Ksp? Did you mean Kfc? A Little Review I. The K’s that we have covered: Kc, Kp, Ka, Kb, Kw II. Solubility: The amount of solute that will dissolve in solution A. Measured in amt./volume B. Example: 40g NaCl/100mL water 0r 2.0x10-4 M (a.k.a molar solubility) III. Solubility rules: some combinations of ions form precipitates: A. Fe(OH)3 or AgCl Solubility Equilibria and Ksp I. Ksp: solubility product constant A. Product means there’s multiplication! B. It measures the solubility of insoluble ionic compounds. C. We are going to see that even insoluble species are slightly soluble II. Writing solubility product expressions: A. You need a balanced chemical equation. B. Write products over reactants. C. The coefficients become exponents. D. Examples: 1. BaSO4 2. Al(OH)3 3. Mg3(PO4)2 E. Solids are not a part of the expression! III. Ksp values are usually pretty small (see p. 427)
Ksp? Did you mean Kfc? IV. Solving for Ksp A. Find the molar solubilities (s) of each species. B. Plug into the expression and solve! C. Examples: 1. The solubility of copper (I) bromide is 2.0x10-4M. Calculate Ksp.
2. The solubility of iron (II) hydroxide is 2.32x10-6 M. Calculate Ksp.
V. Finding molar solubility from Ksp A. Write the balanced chemical equation. B. Write the expression. C. Plug in an “s” for each molar solubility D. Solve for s! E. If needed, plug back into ICE chart to find concentrations.
1. What is the molar solubility of silver sulfide? (Ksp = 1.6x10-49)
2. What is the molar solubility of bismuth (III) sulfide? (Ksp = 1.1x10-73)
VI. Therefore, given s, you can solve for Ksp and given Ksp, you can solve for s. A. General expressions based on formulas: AB A2B AB3 A2B3 B. Expression is dependent of the amount of ionic constituents. VII. Watch out for hydroxides! A. When hydroxides are in water, you need to watch out for the [OH-] from the autoionization of water. B. Because molar solubilities are so small, the [OH-] is significant. (1.0x10-7M) C. Ex: What is the molar solubility of iron (III) hydroxide? (Ksp = 4 x 10-38)
Ksp? Did you mean Kfc? D. Ex: If the pH of the final solution is 2.0, what is the molar solubility of iron (III) hydroxide? (Ksp = 4 x 10-38)
1. The pH gives you [OH-] VIII. Predicting whether or not a precipitate will form: A. You need to remember Q. 1. This is the reaction quotient. 2. This is found using initial concentrations. 3. This states the position of the reaction. B. If Qsp > Ksp, a ppt will form. C. If Qsp < Ksp, a ppt will not form. D. If Qsp = Ksp, the solution is saturated. E. Ex: 50.0 mL of 0.00025 M Na3PO4 is mixed with 50.0 mL of 0.0025 M BaCl2. Will a precipitate form?
IX. What if TWO precipitates can form? A. Using Ksp and initial concentrations, solve for the concentration needed for each precipitate to form. B. The lower concentration will form first. C. There are many variations to this type of problem. D. Ex: Sodium chloride is added to a 50 mL beaker that contains a mixture or 0.00015 M lead (II) nitrate and 0.00035 M silver nitrate. What precipitate will form first?
X. The common ion effect revisited. A. Re: Le Châtelier’s principle. B. If you add one of the ions that a precipitate dissolves into, the result is a shift to the left C. Another explanation: The Ksp does not change so increasing the concentration of an ion will result in the decrease of the other. D. Ex: A sample of solid AgNO3 is added to a saturated solution of Ag2CrO4. Assuming no volume change, does [CrO42-] increase, decrease or remain the same?
Ksp? Did you mean Kfc? XI. Volume changes A. If the volume of a saturated solution is changed, the concentrations of the dissolved ions do not change. B. Because, if the solution is still saturated, the precipitate will dissolve as the water is added. Thus, the volume to dissolved precipitate remains the same. C. Thus, the concentration of solute particles in a saturated solution are independent of volume. XII. Percent of ion remaining. A. If you have excess unprecipitated ion, take the amount that remains and divide by the initial concentration to find the % of ion remaining. XIII. Dissolving precipitates A. Strong acids can dissolve precipitates in which the anion is the conjugate of an acid. 1. Hydroxides, carbonates and sulfides. 2. Hydrochloric acid is added to a suspension of zinc hydroxide. 3. Nitric acid is added to solid zinc carbonate. 4. Hydroiodic acid is added to zinc sulfide precipitate. B. Ligands can be added to form complex ions (remember?)
