Chem 116 POGIL Worksheet - Week 9 Equilibrium Continued Introduction to Acid-Base Concepts

Chem 116 POGIL Worksheet - Week 9 Equilibrium Continued Introduction to Acid-Base Concepts Why? When a reaction reaches equilibrium we can calculate the concentrations of all species, both reactants and products, by using information about starting concentrations or pressures and the numerical value of the equilibrium constant. Knowing how to set up and solve equilibrium problems for gas-phase systems is essential preparation for applying equilibrium concepts to more complicated systems, such as acid-base chemistry. The mixture of reactants and products can often be altered by applying a stress to the system (changing species concentrations, changing pressures, changing temperature, etc.), and the shift in the position of the equilibrium can be understood and predicted on the basis of LeChatelier s Principle. Learning Objectives Understand the concept of the reaction quotient, Q, as a means of determining whether or not a system is at equilibrium, and if not, how a system must proceed to reach equilibrium Understand how to set up a calculation of all species concentrations at equilibrium, given initial concentrations and the value of the equilibrium constant Understand how various stresses cause a shift in the position of equilibrium on the basis of LeChatelier s Principle Understand the Brønsted-Lowry theory of acids and bases Understand the concepts of conjugate acid-base pairs Understand the role of solvent protolysis in acid-base chemistry Understand ph and the ph scale Success Criteria Be able to set up and solve for all species using K c or K p Be able to predict the direction of a reaction on the basis of Q Be able to check equilibrium calculation results, using a Q calculation Be able to apply LeChatelier s Principle to determine the direction a system at equilibrium must shift to reach a new equilibrium Be able to calculate concentrations once equilibrium is reestablished after a stress that causes a shift from an original equilibrium Be able to analyze acid-base reactions in terms of conjugate acid-base pairs and proton transfer Be able to write the formula of an acid s conjugate base or a base s conjugate acid Be able to use K w to calculate concentrations of H 3 O + and OH in acidic or basic solutions Be able to calculate ph and poh Be able to calculate concentrations of hydronium ion and hydroxide ion in solutions of pure strong acid or strong base in water Prerequisite Have read all of Chapter 15 and sections 16.1-16.5

Information (The Reaction Quotient, Q) We can calculate the ratio of concentrations or pressures of reactants and products, like K c and K p, at any time in the course of a reaction. When the system is not at equilibrium, the ratio of the product concentrations raised to their stoichiometric coefficients to the reactant concentrations raised to their stoichiometric coefficients is called the reaction quotient and given the symbol Q. The form of Q is the same as the form of K c or K p, but the values for products and reactants are not presumed to be the values at equilibrium. The value of Q relative to K c or K p indicates the direction in which the reaction must run to achieve equilibrium. If Q < K, then the reactant concentrations or pressures are too high and the product concentrations or pressures are too low, relative to what they would be at equilibrium. To achieve equilibrium, the reaction must run in the forward direction (shift right), using up reactants and forming more products. Conversely, if Q > K, then the reactant concentrations are too low and the product concentrations are too high, relative to what they would be at equilibrium. To achieve equilibrium, the reaction must run in the reverse direction (shift left), using up products and reforming more reactants. If Q = K, then the system is already at equilibrium. Key Question 1. For the reaction H 2 (g) + I 2 (g) º 2 HI(g) at 425 o C, K c = 54.8. Are the following mixtures of H 2, I 2, and HI at 425 o C at equilibrium? If not, in which direction must the reaction proceed to achieve equilibrium? a. [H 2 ] = [I 2 ] = 0.360 mol/l, [HI] = 2.50 mol/l b. [H 2 ] = 0.120 mol/l, [I 2 ] = 0.560 mol/l, [HI] = 2.20 mol/l c. [H 2 ] = 0.120 mol/l, [I 2 ] = 0.219 mol/l, [HI] = 1.20 mol/l Information (Calculating Amounts of All Species in an Equilibrium Mixture) Very often we know the initial concentrations or pressures of reactants and products, and we want to know their values when equilibrium is established. To set up a calculation of these amounts, we generally let the variable x represent the change in an amount of a particular reactant or product that must occur to reach equilibrium. Then, using the stoichiometry of the reaction and, we write algebraic expressions to represent the amounts that will be present at equilibrium. We then substitute these algebraic expressions into the concentration or pressure terms of the equilibrium constant expression. Solving for the value of x allows calculating the numerical values of concentration or pressure for all reactants and products. In setting up the problem, it is useful to write down under each species in the balanced equation the initial amount, the algebraic expression for how each amount will change, and the algebraic expressions for each final amount. It is also useful to know the direction in which the reaction must run in order to reach equilibrium. If this is not obvious (e.g., if only reactants are initially present, the reaction must run to the right), a Q calculation can be done using the initial concentrations of all species. Remember, if Q < K the reaction will run in the forward direction, but if Q > K it will run in the reverse direction.

