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Solution Equilibrium

Solution Equilibrium. Previous Knowledge – 30S Chem – Solutions, Unit 1, and Equilibrium Content – p. 482 - 493. CYU What is solubility? When is a substance soluable ? When is it not? What is a saturated solution? Non-Saturated solution? What factors affect solubility?

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Solution Equilibrium

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  1. Solution Equilibrium Previous Knowledge – 30S Chem – Solutions, Unit 1, and Equilibrium Content – p. 482 - 493

  2. CYU • What is solubility? • When is a substance soluable? When is it not? • What is a saturated solution? Non-Saturated solution? • What factors affect solubility? • What is a molar solution? • What is ionization? Intro to Solution Equilibrium

  3. Substances chemists believed to be insoluble in water, by experiment, were shown to be slightly soluble. • With sensitive instruments, chemists we able to show that substances that would not dissolve in water actually do dissolve to a very small extent. • Dissolving is a dynamic equilibrium between the ions formed and the substance in its whole or non-ionized form Ksp or Solubility Product Constant

  4. As we learned in Acids and Bases, a strong acid or base is one that ionized 100%. In other words, all ionized. • The equilibrium favours the formation of product. • Weak acids and bases do not ionize 100% and only partially, therefore, putting only a little hydrogen ion or hydroxide ion into solution. • The equilibrium favours the formation of reactant. Ksp or Solubility Product Constant

  5. For salts, when a solution is saturated, there is a point when the production of ions (dissolving) is in equilibrium with the ions coming back together to form the salt. • This is called dynamic equilibrium Animation • The more ions that form (ie. Very soluable), the greater the equilibrium constant (Keq is greater than 1) • The less ions that form (ie. Low solubility), the lower the equilibrium constant (Keq is less than 1) Ksp or Solubility Product Constant

  6. Very soluable substance, dissociate 100% Ex. NaCl Na+1 + Cl-1 (100 % ions until it is saturated) • Slightly soluable substance, does not dissociate to the same extent (ie. Less ions in solution that NaCl, at equilibrium) Ex. AgCl Ag+1 + Cl-1 • Notice the arrows, which is favoured at equilibrium • But some does dissolve, just a very small amount goes to ions. Ksp or Solubility Product Constant

  7. When a sparingly soluble ionic solid is dissolved in water to form a saturated solution the general equilibrium equation is: AaBb(s) aA+ (aq) + bB¯ (aq) • Where A is a positively charged ion and B is a negatively charged ion. • For very soluable substances, we know that 100% will form ions at equilibrium Ksp or Solubility Product Constant

  8. At a given temperature, the equilibrium law for this reaction is given as: Keq = [A+]a[B-]b [AaBb] • However, the term Keq[AaBb] can be replaced by a new constant, Ksp, called the solubility product. Ksp = [A+]a[B¯]b • The solubility product constant is the product of ion concentrations in a saturated solution. The solubility product constant takes into account the presence of the solid. Ksp or Solubility Product Constant

  9. Ksp is the same as Keq and Ka / Kb, except it is the measure of the ion concentration at equilibrium (Saturation) • For highly soluable salts, Ksp will be greater than 1 • We don’t care about these salts, because for the most part, we can put a large amount into solution, and ions will form. • The slightly soluable salts are the ones we are concerned about • For slightly soluable salts, Ksp will be less than 1 • Why are the slightly soluable salts a concern when we look at nature? Precipitates Ksp or Solubility Product Constant

  10. Assignment • You must come up with one example of how slightly soluable salts would affect us in our daily lives and how they are used? • Examples – City of Winnipeg Water, Kidney stones, osteoporosis, tooth decay and prevention, limestone caves, Lake Winnipeg • Due next class Ksp or Solubility Product Constant

  11. Example 1 • Write the dissociation equation and the expression for the solubility product constant for calcium hydroxide.(Ksp = 1.3 x 10-6) Solution Write the chemical formula for calcium hydroxide, Ca(OH)2. Write the dissociation equation, remembering that the subscript indicates two hydroxide ions dissociate for every calcium ion: Ca(OH)2(s) Ca2+(aq) + 2 OH¯(aq) When writing the solubility product expression, the molar coefficient in the dissociation equation becomes the exponent. Ksp = [Ca2+][OH¯]2 Ksp or Solubility Product Constant

  12. Example 2 • Write the solubility product expression for lead (II) phosphate (Pb3(PO4)2). • Example 3 • What are the concentrations of ions for a saturated solution of calcium hydroxide? (Ksp = 1.3 x 10-6) • Example 4 • What is the Ksp value for barium sulfate, if at equilibrium, barium sulfate has a concentration of 3.87 x 10-5 M? • Example 5 • If a saturated solution of magnesium fluoride has a Mg+2 ion concentration of 1.17 x 10 –3 M, what is the Ksp?

  13. Solubility and solubility product are two different terms. • Solubility is the maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature. Solubility has an infinite number of possible values, depending on temperature and other solutes present. • Solubility product is an equilibrium constant and has only one value for a given solid at a particular temperature. Ksp or Solubility Product Constant

  14. Example 6 • The solubility of PbF2 is 0.466 g/L. What is the value of the solubility product? Step 1: Determine the number of moles/L of PbF2 Mols = mass/ MM Mols = 0.466g/245.20g/mol So, the concentration of PbF2 is 1.90 x 10-3 mol/L. Step 2: Write the dissociation equation. PbF2(s) Pb2+(aq) + 2 F¯(aq)

  15. Step 3: Calculate ion concentrations. • The concentration of PbF2 an Pb2+ ions will be equal since their stoichiometry is 1:1. • So, [Pb2+] = 1.90 x 10-3 mol/L • From the stoichiometry, • [F¯] = 2 x [Pb2+] = 3.80 x 10-3 mol/L • Step 4: Write the solubility product expression and calculate. • Ksp = [Pb2+][F¯]2 • Ksp = (1.90 x 10-3)(3.80 x 10-3)2 • Ksp = 2.74 x 10-8 • Therefore, the solubility product of PbF2 is 2.74 x 10-8.

