K sp > K trial 1.0 > 0.50 Therefore this system must shift to the RIGHT to produce more products in order to achieve equilibrium

Transcription

1 Chemistry 12 UNIT 3 SATURATION and PRECIPITATION SATURATED: -the state of a solvent in which no more solute will dissolve. The solubility of a solid will be limited: The solid will dissolve until the concentration of the solute reaches a maximum in the solution. This is SATURATION. IMPORTANT NOTE: A substance forms a saturated solution if the dissolved substance is in Equilibrium with some of the undissolved substance. IN OTHER WORDS: To translate and relate this to UNIT 2: SATURATED in Unit 3 is the same as EQUILIBRIUM in Unit 2. CONSIDER: A + B C + D I 2.0 M 1.0 M 1.0 M 1.0 M C E Given that Keq for this equilibrium reaction is 1.0, determine whether the inital data given is at equilibrium. K trial = (1.0)(1.0) / (2.0) (1.0) = 0.50 If it is not, determine which way the equilibrium must shift in order for this system to reach equilibrium. K sp > K trial 1.0 > 0.50 Therefore this system must shift to the RIGHT to produce more products in order to achieve equilibrium

2 NOW CONSIDER: AlPO 4 (s) Al 3+ (aq) + PO 4 3- (aq) I X 0.50 M 0.50 M C E Given that Ksp for this equilibrium reaction (i.e. soluble salt in solution) is 6.0 X Calculate K trial and determine in which direction the above system would shift to achieve equilibrium. K trial = (0.50) (0.50) = 0.25 K sp < K trial 6.0 X <0.25 THEREFORE, this would shift to the LEFT to achieve equilibrium Determine whether a Precipitate forms or in other words, determine whether this solution, at its INITIAL state, is UNSATURATED, SATURATED, or SUPERSATURATED. This system is SUPERSATURATED initially since at I, K sp < K trial The system must shift to the left to achieve equilibrium.

3 Ksp versus K trial: Determining Precipitation and Saturation K sp > K trial UNSATURATED SOLUTION: -contains less than the maximum amount of a substance which can dissolve -therefore more solute may be dissolved in the solution -no undissolved solid present -not at equilibrium -a shift to the right is still possible NO PRECIPITATE K sp = K trial K sp < K trial SATURATED SOLUTION: -maximum amount of substance dissolved in solution -some undissolved material present -equilibrium exists between dissolved and undissolved material rate of ions forming (dissociation) is equal to rate of crystallization (or precipitation) EQUILIBRIUM = SATURATION SUPERSATURATED SOLUTION: -excess salt (more than the maximum amount of dissolved salt in saturated solution) crystallizes out of solution -ppt continues to form until all excess ions have been removed from solution and the value of K trial has been reduced to the value of Ksp (Equilibrium has shifted to the left) NO PRECIPITATE OR The minimum possible precipitate forms (in equilibrium with the amount of dissolved ions in solution) PRECIPITATE FORMS

4 SOLUBLE vs. LOW SOLUBLE SOLUBLE High Ksp - this can be a dangerous generalization - best to base the level of solubility on "s" Solubility > 0.1 M at RTP Easily dissolved in water (i.e. easily forms ions in water) Ions compatible as separate particles Low bonding drive LOW SOLUBLE Low Ksp - this can be a dangerous generalization - best to base the level of solubility on "s Solubility < 0.1 M at RTP Relatively insoluble in water (does not readily form ions in water) Ions incompatible as separate particles High bonding drive CONSIDER: A molecule, AB, with a Ksp = 25 Another molecule, A 3 B 2, with a Ksp = 25 For AB, Ksp = s 2 25 = s 2 s = 5 For A 3 B 2, Ksp = 108s 5 25 = 108s 5 s = 0.75 AHA! The molecule AB is more soluble!

5 Using your DATA BOOKLET USE THE SOLUBILITY TABLE IN YOUR DATA BOOKLET TO DETERMINE WHICH ARE INSOLUBLE IN WATER: NaOH CaSO 4 PbCl 2 KCl CaBr 2 USING THE SOLUBILITY TABLE: mostly soluble mostly low solubility SULPHATE SULPHIDE PLEASE NOTE: Compounds containing alkali metals, or H+, NH4 +, NO3 - will be soluble in water. LOW SOLUBLE means a precipitate forms.