Knowing this at the beginning helps in setting up the algebraic expressions for the changes that must take place to reach equilibrium. For example, consider the equilibrium, I 2 (g) + Br 2 (g) º 2 IBr(g), for which the equilibrium constant K c = 280 at 150 o C. Suppose 0.500 mol of I 2 (g) and 0.500 mol of Br 2 (g) were placed in a one-liter vessel at 150 o C. What will the concentrations of all species be once equilibrium is established? Because we have no IBr initially present, the reaction must go to the right in order to reach equilibrium. To write algebraic expressions for the equilibrium concentrations, we could let x equal the concentration of I 2 that is consumed in forming the equilibrium concentration of IBr. But for every mole of I 2 that reacts, a mole of Br 2 must also react, so x also represents the amount of Br 2 consumed. At the same time, for every mole of I 2 and Br 2 that are consumed, two moles of IBr appear as product, so its concentration at equilibrium will be 2x. Thus, we can write algebraic expression for the equilibrium concentrations as follows: I 2 + Br 2 º 2 IBr Initial 0.500 0.500 0 Change x x 2x Equilibrium 0.500 x 0.500 x 2x Substituting these algebraic expression into the K c expression, we obtain We could expand the denominator and rearrange the resulting expression to solve it for x as a quadratic equation. (Only one root will make sense for the given problem, and the other will be rejected.) However, in this case, the numerator and denominator are perfect squares, so it would be faster to take the square root of both sides and then solve for x. Taking the square root of both sides we obtain Rearranging, 8.36 666 16.7 332 x = 2x 18.7 332 x = 8.36 666 x = 0.446 619

We can now substitute x = 0.446 619 into our algebraic expression to obtain numerical values for each of the concentrations. [I 2 ] = [Br 2 ] = 0.500 x = 0.500 0.446 619 = 0.053 381 = 0.053 mol/l [IBr] = 2x = (2)(0.446 619 ) = 0.893 238 = 0.893 mol/l Key Questions 2. For the reaction (NH 3 )B(CH 3 ) 3 (g) º NH 3 (g) + B(CH 3 ) 3 (g) at 100 EC K p = 4.62 atm. If the partial pressures of NH 3 (g) and B(CH 3 ) 3 (g) in an equilibrium mixture at 100 EC are both 1.52 atm, what is the partial pressure of (NH 3 )B(CH 3 ) 3 (g) in the mixture? 3. At 425 o C, 1.00 mol of H 2 (g) and 1.00 mol of I 2 (g) are mixed in a one liter vessel. What will be the concentrations of H 2 (g), I 2 (g), and HI(g) at equilibrium. K = 54.8 at 425 o C for the reaction H 2 (g) + I 2 (g) º 2 HI(g) Information (Using Q to Check Your Results) In general, equilibrium calculations involve a lot of mathematical manipulation, and it is easy to make a mistake. How can you know when you have gone astray? The best way is to substitute your found values into the expression for Q and compare the calculated value with the given K. Because of rounding, do not expect an exact match. However, if Q K by a wide margin, check your work. In our example calculation of the equilibrium concentrations of I 2 (g), Br 2 (g), and IBr(g), a Q c calculation with the values we found gives the stated value of K c. Thus, we have confidence that the calculation is correct. When doing this kind of check, realize that arithmetic loss of significant figures and rounding differences may cause the calculated value of Q to be somewhat different from K, but the difference should not be great. Key Questions 4. Check the values you found in Key Question 3 by calculating the value of Q. Does your value agree with K c = 54.8?

