Solubility of KHT and Common ion Effect

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Solubility of KHT and Common ion Effect v010516 You are encouraged to carefully read the following sections in Tro (3 rd ed.) to prepare for this experiment: Sec 16.5, pp 783-788 (Solubility Equilibria and the Solubility Product Constant). Objectives: You will observe the common ion effect on the Ksp and molar solubility of a slightly soluble salt, as determined from the hydrogen ion concentration. Background: Ksp: Ksp is the solubility product constant for an ionic compound. Recall in our textbook that a series of rules are given in Table 4.1, p. 161, to determine whether an ionic compound was soluble or insoluble in water. The solubility product constant is the numerical value that explains the entries in the table. The larger the Ksp value the greater the degree of dissociation of the ionic compound in water. The Ksp of an ionic compound can be calculated using the equilibrium expression for the dissociation of that compound and the molar solubility of the compound. The molar solubility is the solubility of the compound in moles per liter. Example: Dissociation of Silver Chloride AgCl (s) Ag + (aq) + Cl - (aq) Ksp = [Ag + ][Cl - ] [Ag + ] [Cl - ] Initial 0 0 Change +x +x Equilibrium x x In the table, x = [Ag + ] = [Cl - ] = molar solubility of AgCl, the concentration of dissolved AgCl in a saturated solution of this solid. Since the Ksp for AgCl is 1.77 x 10-10, then: Ksp = [Ag + ][Cl - ] = 1.77 x 10-10 = (x)(x) 1.77 x 10-10 = x 2 1.33 x 10-5 M = x = molar solubility of AgCl in pure water Common Ion Effect: The Common Ion Effect is observed when an ionic compound is dissolved in a solution that already contains one of the ions found in the salt. Since Ksp is a constant that depends on the temperature of the solution but not on the concentrations of the components in solution and following Le Châtelier s Principle, then the presence of a common ion causes the solubility of the ionic compound to be lowered compared to that found in pure water. Example: Dissociation of Silver Chloride in 0.1 M NaCl NaCl is a very soluble salt and will dissociate completely into Na + (aq) and Cl - (aq). The concentration of each ion will be equal to 0.1 M. The presence of the Cl - (aq) ion means, that when AgCl is added to a solution containing the NaCl, the initial concentration of the Cl - (aq) ion is 0.1 M. AgCl (s) Ag + (aq) + Cl - (aq) Ksp = [Ag + ][Cl - ] [Ag + ] [Cl - ] Initial 0 0.1 Change +y +y Equilibrium y 0.1 + y

In the table, y represents the molar solubility (not in water) but now in 0.1 M NaCl(aq). Additionally, Ksp for AgCl is still 1.77 x 10-10 since it is a constant. So: Ksp = [Ag + ][Cl - ] = 1.77 x 10-10 = (y)(0.1 + y) 1.77 x 10-10 = 0.1(y) + y 2 Rearrangement: 0 = y 2 + 0.1y - 1.77 x 10-10 Using the Quadratic Equation: 1.76 x 10-9 = y = molar solubility of AgCl in 0.1 M NaCl. Note that y, solubility in NaCl(aq), is much lower than the solubility in pure water (x from above) as predicted by LeChatelier s principle. In this laboratory, you will observe the effect of the presence of a common ion on the molar solubility and Ksp of potassium hydrogen tartrate, or KHT. KHT(s) dissociates into the potassium ion, K + (aq), and hydrogen tartrate ion, HT - (aq) in solution. The HT - concentration can then be determined by titrating it against the hydroxide ion, OH -. HT - (aq) + NaOH H2O (l) + NaT - (aq) Then by using an ICE chart along with the solubility of KHT (which equals the concentration of HT - since the mole ratio of KHT:HT - is 1:1), Ksp can be calculated for KHT at the temperature for the experiment. KHT (s) K + (aq) + HT - (aq) Ksp =??? Titration of a Saturated Potassium Hydrogen Tartrate (KHT) Solution Against Standardized NaOH Preparation of the Saturated KHT Solution for Titration 1. Gather: a. 3 125-mL Erlenmeyer flasks (washed and rinsed with and distilled water) b. 1 100-mL graduated cylinder (washed and rinsed with distilled water) c. 1 250-mL beaker (washed and rinsed with distilled water) d. 1 funnel (washed and dried) e. 1 25-mL volumetric pipet (washed) f. Pipet bulb g. Stirring rod (washed and dried) h. 1 piece of filter paper i. 1 250-mL Erlenmeyer flask (washed and dried) 2. Weigh 1.4 g of KHT. 3. Transfer the KHT into a clean 250-mL beaker. 4. Using graduated cylinders, add the volumes of NaCl and KCl indicated by your instructor to the KHT. 5. Stir the mixture with a stir rod for approximately 10 minutes. Solution # Volume of 0.10 M NaCl in ml Volume of 0.10 M KCl in ml 1 100.0 0.0 2 90.0 10.0 3 80.0 20.0 4 70.0 30.0 5 60.0 40.0 6 50.0 50.0 7 40.0 60.0 8 30.0 70.0 9 20.0 80.0 10 10.0 90.0 11 0.0 100.0

