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`AVRSB AP ChemistryAP Chemistry([email protected])The Solubility Product Constant KspSalts differ in their solubilities. In general, compounds of alkali metals are soluble in water. Many ionic compounds however, are insoluble. (See solubility chart) Most &quot; insoluble&quot; salts will dissolve to some extent in water. These salts are said to be slightly soluble in water. For example, when the &quot;insoluble&quot; salt AgCl is mixed with water, a very small amount of it dissolves. AgCls  Ag+aq + Cl-aq Because AgCl is a solid it is left out of the equilibrium expression, Ksp = [Ag+][Cl-] Ksp is equal to the product of the concentration terms each raised to the power of the coefficients of the substance in the dissociation equation. Ksp for AgCl at 25 degrees is 1.8 x 10-10 (from solubility table) Ex. What is the concentration of silver chloride ions in a saturated silver chloride solution at 25 degrees? Ksp = 1.8 x 10-10 AgCls  Ag+aq + Cl-aq Ksp = [Ag+][Cl-] = 1.8 x 10-10 Since [Ag+]= [Cl-] [Ag+]2 = 1.8 x 10-10 [Ag+] = 1.3x10-5 The equilibrium constant Ksp called the solubility constant indicates how soluble a substance is. If the Ksp is low, the solid is not very soluble. If the Ksp is high, the solid is soluble. As long as the product of the concentration of the ions does not exceed the Ksp value, no precipitate forms.Mr. Richards AP ChemistryAVRSB AP ChemistryThe Common Ion Effect What would happen if silver nitrate was added to the above solution? Immediately after the addition, the product of the [Ag+][Cl-] would be greater than the Ksp value (this is because the AgNO3 will dissociate and add more Ag+) LeChatelier's principle applies here, the stress of the additional Ag+ could be relieved by shifting the reaction to the left. This would create additional AgCl solid (precipitate). This process would continue until the product of the [Ag+]x [Cl-] once again equaled 1.8 x 10-10. In this example, the silver ion is called a common ion, because it is found in both salts. Adding silver nitrate to a saturated solution of silver chloride causes the solubility of AgCl to decrease. The lowering of the solubility of a substance by the addition of a common ion is called the common ion effect. Ksp can be used to predict whether a precipitate will form when two solutions are mixed together. If the ion product concentration is greater than the Ksp value a precipitate will form. If it equals the Ksp the solution is saturated and no precipitate forms, if it is less no precipitate forms and the solution is unsaturated. Predicting Precipitate Formation When the product of the concentration of the ions exceeds the value of Ksp they cannot exist in equilibrium anymore. They will form a precipitate in order to reduce the concentrations of the ions in solution back to the equilibrium value.Mr. Richards AP ChemistryAVRSB AP ChemistryTo determine whether the Ksp is exceeded, just substitute the ion concentrations into an expression similar to the Ksp expression and determine an experimental Trial Ion Product. (T.I.P.) If the T.I.P. exceeds the Ksp then a precipitate forms. If the T.I.P. is larger than the values of Ksp then the ions are above the saturation line and the ions are in a super-saturated situation. The ions will precipitate until their individual concentrations multiplied together equal the value of Ksp. If the T.I.P. does not exceed the value of Ksp then the ions are in an unsaturated situation, which means there aren't enough ions dissolved to form a precipitate. If the T.I.P. equals the value of Ksp then the solution is saturated. The solution can hold no more dissolved ions without a precipitate forming. Problem: Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO3 is added to a beaker containing 1 L of tap water with a Cl- ion concentration of 1.0 x 10-5 mol/L? [Cl-] = 1.0 x 10-5 mol/L [Ag+] calculation. Start with the given info, 1 mL of 0.1 mol/L Set up a ratio that states, if there are 0.1 moles of Ag+ in 1000 mL how many moles will be in 1 mL? 0.1 mol = x 1000mL 1 mL x = 0.0001 moles of Ag+ are present. (This is how many moles there are, this is not the concentration) The [Ag+] will be this many moles of Ag+ dissolved in the volume of solution in this question. Therefore [Ag+] = 0.0001 mole = 0.0001 mole = 1.0 x 10-4 mol/L 1000 mL + 1 mL 1.001 LT.I.P. = [Ag+][Cl-] = (1.0 x 10-4)(1.0 x 10-5) = 1.0 x 10-9 The Ksp for AgCl is 1.8 x 10-10 therefore T.I.P &gt; Ksp so a ppt. will form. 50 mL of 0.0010 M NaCl is mixed with 50 mL of 0.00020 M AgNO3. Will a precipitate form? The Ksp of AgCl is 1.6 x 10-10 If a precipitate forms, it will either be AgCl or NaNO3 The Ksp of AgCl(s) is lower. (All nitrates are soluble) Find the concentration of Ag+ and Cl- after the 2 solutions are mixed [Ag+] =original # moles =volume x conc =0.050 L x 0.00020M = 1.0 x 10-5 total volume total volume 0.10 L 0.1Mr. Richards AP ChemistryAVRSB AP Chemistry[Ag+] =1.0 x 10 ­4 M [Cl-] = 0.050 L x 0.0010M = 0.00005 mol = 5.0 x 10 ­4 M 0.10 L 0.10 L [Cl-] = 5.0 x 10 ­4 M Q = (Ag+) (Cl -) = ( 1.0 x 10 ­4) ( 5.0 x 10 ­4) = 5.0 x 10 ­9 This product is greater than the Ksp value. A precipitate will form STEPS TO FOLLOW 1. Determine which ions will form what compound 2. Calculate the number of moles of each ion (n = C x V) 3. Determine the total volume of the solutions after mixing 4. Calculate the concentration of each ion (C = n / V) 5. Substitute the concentrations of the ions into the Ksp expression to solve for Q 6. Compare Q and K Given the initial concentration, we can predict which way a reaction will go Comparing Q and K does this Q &gt; K A precipitate forms and a saturated solution exists Reaction goes to left Q = K A saturated solution exists Equilibrium is established Q &lt; K No precipitate forms and an unsaturated solution exists. Reaction goes to rightMr. Richards AP Chemistry`

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