6 DOES A PRECIPITATE FORM AND WHAT IS IT? FOR EACH OF THE FOLLOWING QUESTIONS, you should be able to: i) Write the dissociation equations for each given formula ii) Use solubility table to determine precipitates in the products formed iii) Write the Formula equation (Show the double replacement reaction) iv) Write the Complete ionic equation (with spectator ions) v) Write the Net Ionic equation (Precipitation equation) a) NaOH and HCl b) Bi(NO3) 2 and NaOH c) Pb(C 2 H 3 O 2 ) 2 and K 2 SO 4 d) CuSO 4 and FeCl 3 e) FeSO 4 and (NH 4 ) 2 S f) K 2 CO 3 and Sr(NO 3 ) 2 g) NaCl and KOH h) NaI and AgNO 3 i) Al 2 SO 3 and CaCl 2 j) Na 2 SO 4 and CaCl 2 k) K 2 SO 4 and Ba(C 2 H 3 O 2 ) 2 l) MgSO 4 and CaBr 2

7 UNIT 3 SAMPLE CALCULATIONS: 1. Recall previous example: AlPO4 (s) Al3+ (aq) + PO4 3- (aq) I X 0.50 M 0.50 M R E a) Given that Ksp for this equilibrium reaction (i.e. soluble salt in solution) is 6.0 X 10-13, Calculate the K trial. b) Considering the K trial that you have calculated, determine in which direction the above equilbrium would shift. OR IN OTHER WORDS, determine if a precipitate would form. c) What would the new equilbrium concentrations of each ion be? OR IN OTHER WORDS, what is the concentration of each ion in the saturated solution? d) HOW MUCH PRECIPITATE FORMS if there was originally 0.25 moles of AlPO4(s) dissolved in 500. ml of solution. 2. The ion concentrations in 2.00 L of 0.32 M K3PO4 are: 3. Which is the least soluble in water: a) CaS b) Fe(OH)3 c) KMnO4 d) NH4HC2O4 4. A solution contains two cations, each having a concentration of 0.20 M. When an equal volume of 0.20 M OH- is added, these cations are removed from the solution by precipitation. These ions are: a) Ba2+ and K+ b) Sr2+ and Na+ c) Sr2+ and Mg2+ d) Mg2+ and Ca2+ 5. The solubility of Mn(IO3)2 is 4.8 x What is the value of the Ksp? 6. Write the net ionic equation representing the reaction that occurs when 50.0 ml of 0.20 M ZnSO4 and 50.0 ml of 0.20 M BaS are combined. 7. When 1.00 g of Mg CO3 is added to 2.0 L of water, some, but not all, will dissolve to form a saturated solution. Calculate the mass of solid that remains undissolved. ANSWERS IN CLASS. Will NOT be posted on Website.

8 8-2 Unit 3 Sample Calculations ANSWERS

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11 Worksheet 3.2: Ksp Calculations 1. Calculate the solubility product of the following: a) Mg(OH) 2 whose solubility is 1.4 x 10-4 M b) La(IO 3 ) 3 whose solubility is 6.9 x 10-4 M 2. The solubility of PbSO 4 in water is 3.8 x 10-2 g/l. Calculate the Ksp of PbSO [Ag + ] = 2.2 x 10-4 M in a saturated solution of Ag 2 C 2 O 4. Determine the solubility product of the compound. 4. SrF 2 has a Ksp = 2.18 x Determine the concentrations of the strontium and fluoride ions in a saturated solution at that temperature. 5. How many grams of PbI 2 will dissolve in 250. ml of water if its solubility product equals 1.7 x 10-5? ANSWERS 1a) Ksp = [Mg +2 ] [OH - ] 2 = (1.4 x 10-4 ) (2.8 x 10-4 ) 2 = 1.1 x b) La(IO3)3 Ksp = 6.4 x Ksp of PbSO4 = 1.7 x Ag2C2O4 Ksp = 5.3 x [Sr +2 ]=1.76x10-3 M and [F - ] = 3.52x10-3 M g of PbI2

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