Aqueous Balance: Equilibrium

Activity 4 Aqueous Balance: Equilibrium GOALS In this activity you will: Determine ph and understand its meaning. Learn the basic principles behind equilibrium and the law of mass action. Calculate a solubility product constant (K sp ) and an equilibrium constant (K eq ) from experimental data. Learn how to shift the equilibrium concentrations by applying LeChatelier s stressors to an aqueous chemical equilibrium. What Do You Think? Water, water, everywhere, And all the boards did shrink; Water, water, everywhere, Nor any drop to drink. The Rime of the Ancient Mariner by Samuel Taylor Coleridge (1772 1834) If there was water everywhere, why was there no water to drink? What is de-ionized water? Record your ideas about these questions in your log. Be prepared to discuss your responses with your small group and the class. Investigate In this investigation you will learn about chemical equilibrium. Your main objective is to understand enough so that you can better calculate the concentration of a solute and predict the effect of changing the equilibrium on other solutes and the purity of water. 877

H 2 Woes Sample solution ph [H ] (M) [OH ] (M) de-ionized water 7.0 1.0 10 7 1.0 10 7 0.10 M HCl(aq) 0.10 M NaOH(aq) your water sample Safety goggles and a lab apron must be worn at all times in a chemistry lab. Part A: ph and the Equilibrium of Water 1. Copy the table above in your log. 2. Pour a small amount of each sample into a small beaker. If you are using a ph probe, be sure to pour enough of each sample to cover the tip of the probe. 3. Using the method recommended by your teacher, determine the ph of the 0.10 M HCl, 0.10 M NaOH, and your water sample. If you are using a ph probe, be sure to completely rinse the probe with distilled water as demonstrated by your teacher. a) Record your ph values in your log table. 4. Calculate the concentration of hydrogen ions, [H ], in each solution. Note that [H ] 10 ph. a) Record these molar concentration values in the table in your log. 5. Water is very stable and very little dissociation takes place. One in 10 million (10 7 ) water molecules dissociate into H and OH ions. The dissociation equation or equilibrium equation is: HOH H (aq) OH (aq). Determining the value of the equilibrium constant, K w : K w [H ][OH ] (1 10 7 M)(1 10 7 M) 1 10 14 Another way that you can look at this is: pk w ph poh 14 If you know the ph or the [H ] then you can determine poh or [OH ]. As an example, if the [H ] 1 10 3 M you can substitute this value into the equation (K w [H ][OH]) and show that [OH] 1 10 14 1 10 3 1 10 11 M From this you can then say that the poh 11. a) Using K w, calculate the concentration of hydroxide ions, [OH ], in each solution in your table. Record these molar concentration values in your log. 878

Activity 4 Aqueous Balance: Equilibrium A 0.10 M B 0.010 M C 0.0010 M D 0.00010 M E 0.000010 M Na 2 C 2 O 4 (the original solution) 9 drops distilled water 1 drop of solution A 9 drops distilled water 1 drop of solution B 9 drops distilled water 1 drop of solution C 9 drops distilled water 1 drop of solution D Part B: Determining the Value of the Solubility Product of Calcium Oxalate 1. Obtain 0.10 M solutions of both Na 2 C 2 O 4 (sodium oxalate) and Ca(NO 3 ) 2 (calcium nitrate). 2. Dilute these concentrations serially by a factor of 10 each time using a multi-well microplate. To dilute by a factor of 10, you add 9 parts distilled water to 1 part of the solution to be diluted. A serial dilution means diluting each subsequent solution by a factor of 10, leaving you with a microplate with the following concentrations of each solute, labeled Na 2 C 2 O 4 A E and Ca(NO 3 ) 2 A E. 3. Mix the two solutions (A with A, B with B, and so on) in separate wells and check for formation of a CaC 2 O 4 precipitate. a) For which mixtures was no precipitate observed? This means then the solution is now too dilute for any solid CaC 2 O 4 (calcium oxalate) to exist in equilibrium with the solution. 4. To approach the experiment from the opposite direction, mix the two solutions that are more diluted (E with E, D with D, and so on) until you observe the barest amount of precipitate. a) Once you discover the concentrations to mix to get a small amount of CaC 2 O 4, record these concentrations in your log. b) Calculate the concentrations of C 2 O 4 2 (aq) and Ca 2 (aq) using the mixture of the two solutions that gave the first noticeable precipitate. Note that when you mix one drop of sodium oxalate with one drop of calcium nitrate, the concentrations of the calcium and oxalate ions will be cut in half. For example, if one drop of 0.10 M oxalate ion is mixed with one drop of 0.10 M calcium ion, the concentration of each of the ions will now be 0.05 M. c) The reaction can be written Na 2 C 2 O 4 (aq) Ca(NO 3 ) 2 (aq) 2NaNO 3 (aq) CaC 2 O 4 (s) The total ionic equation for the reaction that shows all of the ions is: 2Na (aq) C 2 O 2 4 (aq) Ca 2 (aq) 2NO 3 (aq) 2Na (aq) 2NO 3 (aq) CaC 2 O 4 (s). Notice that the Na and NO 3 ions appear in the same form before and after the reaction takes place. Any such ions that are not changed by a reaction are called spectator ions. They may be omitted in writing what is called the net ionic equation. Cancel the Na and NO 3 spectator ions that appear on both sides of the total ionic equation to obtain the net ionic equation: Ca 2 (aq) C 2 O 4 2 (aq) CaC 2 O 4 (s). Write this net ionic equation in your log. 879

