1. Solubility Product and Common ion effect Experiment #9

2. What are we doing in this experiment? Determine the molar solubility and solubility product constant (Ksp) of potassium hydrogen tartarate (KHT). Study the effect of common ion on the Ksp of KHT and the molar solubilites of its ions.

3. Remember!! In this experiment, we are dealing with compounds that are very slightly soluble that they are called “Insoluble compounds”. Our bones and teeth are mostly calcium phosphate, Ca 3 (PO 4 ) 2, a very slightly soluble compound.

4. Solubility product In general, the solubility product expression for a compound is the product of the concentration (molar solubility) of its constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. The quantity is constant at constant temperature for a saturated solution of the compound.This statement is called the solubility product principle M y X z (s) yM Z+ (aq) + zX Y- (aq) Solubility product constant Molar solubility of the ions

5. Solubility product Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2- (aq) Solubility product constant Molar solubility of the ions Remember that we are dealing with molar solubilities and not concentration. For a saturated solution, molar solubility is equal to molar concentration.

6. Different types of solution Unsaturated solution: More solute can be dissolved in it. Saturated solution: No more solute can be dissolved in it. Any more of solute you add will not dissolve. It will precipitate out. Super saturated solution: Has more solute than can be dissolved in it. The solute precipitates out.

7. Molar solubility Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2- (aq) Let us say, we try to dissolve 1 g of Bi 2 S 3 in 1 L of water. If only 8.78×10 -13 g out of the 1.0 g dissolves, we can make the following conclusions: 1. The solution is saturated with Bi 2 S 3. 2. If we filter out the undissolved Bi 2 S 3, the amount of solute that dissolved (soluble) in 1. 0 L of water is 0.0025 g. So we can say, the solubility of Bi 2 S 3 is 8.78 ×10 -13 g per liter

8. Molar solubility Molar solubility is solubility in moles per liter

9. Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2- (aq) Solubility product constant Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2- (aq) If we wanted to figure out the K sp of Bi 2 S 3, then we need to know the molar solubilities of Bi 3+ and S 2-. The molar solubilities of the ions are usually figured out from the solubility of the parent compound.

10. Solubility product constant Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2- (aq) If the molar solubility of Bi 2 S 3 is “s”, the molar solubility of Bi 2+ is “2s” and the molar solubility of S 2- is “3s”. s 2×s 3×s This is because, there are 2 ions of Bi 3+ produced for Each molecule of the parent, Bi 2 S 3 and 3 ions of S 2- produced for each molecule of the parent.

11. Solubility product constant Since we already know the value of molar solubility for Bi 2 S 3, which is 1.708×10 -15 M

12. How to find the molar solubility if we know is K sp ? Find the molar solubility of Ca (OH) 2, if the K sp of Ca(OH) 2 is 7.9 ×10 -6. Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) Let the molar solubility of Ca(OH) 2 be “s”. So, the molar Solubility of Ca 2+ should be “1s” and the molar solubility of OH - should be “2s”. s 1×s 2×s This is because, there are 1 ion of Ca 2+ produced for each molecule of the parent, Ca(OH) 2 and 2 ions of OH - produced for each molecule of the parent.

13. How to find the molar solubility if we know is K sp ? Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) s 1×s 2×s

14. How to find the molar solubility if we know is K sp ?

15. How to find the molar solubility if we know is K sp ? So the molar solubility of Ca(OH) 2 = s = 1.25 ×10 -2 M The molar solubility of Ca 2+ = [Ca 2+ ] =1s = 1.25 ×10 -2 M The molar solubility of OH - = [OH - ] = 2s = 2×1.25 ×10 -2 M 2.50 ×10 -2 M

16. Is it possible to find the pH of the Ca(OH) 2 solution at 25  C? Yes We know that [OH - ] = 2.5 ×10 -2 M

17. Is it possible to find the pH of the Ca(OH) 2 solution at 25  C? The Ca(OH) 2 solution is basic.

18. Experiment- To determine the K sp of Potassium Hydrogen Tartarate, KHT KHT is also called cream of tartar H-C-OH COOH COOK KHT KHT (s) K + (aq) + HT - (aq) s 1×s 1×s If we want to determine the K sp of KHT, we need to know the molar solubilities of K + and HT -. Also remember that K sp is measured for a saturated solution.

19. How do we determine [K + ] and [HT - ]? Firstly prepare a saturated solution of KHT. 3.0 g of KHT in 200 ml of water. Filter out the undissolved KHT using gravity filtration. Now we have a saturated solution of KHT. H-C-OH COOH COOK KHT KHT (s) K + (aq) + HT - (aq) s 1×s 1×s HT - can act an acid, so if we titrate it with a known concentration of base (NaOH), we can find the [HT - ]

20. How do we determine [K + ] and [HT - ]? Once we know the concentration of HT-, based on 1 to 1 molar relationship between K + and HT -, [K + ]= [HT - ] NaOH is hygroscopic, so the NaOH solution needs to be standardized by using KHP

21. LeChatelier’s Principle If a stress (change of condition) is applied to a system at dynamic equilibrium,the system shifts in the direction that reduces the stress. Common ion effect Suppression of ionization of a weak electrolyte by the presence in the same solution of a strong electrolyte containing one of the same ions as the weak electrolyte.

22. About Common ion effect Common ion effect is a special case of LeChatelier principle Addition of a common ion is equivalent to adding a stress to the system. The system responds to the stress by reducing the solubility of one of the ions and keeping the K sp constant.

23. Calculate the molar solubility of lead iodide PbI 2, from its K sp in water at 25  C PbI 2 (s) Pb 2+ (aq) + 2I - (aq) s 1×s 2×s

25. Calculate the molar solubility of lead iodide PbI 2, in 0.1 M NaI solution PbI 2 (s) Pb 2+ (aq) + 2I - (aq) s 1×s 2×s NaI (s) Na + (aq) + I - (aq) 0.1M 0.1M Common ion

26. Calculate the molar solubility of lead iodide PbI 2, in 0.1 M NaI solution Because the K sp of PbI 2 is really small, the solubility s is going to be really small. Hence we can make a simplification.

27. A comparison of solubility of PbI 2 With and without common ion With common ion Without common ion s= 1.3 × 10 -3 M s= 7.9 × 10 -7 M Solubility decreases because of the presence of common ion

28. To study the effect of common ion on the solubility of KHT KHT (s) K + (aq) + HT - (aq) s 1×s 1×s KCl (s) K + (aq) + Cl - (aq) 0.1 M 0.1 M HT - can act an acid, so if we titrate it with a known concentration of base (NaOH), we can find the [HT - ] Common ion