  16. Example 7 • The Ksp of magnesium hydroxide is 8.9 x 10-12. What will be the equilibrium concentrations of the dissolved ions in a saturated solution of Mg(OH)2? • Example 8 • What is the hydroxide concentration in a saturated solution of iron (III) hydroxide? • (Ksp = 4.00 x 10-38) Answer = 5.89 x 10-10 M • Example 9 • What mass of silver (I) sulfate could be dissolved in 500 mL of water? (Ksp = 1.20 x 10-5) • Answer = 2.18 g

  17. Insoluble compounds are generally described as those that precipitate upon mixing equal volumes of solutions which are 0.10 mol/L in the respective ions. • A precipitate is the solid that forms when two solutions are mixed. A precipitate will usually cause a mixture to appear cloudy. • In previous chemistry courses you have learned the general rules for determining combinations of ions that have a low solubility. Solubility Rules and Precipitates

  18. Example • Which of the following ions will precipitate with a solution of sodium chloride? • Copper (II) ion Lead (II) ion Silver ions Sulfide ions • Use the solubility rules to figure out the answer. Solubility Rules and Precipitates

  19. Not all reactions produce a precipitate. If the volume of solution is large enough, and the amount of solute is small enough no precipitate will form. • We will use the reaction quotient, Q, to predict whether a precipitate forms. • The form of the reaction quotient is the same as the Ksp, just as it was with Keq and Ka/b values. • By comparing the value of the reaction quotient with the solubility product, we can determine if an aqueous solution is saturated or unsaturated. Solubility Rules and Precipitates

  20. If, • Q = Ksp, the solution is just saturated and no precipitate forms. • Q>Ksp, the solution is saturated and a precipitate forms. • Q<Ksp, the solution is unsaturated and no precipitate forms. • We determine the Q value, from comparing our experimental concentrations to the Ksp value for the saturated solution Solubility Rules and Precipitates

  21. Example • The Ksp of lead (II) chloride is 1.6 x 10-5. If 0.57 g of lead (II) chloride are added to 1500 mL of water, is the solution saturated? Assume no volume change. Solubility Rules and Precipitates

  22. Things to consider for determining precipitates: • The initial solutions must be soluable, or you do not have a solution • The chemical reaction is a double displacement reaction • The concentrations of the ions are what the stoichiometry states by the dissociation of the compound • When solutions are mixed, the volume changes, and so does the concentration of the ion • The net ionic equation will tell you what the Ksp expression should be • Find the Q and compare to the literature Ksp value Solubility Rules and Precipitates

  23. Example • If 20.0 mL of a 0.0010 mol/L silver nitrate solution is mixed with 20.0 mL of a 3.0 x 10-5mol/L potassium bromide solution, does silver bromide (Ksp = 5.0 x 10-13) precipitate? Assume the volumes are additive. • Example 2 • 100 ml of a 0.01 M potassium hydroxide solution is combined with 50 ml of a 0.1 M copper (II) nitrate solution. Will a precipitate form? • (Ksp for the precipitate = 1.60 x 10 –19) Solubility Rules and Precipitates

  24. When an ionic compound dissolves in pure water, the initial concentration of each ion is zero. • However, if an ionic compound dissolves in a solution that has an ion in common with the compound, this is not the case. • Even though the starting concentrations may not be zero, the product of the ions must still equal the solubility product constant. • For example, how would the solubility of silver chloride in pure water change if we try dissolving it in tap water? Common Ion Effect

  25. Let's write out the dissociation equation: AgCl(s) Ag+(aq) + Cl¯(aq) Tap water often has chlorine added to kill bacteria. The chlorine exists as chloride ions, so when we dissolve silver chloride in tap water, chloride ions are present. According to Le Chatelier's Principle, Adding more chloride ions to a saturated solution would cause the equilibrium to shift to the left to use up the excess product. This would result in more solid formed, and a decreased solubility. Common Ion Effect

  26. Le Chatelier's Principle predicts that the solubility of an ionic solid in a solution containing a common ion decreases its solubility. • Let's see if this is supported by the calculations. • Example • Determine the solubility of silver chloride in pure water and in a solution of 0.10 mol/L sodium chloride. The Ksp of AgCl is 1.7 x 10-10. Common Ion Effect

  27. Step 1 • Find the solubility of the ions in pure water at saturation • Step 2 • Set up an ICE Chart for the addition of the NaCl solution. What is the common ion? What is its concentration? • Step 3 • Substitute the values from your ICE chart into your Ksp equation. • Step 4 • Solve for x. Since x is small, disregard as it will have little impact on the final answer. • Step 5 • Find the value of solubility of the NaCl, using the chloride ion value Common Ion Effect

  28. The Ksp of lead (II) chloride, PbCl2, is 1.6 x 10-5. What is the solubility of lead (II) chloride in a 0.10 mol/L solution of magnesium chloride, MgCl2? • MgCl2 is a highly soluable salt, with 100% ionization. Therefore, the solubility of PbCl2 will be less, due to the common ion in solution. Common Ion Effect

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