H 2 Woes AgNO 3 solution will stain your skin. Always use latex gloves when working with a solution of AgNO 3. 5. Every chemical equilibrium has an equilibrium constant expression, also called a mass-action expression. The law of mass action is a result of many experimental observations and states for a general reaction: a moles of reactant A with b moles of reactant B forms c moles of product C, d moles of product D and e moles of product E (aa bb cc dd ee) is expressed as K eq [C]c [D] d [E] e, [A] a [B] b where [C] stands for the molar concentration (M) of C and so on for D, E, A, and B. If a reactant or product is a solid or water, then you do not include it in the mass-action expression because its concentration is constant. Also note the powers of concentration used in the equilibrium constant correspond to the coefficients in front of each substance in the equation. a) In your log, write the mass-action (equilibrium constant) expression for both the total ionic equation and the net ionic equation. Note the states of matter of each species in each equation! b) Are the expressions equal? Explain. c) Calculate your K eq using your concentration values calculated in Step 5, expressing your answer in scientific notation. Have these checked by your teacher before moving onto Step 6. 6. Since the reaction is reversible, it can also be written as: 2NaNO 3 (aq) CaC 2 O 4 (s) Na 2 C 2 O 4 (aq) Ca(NO 3 )(aq) When the reaction is written in the direction of the solid dissolving the equilibrium constant K eq is also called the solubility product (K sp ). a) Calculate your K sp using your concentration values calculated in Step 5, expressing your answer in scientific notation. 7. The agreed-on experimental value of the calcium oxalate K sp at 25ºC is 2.3 10 9. a) How does your value compare to your experimentally determined value? b) If there are differences, offer a few reasons why this may be. Part C: Shifting Equilibrium LeChatelier s Principle LeChatelier s Principle is a qualitative statement about chemical reactions which are in a state of equilibrium. To observe LeChatelier s Principle in action, you will work with the following chemical reaction: Fe(NO 3 ) 3 (aq) + KSCN(aq) FeSCN(NO 3 ) 2 (aq) + KNO 3 (aq) 1. Label 5 small test tubes A, B, C, D, E and place them in a test-tube rack. Add 5.0 ml of distilled water to test tube A. Add 5.0 ml of 0.0020 M KSCN solution to test tubes B E. Add two drops of 0.20 M Fe(NO 3 ) 3 solution to test tube A. a) Describe the appearance of the 5 test tubes in your Active Chemistry log. 2. To test tubes B D, add two drops of 0.20 M Fe(NO 3 ) 3 solution. Mix each solution well with a clean stirring rod. Compare test tube B with test tube A. a) Record any changes observed. Is there any reason to believe a chemical reaction has taken 880