6. Using a 250-mL Erlenmeyer flask, funnel, and filter paper, prepare a gravity filtration apparatus. Don t wet the filter paper and make sure the funnel is dry to avoid adding water to the filtrate. 7. Filter your solution (note: the filtrate must be clear). 8. Using a 25-mL volumetric pipet, withdraw 25 ml of saturated KHT solution and discard in order to rinse the pipet. 9. Using the 25-mL volumetric pipet, transfer 25-mL of the saturated KHT solution into a 125- ml Erlenmeyer flask and record on the data sheet as volume of KHT saturated solution titrated, to the nearest 0.01 ml. 10. Repeat 1 more time into an additional 125-mL flask. 11. Measure the temperature of the two solutions and record the temperature on your data sheet. 12. Add two drops of phenolphthalein indicator to each of the saturated KHT solutions in the 125- ml Erlenmeyer flasks. Preparation of the Standard NaOH Solution for Titration Remember that standardized NaOH is valuable and time consuming to make. Take only what you need from the bottle, and be certain the stock bottle is tightly sealed after you use it. 1. Rinse the buret with a small portion (~5 ml) of the NaOH solution, discard this rinse. 2. Fill the buret with NaOH and open the stopckock to filling the tip of the buret with the NaOH solution. Discard this small amount of solution. Titration 1. Record the indicated molarity of the standardized NaOH solution from the bottle (should be 0.05 M) onto your data sheet. 2. Record the initial buret reading on your data sheet to the nearest 0.01 ml. 3. Titrate the solution with the standardized NaOH solution until the light pink end point persists for at least 30 s. 4. Record the final buret reading on your data sheet to the nearest 0.01 ml. 5. Refill your buret with additional NaOH. 6. Titrate a second 25-mL portion of your filtrate. 7. Continue titrating samples until your instructor has approved two titrations.

Data Run 1 Run 2 Volume of 0.10 M KCl Volume of 0.10 M NaCl Volume of KHT saturated solution titrated Temperature of KHT solution Molarity of standardized NaOH from bottle Initial Buret Reading of NaOH (ml) Final Buret Reading of NaOH (ml) Instructor s initials Volume of NaOH used, ml Results Titration: NaOH + HT - H2O + NaT - Run 1 Run 2 Moles of NaOH required to reach endpoint Moles of HT - Concentration of HT - in saturated solution, M Molar Solubility of KHT, M Average Molar Solubility of KHT, M Concentration of K + before KHT dissolves (don't forget dilution calculation), M Concentration of K + from dissolved KHT, M Concentration of K + in saturated solution, M Ksp of KHT Average Ksp of KHT

Class Results Solution # [K + ] from KCl Avg. [K + ] in saturated solution Avg. [HT - ] (molar solubility) Avg. Ksp 1 2 3 4 5 6 7 8 9 10 11

Post Laboratory Problems 1. A student determined the molar solubility of Mg(OH)2 under a variety of conditions. In each of the following scenarios would the student expect to observe a decreased molar solubility due to the common ion effect? Explain your reasoning. a. Mg(OH)2 was dissolved in a solution of 0.1 M MgCl2. b. Mg(OH)2 was dissolved in a solution of KOH. c. Mg(OH)2 was dissolved in a solution of NaCl. 2. From the Class Results, how was the molar solubility affected by the change in [K + ] concentration from KCl? Explain your answer. 3. From the Class Results, did the Ksp change significantly depending on the [K + ] concentration from KCl? Explain your answer.

4. An experiment (similar to the one performed in this lab) involved dissolving lead (II) chloride, PbCl2, in a 0.10 M solution of lead (II) nitrate, Pb(NO3)2. The chloride ion was then detected using Fajans Method. Fajans Method involves titrating Cl - against silver nitrate to make AgCl. The endpoint of the titration is observed when a dichlorofluorescein indicator changes from yellow to pink. a. Write the reaction for the dissociation of PbCl2(s). b. Write the Ksp expression for the dissociation of PbCl2. c. Develop an ICE Chart for the dissociation of PbCl2 in 0.10 M Pb(NO3)2. d. According to Fajans method the [Cl - ] = 0.00527 M in the saturated solution. Determine the Ksp of PbCl2.

Pre-Laboratory Problems 1. Write the reaction for the dissociation of KHT(s) in water. 2. Write the equilibrium expression for the dissociation of KHT in water. 3. Develop an ICE Chart for the dissociation of KHT in water. 4. During the Weak Acid Laboratory, you determined the HT - concentration for the dissociation of KHT in water to be approximately 0.0387 M. Determine the Ksp for the dissociation of KHT in pure water. 5. Why is it important to add enough KHT to the KCl-NaCl solution in this lab to ensure solid remains?