5. Suppose 0.800 mol H 2 (g), 0.900 mol I 2 (g), and 0.100 mol HI(g) are mixed in a one liter vessel at 425 o C. K c = 54.8 at 425 o C for the reaction H 2 (g) + I 2 (g) º 2 HI(g) a. In which direction must the reaction run (forward or backwards) to achieve equilibrium? b. What are the concentrations of all species at equilibrium? Check your final answers with a Q calculation. Information (LeChatelier's Principle) In 1884 Henri LeChatelier formulated the following principle: If a stress is applied to a system at equilibrium, the system will tend to adjust to a new equilibrium, which minimizes the stress, if possible. The stresses are changes in concentration, pressure, and temperature. If the stress causes a change in the amounts of reactants and products present once equilibrium is reestablished, we say that a shift in the position of the equilibrium has occurred. If we think of a reaction equation in the usual way, with reactants on the left and products on the right, a stress to the system at equilibrium may cause either of the following: Shift to the right Shift to the left more reactant(s) consumed resulting in greater product and lesser reactant concentrations more product(s) consumed resulting in greater reactant and lesser product concentrations Sometimes the stress cannot be alleviated by either kind of shift, in which case the original equilibrium is maintained. The effects of each kind of stress on a system at equilibrium are summarized below. Concentration change: T K c or K p remains the same. T Increasing reactant concentrations or decreasing product concentrations causes a shift right (more product forms). T Increasing product concentrations or decreasing reactant concentrations causes a shift left (more reactant forms).

Pressure change: T K c or K p remains the same. T Only changes that affect the partial pressures of reactants and/or products can cause a change. For example, adding an inert gas has no effect. T Increasing pressure causes a shift to the side with the lower sum of coefficients on gas species. For example, increasing the pressure on an equilibrium mixture for the reaction, N 2 (g) + 3 H 2 (g) º 2 NH 3 (g), causes a shift right. T If the sum of coefficients on gas species is the same on the left and right, changing the pressure has no effect. For example, for an equilibrium mixture for the reaction, H 2 (g) + I 2 (g) º 2 HI(g), changing the pressure has no effect on the position of the equilibrium. Temperature change: T K c and K p change values! T Raising the temperature drives the endothermic process. For example, for the reaction, N 2 O 4 (g) º 2 NO 2 (g), H o = +58.0 kj/mol. Therefore, raising the temperature on an equilibrium mixture will favor formation of more NO 2 (g) T Lowering the temperature drives the exothermic process. For example, for the reaction, N 2 O 4 (g) º 2 NO 2 (g), H o = +58.0 kj/mol. Therefore, lowering the temperature on an equilibrium mixture will favor formation of more N 2 O 4 (g), because the reverse reaction is exothermic. Adding a catalyst T A catalyst has no effect on the position of the equilibrium, just how fast it gets there.