Activity 4 Aqueous Balance: Equilibrium place? Why or why not? b) In the equation on the previous page, which of the four substances do you think is responsible for the change? Why do you think this substance is the one? How could you prove this? c) Why is it important to compare test tubes B E with test tube A? 3. To test tube C, add a small crystal of solid KSCN. Mix well with a clean stirring rod. Compare test tube C with test tube B. a) Record any changes observed. Is there any evidence that the equilibrium shifted either to the left (reactants) or to the right (products)? If so, which way did it shift? 4. To test tube D, add 5 drops of the 0.20 M iron (III) nitrate solution. Mix well with a clean stirring rod. Compare test tube D with test tube B. a) Record any changes observed. Is there any evidence that the equilibrium shifted either to the left (reactants) or to the right (products)? If so, which way did it shift? b) Is this consistent with what you observed in Step 3? Why or why not? 5. Disodium hydrogen phosphate (Na 2 HPO 4 ) can be used to remove Fe 3+ from solution as a precipitate. Add a crystal of this salt to test tube E. Compare test tube E with test tube B. a) Record any changes observed in test tube E. Is there any evidence that the equilibrium shifted either to the left (reactants) or to the right (products)? If so, which way did it shift? b) Is this consistent with what you observed in Steps 3 and 4? Why or why not? 6. Why is it important to compare test tube B with each of the other test tubes, C E? 7. Predict whether this reaction is exothermic or endothermic. Consider whether you think the addition of heat to the equilibrium in test tube B will push the reaction toward reactants, products, or have no effect. What would be the effect on the color in each case? If exothermic: Fe(NO 3 ) 3 (aq) + KSCN(aq) FeSCN(NO 3 ) 2 (aq) + KNO3(aq) + heat If endothermic: heat + Fe(NO 3 ) 3 (aq) + KSCN(aq) FeSCN(NO 3 ) 2 (aq) + KNO 3 (aq) a) Design an experiment to determine if this reaction is exothermic, endothermic, or neither. Record it in your Active Chemistry log. b) With your teacher s permission, carry out your experiment and, if time permits, report your results to the class. 8. Write the complete ionic equation for this reaction, showing all of the ions. 9. Write the net ionic equation for this reaction by removing ions which appear on both sides of the equation in Step 8. 10. Write the equilibrium constant expression for this reaction using the principles of mass action. 11. Follow your teacher s instructions for cleaning up your lab area. 881

H 2 Woes Dispose of the materials as directed by your teacher. Clean up your workstation. Wash your hands and arms thoroughly after the activity. Chem Words equilibrium: the condition when the forward rate of the reaction is equal to the reverse rate of the reaction and the concentrations of reactants and products are unchanging. aa bb cc dd ee law of mass action: a reaction equilibrium expression where the product of the product concentrations divided by the product of the reactant concentrations is equal to a constant. mass-action expression (equilibrium constant): K eq [C]c [D] d [E] e [A] a [B] b EQUILIBRIUM Overview Some chemical reactions are reversible. If you mix hydrogen gas and iodine gas, hydrogen iodide gas will form. That hydrogen iodide gas will then decompose into hydrogen gas and iodine gas. If confined to a closed container, a state of equilibrium would be reached where the amount of hydrogen gas and iodine gas producing hydrogen iodide would be equal to the amount of hydrogen iodide decomposing into the hydrogen and iodine. An experiment can be performed that begins with equal concentrations of hydrogen and iodine and no hydrogen iodide. After a given time, the system will reach equilibrium and the concentrations of hydrogen, iodine, and hydrogen iodide can be measured. A second experiment can be performed which begins with only hydrogen iodide. After a given time, the system will reach equilibrium and the concentrations of hydrogen, iodine, and hydrogen iodide can be measured. A third experiment can be performed which begins with different concentrations of hydrogen and iodine. After a given time, the system will reach equilibrium and the concentrations of hydrogen, iodine, and hydrogen iodide can be measured. In each of the experiments, the initial concentrations are different. The final concentrations for the hydrogen, iodine, and hydrogen iodide gases in each of these experiments will also be different. However, there is a quantity that is identical in all experiments. This is the mass-action expression, also called the equilibrium constant. This quantity will be the same for the different experiments as long as the temperature is constant. The law of mass action is a result of many experimental observations and states. For a general reaction of a moles of reactant A with b moles of reactant B to form c moles of product C, d moles of product D, and e moles of product E (aa bb cc dd ee) : K eq [C]c [D] d [E] e [A] a [B] b, where [C] stands for the molar concentration (M) of C, and so on for D, E, A, and B. If a reactant or product is in the solid or liquid phase, then you do not include it in the mass-action expression because its concentration is more or less constant. Also, K eq is temperature dependent; therefore, the values of K eq will change for systems at different temperatures. 882

