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Solubility Product and Common ion effect

Experiment #9. Solubility Product and Common ion effect. 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

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Solubility Product and Common ion effect

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  1. Experiment #9 Solubility Product and Common ion effect

  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,Ca3(PO4)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 MyXz (s) yMZ+ (aq) + zXY-(aq) Molar solubility of the ions Solubility product constant

  5. Solubility product Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq) Molar solubility of the ions Solubility product constant 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 Let us say, we try to dissolve 1 g of Bi2S3 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 Bi2S3. 2. If we filter out the undissolved Bi2S3, the amount of solute that dissolved (soluble) in 1. 0 L of water is 0.0025 g. Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq) So we can say, the solubility of Bi2S3 is 8.78 ×10-13 g per liter

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

  9. Solubility product constant Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq) Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq) If we wanted to figure out the Ksp of Bi2S3, then we need to know the molar solubilities of Bi3+ and S2-. The molar solubilities of the ions are usually figured out from the solubility of the parent compound.

  10. Solubility product constant If the molar solubility of Bi2S3 is “s”, the molar solubility of Bi2+ is “2s” and the molar solubility of S2- is “3s”. Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq) s 2×s 3×s This is because, there are 2 ions of Bi3+ produced for Each molecule of the parent, Bi2S3 and 3 ions of S2- produced for each molecule of the parent.

  11. Solubility product constant Since we already know the value of molar solubility for Bi2S3, which is 1.708×10-15 M

  12. How to find the molar solubility if we know is Ksp? Find the molar solubility of Ca (OH)2, if the Ksp of Ca(OH)2 is 7.9 ×10-6. Ca(OH)2 (s) Ca2+ (aq) + 2OH-(aq) s 1×s 2×s Let the molar solubility of Ca(OH)2 be “s”. So, the molar Solubility of Ca2+ should be “1s” and the molar solubility of OH- should be “2s”. This is because, there are 1 ion of Ca2+ 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 Ksp? Ca(OH)2 (s) Ca2+ (aq) + 2OH-(aq) s 1×s 2×s

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

  15. How to find the molar solubility if we know is Ksp? So the molar solubility of Ca(OH)2 = s = 1.25 ×10-2 M The molar solubility of Ca2+ = [Ca2+] =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. COOK H-C-OH H-C-OH COOH Experiment- To determine the Ksp of Potassium Hydrogen Tartarate, KHT KHT is also called cream of tartar KHT (s) K+ (aq) + HT-(aq) s 1×s 1×s If we want to determine the Ksp of KHT, we need to know the molar solubilities of K+ and HT-. Also remember that Ksp is measured for a saturated solution. KHT

  19. COOK H-C-OH H-C-OH COOH 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. 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-] KHT

  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 Ksp constant.

  23. Calculate the molar solubility of lead iodide PbI2, from its Ksp in water at 25C PbI2 (s) Pb2+ (aq) + 2I-(aq) s 1×s 2×s

  24. Calculate the molar solubility of lead iodide PbI2, in 0.1 M NaI solution PbI2 (s) Pb2+ (aq) + 2I-(aq) s 1×s 2×s NaI (s) Na+ (aq) + I-(aq) 0.1M 0.1M Common ion

  25. Calculate the molar solubility of lead iodide PbI2, in 0.1 M NaI solution Because the Ksp of PbI2 is really small, the solubility s is going to be really small. Hence we can make a simplification.

  26. A comparison of solubility of PbI2 With and without common ion Without common ion With common ion s= 7.9 × 10-7 M s= 1.3 × 10-3 M Solubility decreases because of the presence of common ion

  27. 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 Common ion HT- can act an acid, so if we titrate it with a known concentration of base (NaOH), we can find the [HT-]

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