Key Questions 8. For each of the following reactions at equilibrium, predict the effect (if any) the indicated stress would have on the position of the equilibrium. Note whether or not the value of the equilibrium constant changes. a. H 2 (g) + I 2 (g) º 2 HI(g) More HI(g) is added. b. N 2 (g) + 3 H 2 (g) º 2 NH 3 (g) NH 3 (g) is removed as it forms. c. 2 NO(g) + Cl 2 (g) º 2 NOCl(g) Overall pressure is increased. d. 2 NO(g) + Cl 2 (g) º 2 NOCl(g) H o = 75.5 kj/mol Temperature is increased. e. H 2 O(g) + C(s) º H 2 (g) + CO(g) Overall pressure is increased. f. C(s) + O 2 (g) º CO 2 (g) Overall pressure is decreased. g. N 2 O 4 (g) º 2 NO 2 (g) N 2 (g) is added, increasing overall pressure. h. N 2 (g) + 3 H 2 (g) º 2 NH 3 (g) Iron powder is added as a catalyst. 9. An equilibrium mixture of H 2 (g), I 2 (g), and HI(g) in a one-liter vessel at 425 o C is found to have the following concentrations: [H 2 ] = 0.146 mol/l, [I 2 ] = 0.246 mol/l, [HI] = 1.41 mol/l. If 0.59 mol HI is added to the vessel, what will be the concentrations of all species once equilibrium is reestablished? K c = 54.8 at 425 o C for the reaction Information (Brønsted-Lowry Theory) H 2 (g) + I 2 (g) º 2 HI(g) In the Arrhenius theory, an acid is a substance that produces H + ion in solution, and a base is a substance that produces OH ion in solution. This theory is only useful for aqueous (water) solutions. In 1923, Brønsted (Danish) and Lowry (English) independently proposed a new theory that could be applied to other solvent systems. The Brønsted-Lowry Theory uses the following definitions: Acid - a substance that donates protons (H + ) Base - a substance that accepts protons (H + ) In this theory, acid-base reactions are seen as proton transfer reactions. For example, consider the reaction between the hydronium ion and ammonia in water: H 3 O + (aq) + NH 3 (aq) º H 2 O(l) + NH 4 + (aq)

This involves transfer of a proton from the hydronium ion to the ammonia molecule to produce water and the ammonium ion. This makes H 3 O + an acid and NH 3 a base. In general terms, all Brønsted-Lowry acid-base reactions fit the general pattern HA + B º A + HB + acid base When an acid, HA, loses a proton it becomes its conjugate base, A, a species capable of accepting a proton in the reverse reaction. HA º A + H + acid conjugate base Likewise, when a base, B, gains a proton, it becomes its conjugate acid, BH +, a species capable of donating a proton in the reverse reaction. B + H + º HB + base conjugate acid The Brønsted-Lowry concept of conjugate acid base pairs leads to the idea that all acid-base reactions are proton transfer reactions. The generic reaction between an acid HA and a base B can be viewed as HA º A + H + H + + B º HB + HA + B º A + HB + acid 1 base 2 base 1 acid 2 Species with the same subscripts are a conjugate acid-base pairs. Key Questions 10. In the spaces provided, write the formulas of the conjugate bases of the given Brønsted- Lowry acids. Remember to write the proper charge, if any, for the conjugate base in each case. Acid HNO 3 NH 4 + HSO 4 H 2 SO 4 HC 2 H 3 O 2 Conjugate base

11. In the spaces provided, write the formulas of the conjugate acids of the given Brønsted- Lowry bases. Remember to write the proper charge, if any, for the conjugate acid in each case. Base OH NH 3 HS SO 4 2 CH 3 NH 2 Conjugate acid 12. Using the notation acid 1 /base 1 and base 2 /acid 2 (as shown above), identify the Brønsted-Lowry conjugate acid-base pairs in the following reactions. HS (aq) + HC 2 H 3 O 2 (aq) º H 2 S + C 2 H 3 O 2 HF(aq) + PO 4 3 (aq) º F (aq) + HPO 4 2 (aq) HNO 2 (aq) + H 2 O(l) º NO 2 (aq) + H 3 O + CH 3 NH 2 (aq) + H 2 O(l) º CH 3 NH 3 + (aq) + OH (aq) Information (Water s Autoprotolysis) Pure water has an equilibrium with H 3 O + and OH called autoprotolysis or autodissociation. H 2 O + H 2 O º H 3 O + + OH base 1 acid 2 acid 1 base 2 for which we might write the following equilibrium expression: K c / K w = [H 3 O + ][OH ] We omit the concentration of H 2 O(l) from the expression, because it is a pure liquid. We can also ignore the concentration of H 2 O for dilute solutions, because [H 2 O] is virtually constant. At a temperature of 25 o C K w has a measured value of 1.0 x 10 14 : K w = [H 3 O + ][OH ] = 1.00 x 10 14 M 2 at 25 o C K w is called the ion product or dissociation constant of water. The units on K w are generally not written, but they are implicitly M 2, as shown above.