Activity 4 Aqueous Balance: Equilibrium Sample Problem 1 Nitrosyl chloride, NO 2 Cl(g), is in equilibrium with NO 2 (g) and Cl 2 (g): 2NO 2 Cl(g) 2NO 2 (g) Cl 2 (g) The concentrations of the three substances at equilibrium are [NO 2 Cl] 0.00106 M, [NO 2 ] 0.0108 M, and [Cl 2 ] 0.0054 M. Find the equilibrium constant using: K eq [NO 2] 2 [Cl 2 ] [NO 2 Cl] 2 (0.0108)2 (0.0054) (0.00106) 2 0.56. The Equilibrium of Water To understand the ph of water, you must remember that the water will be in equilibrium with a small amount of H and OH. So, for the reaction H 2 O(l) H (aq) OH (aq), the equilibrium constant is defined as K eq [H ] [OH ]. Note that since the concentration of [H 2 O(l)] is too large (55.5 M) it is considered a constant, and is omitted from this expression. This equilibrium constant is also called the ionization constant of water and is given the special symbol K w : K w [H ][OH ] 1.0 10 14 You can use this expression to solve mathematically for the ph of de-ionized water using the following steps. First, in de-ionized water the concentration of hydrogen ions is equal to the concentration of hydroxide ions or [H ] [OH ] 1 10 7 K w [H ][OH ] 1.0 10 14 Then, remembering ph log 10 [H ], you substitute 1.0 x 10 7 for the hydrogen ion concentration and find the ph is 7 and the poh log 10 [OH ] 7 as well. In this activity, you explored the ph of your natural water sample, as well as solutions of hydrochloric acid (HCl(aq)) and sodium hydroxide (NaOH(aq)), and determined the concentrations of hydrogen ions and hydroxide ions in your water sample. 883

H 2 Woes Chem Words solubility product: the equilibrium constant for the dissolution of an ionic substance in water. K sp [A ] a [B ] b Calculating the Solubility Product (K sp ) In a saturated solution of an insoluble ionic compound, a constant called the solubility product can be used to determine the equilibrium concentrations of ions in solution. In this part of the activity, you determined the value of the solubility product for a solution of calcium oxalate (CaC 2 O 4 ).You wanted to form only the smallest amount of CaC 2 O 4 because forming additional precipitate would change the ion concentrations from their initial values and require you to measure the amount of precipitate in order to know how much of the ions remain in solution at equilibrium. For dissolution of an ionic solid, the equilibrium constant is called the solubility product, and it is simply the product of the concentrations of all the ions that become solvated. For this type of equilibrium constant, knowledge of its value is enough information to tell you the equilibrium concentrations. This makes calculation of the equilibrium ion concentrations very straightforward if you prepare a saturated solution of a single salt. AB(s) A (aq) B (aq), so K sp [A ] [B ]. The concentrations are found using [Ca 2 ][C 2 O 4 2 ] [Ca 2 ] 2 K sp. You can solve for the Ca 2 (aq) concentration by taking the square root of both sides of this last equation. So [Ca 2 ] K sp. Note that the solution must be saturated in order to assure that the equilibrium between dissolved and undissolved CaC 2 O 4 is established. However, that was not how you prepared the solution containing Ca 2 (aq) and C 2 O 4 2 (aq). In your preparation, it was not essential that [Ca 2 ] and [C 2 O 4 2 ] be equal since they came from separate solutions. The sample problem on the next page shows you how to use the solubility product constant, like the one you calculated, to determine the solubility of an ionic compound. This calculation can lead you to the solubility of any ionic compound if you know the K sp for a specific temperature in an aqueous system. Tables of K sp values can be found in numerous references. Values at 25ºC can be found in the appendix of this text. 884

Activity 4 Aqueous Balance: Equilibrium Sample Problem 2 What is the solubility of CaSO 4? The solubility is determined by measuring the amount of compound dissolved per liter of solution when equilibrium exists between dissolved and solid compound. K sp for CaSO 4 is given in the appendix solubility chart as 2.4 10 5. Now K sp [Ca 2 ][SO 2 4 ] [Ca 2 ] 2 because [Ca 2 ] [SO 2 4 ]. Hence [Ca 2 ] 2 2.4 10 5. Find the square root of both sides to find [Ca 2 ] 4.9 10 3 M. Because one mol of Ca 2 goes into solution for each mole of CaSO 4 dissolved, the solubility of CaSO 4 is 4.9 10 3 mol/l of solution. The solubility in grams per liter is readily found by multiplying the moles of CaSO 4 times its formula mass of 136.1 g/mol. So the solubility is 4.9 10 3 mol CaSO 4 L 136.1 g CaSO 4 1 mol CaSO 0.67 g CaSO 4 4 L. LeChatelier s Principle and Equilibrium Constants (K eq ) Le Chatelier s Principle states that when a change in concentration, temperature, or pressure (a stress) is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of that change. To observe LeChatelier s Principle, you worked with the following chemical reaction which was in a state of dynamic equilibrium. Fe 3+ (aq) + SCN (aq) FeSCN2+ (aq) (colorless) (orange-red) Chem Words LeChatelier s Principle: when a reaction at equilibrium is stressed by a change in concentration of a reactant or product, the reaction will shift in the direction to relieve the stress and establish a new equilibrium. As you observed, an increase of SCN ion concentration on the reactant side increased the forward rate, forming more of the orange-red complex ion (FeSCN 2+ ). A simplified expression for the forward rate of the reaction shows why this would happen. Rate forward = k[fe 3+ ][SCN ] where k = the specific rate constant for the forward reaction. 885

