LARGEK sp s indicate a more soluble salt.

Transcription

1 UNIT 3B SOLUBILITY CALCULATIONS Chemistry 12 Solubility Equilibrium Calculations If a salt is only slightly soluble in water, it can easily form a saturated solution = equilibrium Just like for any equilibrium a K eq expression can be written For the dissociation of low solubility salts, the K eq is called K sp Ex. The salt PbI 2(s) Very little PbI 2(s) is required to form a saturated aqueous solution: PbI 2(s) Pb 2+ (aq) + 2I - (aq) The equilibrium expression corresponding to the dissociation equation above is called the solubility product expression (K sp ): Read the questions carefully!!!! If you are asked to predict if a compound has a low solubility, just look at the SOLUBILITY OF COMMON COMPOUNDS IN WATER table (like before) If you have to compareor calculate solubilities, use the SOLUBILITY PRODUCT CONSTANTS AT 25 o C (pg. 5 in your Data Booklet). LARGEK sp s indicate a more soluble salt. smallk sp s indicate a less soluble salt. 1

2 Chart problems Two types of problems to start with: 1. Find a K sp given the solubility of a compound. 2. Find the concentrations of the species given the K sp. For everyk sp problem you should: Write out the equilibrium equation showing the dissolvingof the salt. Write out the solubility product expression. 2

3 Homework: Read: Pages Do: #40-55 Predicting if a Precipitate Will Form When we have two solutions containing ions, one can calculate a trial value for K sp to see if a precipitate will form. If we mix Ag + (aq)and Cl - (aq), we have to consider whether there is a sufficient concentration of both ions to establish the equilibrium: AgCl (s) Ag + (aq)+ Cl - (aq) We can then define the ION PRODUCT or TRIAL ION PRODUCT (TIP) or TRIAL K sp as: Q = [Ag + ] START [Cl - ] START The values of K sp and Q for a sparingly soluble salt are defined as follows: Q = the product of the ion concentrations which actually exist in solution. the product of the ion concentrations K sp = required to establish a solubility equilibrium. If we mix two solutions there are three possible outcomes: CASE 1: Q < K sp In this case, there are not enough ionsin solution to form a precipitate. a precipitate will not form if Q < K sp 3

4 CASE 2: Q = K sp In this case there are exactly enough ions to form a saturated solution. There will not be any precipitate forming at this point. a saturated solution is formed if Q = K sp CASE 3: Q > K sp In this case there is an excess of ions, over the amount needed to form a precipitate. A precipitate will form and will continue to form until the excess ions have been removed from solution and the value of Q will be reduced down to the value of K sp. a precipitate willform if Q > K sp 4

5 Homework: Read: Pages Do: #56 69 Study for quiz!!! The Application of Solubility Principles to Chloride Titrations Remember titrations? (we talked about it once in chem 11!) In a Chlorine titration, we are dealing with the following equation: Ag + (aq)+ Cl - (aq) AgCl (s) Some solution containing an unknown concentration of Chlorine ions is titrated with a Silver solution (usually AgNO 3(aq) ) of a known concentration. The precipitate formed is AgCl (s). 5

6 We now have a problem The precipitate formed is white. when do we stop the titration? The colouror its intensity does not change when the titration is complete. We have no idea when we ve reached the equivalence point(when the stoichiometry of the reaction is satisfied). or when mol Ag + added = mol Cl - present In comes a Chromate indicator!!! As the Ag + (aq)solution is added, it precipitates with the Cl - (aq)ion first to form AgCl (s). (K sp for AgCl= 1.8x10-10 vs. K sp for AgCrO 4 = 1.0x10-5 ) The colourof the precipitate will be milky yellow (white AgCl (s) combined with the yellow chromate indicator). As the concentration of the Cl - gets lower and lower, we see a red precipitate start to form. Eventually all the Cl - is used up and any added Ag + will begin to precipitate with the CrO 4. As soon as we see the first appearance of the red Ag 2 CrO 4 precipitate that does not disappear, we stop our titration. At this point all of the Cl- will be gone (precipitated as AgCl). If everything was done correctly, we have reached our endpoint! DO NOT USE K sp s!!! (for titrations!) 6

7 Homework: Read: Pages Lab Do: #

8 Common Ion Effect The lowering/increasing of the solubility of one salt by adding a second salt which will affect the equilibrium established in the saturated solution. A direct consequence of Le Chatelier s principle. Decreasing the Solubility of a Salt Assume we have a saturated solution of CaSO 4 CaSO 4(s) Ca 2+ (aq) + SO 4 (aq) We can increase the amount of solid CaSO 4 in the solution by: 1. Increasing the [Ca 2+ ]: This will cause the equilibrium to shift the solid side (left). 2. Increasing the [SO 4 ]: Again, this causes the equilibrium to shift to the solid side (left). NOTE: We cannot simultaneously increase the [Ca 2+ ] and [SO 4 ] because [Ca 2+ ][SO 4 ] must be a constant. We can only increase the concentration of one ion so as to decrease the concentration of the other ion and cause a precipitate to form. Increasing the Solubility of a Salt We can decrease the amount of CaSO 4(s) that goes into solution by: 1. Adding an ion such as PO 4 3-, which precipitates the Ca 2+ (aq): CaSO 4(s) Equilibrium shifts right Ca 2+ (aq)+ SO 4 (aq) + PO 4 3- (aq) Ca 3 (PO 4 ) 2(s) Concentration of Ca 2+ (aq) decreases. and more CaSO 4(aq) dissolves. 8

9 2. Adding an ion such asag +, which precipitates theso 4 (aq): CaSO 4(s) Ca 2+ (aq) + SO 4 (aq) + Ag + (aq) Ag 2SO 4(s) Concentration of SO 4 (aq) decreases. Equilibrium shifts right and more CaSO 4(aq) dissolves. Removing Hardness From Water What is hard water? Hardness in water results from the presence of Mg 2+ and Ca 2+. The presence of Ca 2+ or Mg 2+ directly: causes a bitter taste in the water leaves white deposits whenever the water is heated or evaporates prevents the proper cleaning action of soaps What affects the soap? The effective cleaning ingredient in most soap is a large organic ion called the stearate ion: C 17 H 35 COO -. Calcium and Magnesium stearate are quite insoluble: K sp = 1.0 x10-12 for Ca(C 17 H 35 COO) 2 K sp = 5.0 x for Mg(C 17 H 35 COO) 2 The presence of Ca 2+ or Mg 2+ in soapy water causes a gray/white curd-like precipitate of Calcium and Magnesium stearate to form. This removes both unwanted ions, but it also uses up some of the soap. Even more soap is then required to build up a sufficient concentration of stearate ion to permit proper cleansing action. 9

10 How can we get rid of the Ca 2+ and Mg 2+ in hard water, inexpensively, so as to soften the water? Permanently Hard Water: Add washing soda (Na 3 CO 3(s) ). Homework: Do: #76-86 Ca 2+ (aq)+ Na 3 CO 3(aq) Water softeners can do this. CaCO 3(s) + 3Na + (aq) Temporarily Hard Water: Contains carbonates that can be removed by heating: Ca 2+ (aq)+ 2 HCO 3 - (aq) + energy <->CaCO 3(s) + H 2 O (l) + CO 2(g) 10