    
12. 12. • The equilibrium condition does not require that the two ion concentrations be equal. Equilibrium will still be established so that the ion product does not exceed the value of Ksp for the system. • If the ion product is less than the value of Ksp at a particular temperature, the solution is unsaturated. • If the ion product is greater than the value for Ksp, solid precipitates.
13. 13. • Unequal quantities of BaCl2 and Na2CO3 are dissolved in water and the solutions are mixed. • If the ion product exceeds the Ksp of BaCO3, a precipitate of BaCO3 forms. • Precipitation continues until the ion concentrations decrease to the point at which equals the Ksp. • The solubility product can be used to predict whether a precipitate forms when two solutions are mixed.
14. 14. 1. Will a precipitate form if 20.0 mL of 0.010 M BaCl2 is mixed with 20.0 mL of 0.0050 M Na2SO4? Given: concentration of BaCl2 = 0.010 M volume of BaCl2 = 20.0 mL concentration of Na2SO4 = 0.0050 M volume of Na2SO4 = 20.0 mL Unknown: whether a precipitate forms Solution: The two possible new pairings of ions are NaCl and BaSO4. BaSO4 is a sparingly soluble salt.
15. 15.   2 2– 4 4BaSO ( ) Ba ( ) SO ( )s aq aq spK 2 2– –10 4[Ba ][SO ] 1.1 10    2 20.010 mol Ba 0.020 L 0.000 20 mol Ba 1 L     2 2– 2–4 – 4 4 0.0050 mol SO 0.020 L 0.000 10 mol SO 1 L mol SO ion:   mol Ba2+ ion:
16. 16. total volume of solution: 0.020 L + 0.020 L = 0.040 L concentration Ba2+ ion in combined solution: 2 –3 20.000 20 mol Ba 5.0 10 mol/L Ba 0.040 L     2– –3 2– 4 4 2– 4 0 co .000 10 mol SO 2.5 10 mol/L SO 0.040 L ncentration SO ion in combined solution :   the ion product: 2 2– –3 –3 –5 4[Ba ][SO ] (5.0 10 )(2.5 10 ) 1.2 10      spK –5 –10 2 2– 4 1.2 10 1.1 10 [Ba ][SO ]     Precipitation occurs.
17. 17. • The solubility product principle can be very useful when applied to solutions of sparingly soluble substances. • It cannot be applied very successfully to solutions of moderately soluble or very soluble substances. • The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together. • Sometimes it is necessary to consider two equilibria simultaneously.