H 2 Woes By adding the Na 2 HPO 4, the Fe 3+ concentration on the reactant side was decreased. This resulted in a decrease in the forward reaction rate and the equilibrium shifted to the left, reducing the intensity of the orange-red color. At equilibrium, the forward rate = reverse rate and so, Rate forward = k f [Fe 3+ ][SCN - ] = Rate reverse = k r [FeSCN 2+ ] Checking Up 1. In your own words, describe a dynamic equilibrium. 2. What is the law of mass action? 3. Define poh. What is its value when the ph is 5? 4. What is a saturated solution? 5. Why are solids and solvents not included in the K sp expression for a solid which is in equilibrium with its ions? 6. In your own words, explain LeChatelier s Principle. or, upon rearranging, k f = K eq = [FeSCN 2+ ] k r [Fe 3+ ][SCN - ] Note that the concentrations of the products are divided by the concentrations of the reactants while k f / k r is reduced to K eq. With this expression, LeChatelier s Principle can be understood mathematically. At equilibrium, the K eq is equal to the fraction on the right side of the equation. When the [SCN - ] is increased by adding a crystal of KSCN, thus disturbing the equilibrium, the right side of the equation momentarily is smaller than the K eq. The equilibrium shifts to attempt to reestablish the original K eq. This is done by excess SCN - reacting with Fe 3+, which lowers the Fe 3+ concentration and increases the size of the fraction on the right side of the equation. In doing this, more FeSCN 2+ is formed, which increases its concentration and again increases the size of the fraction on the right side of the equation. Both effects work to bring the K eq back toward its original value. The new equilibrium will never actually quite reach its exact original value because concentrations have changed, but it will shift in the direction of its original value. In a similar manner, when a crystal of Na 2 HPO 4 is added to the original equilibrium, Fe 3+ is removed. Momentarily, the lower concentration of Fe 3+ will make the fraction on the right side of the equation larger than K eq. This slows the forward rate of reaction which lowers the FeSCN 2+ concentration and increases the SCN - concentration. Both of these changes work to reduce the numerical value of the right-side fraction, bringing it toward the original K eq. As before, it will approach but never equal the original K eq value because the concentrations have been changed. 886

Activity 4 Aqueous Balance: Equilibrium Sample Problem 3 Equilibrium Constant Calculations The expression for the equilibrium constant for the reaction studied is K eq = [FeSCN2+ ] [Fe 3+ ][SCN - ] To calculate the value for K eq, you can see that the concentrations of the three chemical substances must be known. The initial solutions mixed in the activity were 5.0 ml of 0.0020 M KSCN and 2 drops of 0.20 M Fe(NO 3 ) 3. On average, 2 drops is about 0.06 ml. To calculate concentrations after dilution, the relationship, M 1 V 1 = M 2 V 2 is used. In this way, the new initial concentrations can be calculated: [SCN - ]: 0.0020 M 5.0 ml = [SCN - ] 5.06 ml. [SCN - ] remains at 0.0020 M. [Fe 3+ ]: 0.20 M 0.06 ml = [Fe 3+ ] 5.06 ml. [Fe 3+ ] = 0.0023 M. However, you still need the [FeSCN 2+ ] at equilibrium. This is obtained by additional information from another source. A colorimeter or spectrophotometer can measure the intensity of the absorbance of light at specific wavelengths as a function of concentration. Using a colorimeter, a standard curve is determined for the absorbance (A) of the FeSCN 2+ ion at 475 nm versus the concentration of the ion, [FeSCN 2+ ]. A straight line results and the equation for this line is A (475) = 5.0 10 3 [FeSCN 2+ ] (units omitted) If the absorbance of your sample is 5.5, then the [FeSCN 2+ ] is 0.0011 M. But you still are not ready to do the final calculation for K eq. You have the initial concentrations for two of the three ions, Fe 3+ and SCN -, and the equilibrium concentration for FeSCN 2+. Looking once again at the equation, Fe 3+ (aq) + SCN - (aq) FeSCN2+ (aq) it can be seen that for every FeSCN 2+ ion that is generated by the forward reaction, one Fe 3+ and one SCN - ion is used. Therefore, the concentration of the product can be subtracted from the concentrations of the reactants. Using the following table, the equilibrium concentrations for Fe 3+ and SCN - can be calculated. Ions Initial concentration (M) Calculation Equilibrium concentration (M) Fe 3+ 0.0023 0.0023-0.0011 0.0012 SCN - 0.0020 0.0020-0.0011 0.0009 FeSCN 2+ 0.0011 Now you are ready to calculate the K eq. [FeSCN K eq = 2+ ] 1.1 10 = -3 [Fe 3+ ][SCN - ] (1.2 10-3 )(9 10-4 ) = 1.0 10 3 (units omitted) 887