For pure water, [H 3 O + ] = [OH ]. To find the value of this concentration, let x = [H 3 O + ] = [OH ] and substitute into the ion product expression for water: K w = [H 3 O + ][OH ] = 1.0 x 10 14 = x 2 x = 1.0 x 10 7 M = [H 3 O + ] = [OH ] This condition of [H 3 O + ] = [OH ] = 1.0 x 10 7 M defines neutral water. The K w expression must be obeyed for any dilute solution in water. Suppose we add a strong acid, such as HCl, which dissociates completely into H 3 O + and Cl ions. This will increase [H 3 O + ] in the solution, but by LeChatelier s Principle it will cause the autoprotolysis equilibrium to shift left to maintain K w = [H 3 O + ][OH ] = 1.0 x 10 14. This will cause [OH ] to become smaller as more molecular H 2 O is formed in response to the stress of excess H 3 O + ion in the solution. A strong base, such as NaOH, which dissociates completely into Na + and OH ions, would have the reverse effect. This will increase [OH ] in the solution, but it will make [H 3 O + ] smaller as the autoprotolysis equilibrium shifts left to maintain K w = [H 3 O + ][OH ] = 1.0 x 10 14. In general, owing to these shifts in the autoprotolysis equilibrium, a solution is acidic if [H 3 O + ] > 10 7 M and [OH ] < 10 7 M, and a solution is basic if [H 3 O + ] < 10 7 M and [OH ] > 10 7 M. Key Questions 13. Given the following concentrations of H 3 O + or OH, calculate the concentration of OH or H 3 O +, and indicate whether the solution is acidic or basic. [H 3 O + ] [OH ] Acidic or basic? 1.0 x 10 5 M 4.0 x 10 9 M 1.2 x 10 8 M Information (ph and poh) It is often convenient to express [H 3 O + ] and [OH ] by the logarithmic terms ph and poh, respectively, defined as ph = log [H 3 O + ] poh = log [OH ] Note that the symbol "p" in front of any quantity means "negative base-10 logarithm of"; e.g., pk w = logk w. The relationship between ph and poh can be derived from K w. K w = [H 3 O + ][OH ] = 1.0 x 10 14

Taking base-10 logarithms of both sides of this equation, we have log K w = log[h 3 O + ][OH ] = log[h 3 O + ] + log[oh ] = 14.00 Changing the sign across this equation, we have log K w = log[h 3 O + ] log[oh ] = 14.00 If we now recognize that the negative logarithm of any quantity X can be designated with the symbol px, then this equation becomes pk w = ph + poh = 14.00 For neutral water, where [H 3 O + ] = [OH ] = 1.0 x 10 7 M, ph = poh = 7.00. For acidic solutions ph is less than 7, and for basic solutions ph is greater than 7. Note that poh runs in the opposite sense. For acid solutions poh is greater than 7, and for basic solutions poh is less than 7. It is possible to have solutions in which ph or poh are greater than 14 or less than 0 (i.e., negative). Note that the number of decimal places in a logarithm indicates the number of significant figures in the original number. For example, the base-10 logarithm of 1.45 x 10 15 (3 sig. figs.) is 14.839, where the decimal part indicates the three significant figures. The 14 part (left of the decimal) relates to the power of ten, and has no relevance to significant figures. Why is this? Because we cannot take the logarithm of a negative decimal number, the logarithm 14.839 is actually the sum of a negative integer ( 15) and a positive decimal (0.161); i.e., 15 + 0.161 = 14.839. The 15, called the characteristic, relates to the 10 15 of the original number, and the 0.161, called the mantissa, relates to the 1.45 of the original number, which is where the significant digits are. Be sure when calculating ph or poh that the number of decimal places matches the number of significant figures for the concentration of [H 3 O + ] or [OH ]. Key Questions 14. Complete the following table by calculating the missing entries and indicate whether the solution is acidic or basic. Be sure your answers are expressed to the proper number of significant figures. [H 3 O + ] [OH ] ph poh acidic or basic? 5.8 x 10 5 M 6.2 x 10 6 M 11.24 9.70 2.0 M