H 2 Woes What Do You Think Now? At the beginning of this activity you were asked: If there was water everywhere, why was there no water to drink? What is de-ionized water? Go back and reread your answers to the What Do You Think? section. It probably seems much easier to answer the questions now that you better understand chemical equilibrium. What does it mean? Chemistry explains a macroscopic phenomenon (what you observe) with a description of what happens at the nanoscopic level (atoms and molecules) using symbolic structures as a way to communicate. Complete the chart below in your log. MACRO NANO SYMBOLIC What evidence did you see that helped you determine the ph of different solutions? Explain what will happen when calcium ions are added to a saturated calcium oxalate solution. How can you make the equilibrium equation below favor the creation of more products? Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) How do you know? Making specific reference to your data and using terms from the Chem Talk section, explain your observations in Part C of this activity. Why do you believe? How does this concept of equilibrium make sense conceptually? How does it relate to you in your life? Why should you care? How can you use your understanding of equilibria to develop a method for water purification in your Chapter Challenge? 888

Activity 4 Aqueous Balance: Equilibrium Reflecting on the Activity and the Challenge In this activity, you have learned how the concentrations of various substances in equilibrium can often be calculated rather than measured. You related the concentration of one solute to other solutes using equilibrium constants. The equilibrium constants can change, however, if the temperature changes. The techniques used in applying equilibrium constants are the same whether you are dealing with a chemical reaction, a dissolution process, or the dissociation of water. They are extensively used in making predictions about the composition of water in various stages of treatment. You have introduced these three types of equilibrium as though they were isolated processes with no complications from other species in solution. In reality, water chemistry involves multiple equilibria that have to be treated simultaneously. 1. Is the ph of your sample of tap water within the Secondary Drinking Water Standard ph range of 6.5 8.5? 2. How is the solubility product, K sp, defined for the ionic substance CaSO 4? 3. Use the solubility product, K sp, of MgCO 3 (3.5 10 8 ) to calculate the concentrations of Mg 2 (aq) and CO 3 2 (aq) in a saturated solution of MgCO 3. 4. MgF 2 has limited solubility in water, whereas MgCl 2 and NaF are highly soluble. Suppose that a student adds MgCl 2 (s) to a 0.001 M solution of NaF. Use the solubility product of MgF 2 (6.6 10 9 ) to find the concentration of Mg 2 (aq) if just enough MgCl 2 is added to precipitate a tiny amount of MgF 2. 5. Write an expression for the equilibrium constant, K eq, for the reaction N 2 O 4 (g) 2NO 2 (g). 6. A mixture of N 2 O 4 (g) and NO 2 (g) at equilibrium has concentrations of 0.0457 M for NO 2 and 0.448 M for N 2 O 4. What is the equilibrium constant for the reaction shown in the previous question? 7. How do you think you would calculate the equilibrium constant for the reaction CO 2 (aq) H 2 O(l) HCO 3 (aq) H (aq)? What concepts might tie into a problem like this? 8. In Activity 3, you observed the effect of temperature on solutions in equilibrium with undissolved solids. How is it possible for solubility to increase or decrease if the solubility product is a constant? 889