Information (Solutions of Strong Acids and Bases) When a strong acid, HA, dissolves in water it dissociates completely into H 3 O + and A ions, leaving no molecular HA in the solution: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Thus, when we talk about the concentration of a strong acid solution, say 0.10 M HCl, we are usually referring to the number of moles per liter used to make the solution, not what is actually present in the solution. Because of dissociation, there is virtually no molecular HCl in a 0.10 M HCl solution! We can define the analytical concentration of the acid HA, symbolized C HA, as the number of moles per liter of acid used to make the solution. Thus, for 0.10 M HCl, C HCl = 0.10 M. Then, using bracket symbols, we can refer to actual concentrations in the solution as [HA], [H 3 O + ], and [A @ ]. For moderate analytical concentrations of strong acid (C HA >> 10 7 M), we can generally assume that the acid s dissociation supplies virtually all the H 3 O + in the solution; i.e., water s contribution is insignificant. For a given analytical concentration of a pure strong acid at moderate concentration, we would have the following values initially and after dissociation: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Initial C HA -0 0 After dissociation -0 C HA C HA In other words, the concentrations of hydronium ion and conjugate base in the solution are equivalent to the number of moles per liter of acid added to make the solution; i.e., [H 3 O + ] = [A ] = C HA. Because of complete dissociation, [HA] = 0. For 0.10 M HCl, we would have [HCl] = 0, [H 3 O + ] = [Cl ] = 0.10 M. In similar fashion, a strong base dissociates completely on dissolving in water. Most strong bases are ionic hydroxide compounds, with the general formula M(OH) n, where n = 1, 2, 3, depending on the charge on the cation, M n+. The number of moles per liter of ionic base used to make the solution, the analytical concentration, can be designated. Analogous to the strong acid case, for solutions of pure strong base at moderate analytical concentration ( >> 10 7 M), we can generally assume that the base s dissociation supplies virtually all the OH ion in the solution; i.e., water s contribution is insignificant. For a given analytical concentration of a pure strong base at moderate concentration, we would have the following values initially and after dissociation: M(OH) n (aq) M n+ (aq) + n OH (aq) Initial 0-0 After dissociation -0 n In other words, the concentration of hydroxide ion and cation in the solution are equivalent to the number of moles per liter of base used to make the solution, factoring in the stoichiometry of the

dissociation; i.e., [OH ] = n, [M n+ ] =. Because of complete dissociation, [M(OH) n ] = 0. For any acid solution, once [H 3 O + ] is determined, [OH ] = K w /[H 3 O + ]. Likewise, for any base solution, once [OH ] is determined, [H 3 O + ] = K w /[OH ]. Key Questions 15. Determine the concentrations of H 3 O + and OH in the following solutions of strong acids or bases in water. [Caution: Think about how the acid or base dissociates in water.] Solution [H 3 O + ] [OH ] 2.5 x 10 3 M HCl 0.0010 M NaOH 7.5 x 10 5 M Ca(OH) 2