Question 1 (c)

Using Ka to Calculate pH Knowing the value of Ka and the initial concentration of a weak acid, we can calculate the concentration of H+(aq) in a solution of the acid. Lees calculate the pH at 25 oc of a 0.30 M solution of acetic acid (CH3COOH), the weak acid responsible for the charac- teristic odor and acidity of vinegar. 1. Our first step is to write the ionization equilibrium: CH3COOH(aq) H+(aq) + CH3COO (aq) 16.28 Notice that the hydrogen that ionizes is the one attached to an oxygen atom. 2. The second step is to write the equilibrium-constant expression and the value for the equilibrium constant. Taking Ka = 1.8 X 10 5 from Table 16.2, we write 5 = 1.8 x 10 CH3COOH 16.29 3. The third step is to express the concentrations involved in the equilibrium reac- tion. This can be done with a little accounting, as described in Sample Exercise 16.10. Because we want to find the equilibrium value for l, let's call this quan- tity x. The concentration of acetic acid before any of it ionizes is 0.30 M. The chem- ical equation tells us that for each molecule of CH3COOH that ionizes, one H+(aq) and one CH3COO (aq) are formed. Consequently, if x moles per liter of H+(aq) form at equilibrium, x moles per liter of rnust also form and x moles per liter of CH3COOH rnust be ionized:

Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) CH3COOH(aq) 0.30 —x (0.30 - x) H+(aq) x CH3COO (aq) x 4. The fourth step is to substitute the equilibrium concentrations into the equilibrium- constant expression and solve for x: 5 = 1.8 x 10 CH3COOH1 0.30 — x 16.30 This expression leads to a quadratic equation in x, which we can solve by using either an equation-solving calculator or the quadratic formula. We can simplify the problem, however, by noting that the value of Ka is quite small. As a result, we anticipate that the equilibrium lies far to the left and that x is much smaller than the initial concentration of acetic acid. Thus we assume that x is ne li ible relative to 0.30 so that 0.30 — x is essentiall e ual to 0.30. We can (and should\!) check the validity of this assumption when we finish the problem. By using this assumption, Equation 16.30 becomes — @ = 1.8 X 10 5 0.30

Solving for х, we have х2 = х 10-5) = 5.4 х 10 —6 -6 = 2.3 з 5.4 х 10 х 10 = х = 2.3 х 10 зм рн = —log(2.3 х 10-3) = 2.64

Now we check the validity of our simplifying assumption that 0.30 — x 0.30. The value of x we determined is so small that, for this number of significant figures, the assumption is entirely valid. We are thus satisfied that the assumption was a reasonable one to make. Because x represents the moles per liter of acetic acid that ionize, we see that, in this particular case, less than 1% of the acetic acid molecules ionize: 0.0023 M Percent ionization of CH3COOH — x 100% = 0.77% 0.30 M As a general rule, if x is more than about 5% of the initial concentration value, it is better to use the quadratic formula. You should always check the validity of any simplifying assumptions after you have finished solving a problem. We have also made one other assumption, namely that all of the in the solution comes from ionization of CH3COOH. Are we justified in neglecting the autoionization of H20? The answer is yes—the additional H+ due to water, which would be on the order of 10 7 M, is negligible compared to the H+ from the acid (which in this case is on the order of 10 3 M). In extremely precise work, or in cases involving very dilute solutions of acids, we would need to consider the autoionization ofwater more fully.

Question 1 (e)

25.00 0.100 M equivalence point 1/2 equivalence point 0.100 M NaOH added

Solution containing weak acid Neutraliza tion reaction: HA decreases A — increases HA + OW A - + H20 Calculate new values of HA and A -1 Add strong base Use Ka, HA , and A—I to calculate H +1 Equilibrium calculation pH Stoichiometry calculation A Figure 17.10 strong base. Procedure for calculating pH when a weak acid is partially neutralized by a

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081 00 • o) 00 • hl SOO-O) s- -z h1S00 -O

Question 2 (b)

met\* (Z — I) 2 8h2 or (dividing both sides by n to convert Eto t): 1 1 22 — (Z— (2.48. 1015 1) 8h3E2 4

1 1 1 Wavelength R Rydberg constant Z z- Atomic number (3 Transition n Integer : Til <

The positive charge holding the electron is greater for He+, which has a 2+ nucleus, than for H with its 1+ nucleus. The stronger attraction means that it requires more energyl pt for the electron to move to higher energy levels. Therefore, transitions from high energy states to lower states will be more energetic for He + than for H. Note: Other arguments accepted, such as, "E is proportional to Z 2. Since Z = 2 for He and Z = 1 for H, all energy levels in He + are raised (by a factor of 4)." Other accepted answers must refer to the increased gharge on the He+ nucleus, and NOT the mass.

Question 3 (b)

Reaction Orders: The Exponents in the Rate Law The rate law for most reactions has the form Rate = kreactant 1 lmrreactant 21M ... 14.8 The exponents m and n are called reaction orders. For example, consider again the rate law for the reaction of NH4+ with N02 Rate = Because the exponent of NH4+ is 1, the rate is first order in NH4+. The rate is also first order in N02— . (The exponent 1 is not shown in rate laws.) The overall reaction order is the sum of the orders with respect to each reactant represented in the rate law. Thus, for the NH4+ — N02— reaction, the rate law has an overall reaction order of 1 + 1 = 2, and the reaction is second order overall. The exponents in a rate law indicate how the rate is affected by each reactant con- centration. Because the rate at which NH4+ reacts with N02— depends on C NH4+ raised to the first power, the rate doubles when NH4+l doubles, triples when r NH4+l triples, and so forth. Doubling or tripling r N02— likewise doubles or triples the rate. If a rate law is second order with respect to a reactant, r A 12, then doubling the concentration of that substance causes the reaction rate to quadruple because 212 = 4, whereas tripling the concentration causes the rate to increase ninefold: 31 The following are some additional examples of experimentally determined rate laws: 2 N205(g) 4 N02(g) + 02(g) Rate = kCN205 H2(g) + 12(g) 2 Hl(g) Rate = 14.9 14.10 CHC13(g) + C12(g) CC14(g) + HC1(g) Rate = 14.11 Although the exponents in a rate law are sometimes the same as the coefficients in the balanced equation, this is not necessarily the case, as Equations 14.9 and 14.11 show. For any reaction, the rate law must be determined experimentally. In most rate laws, reac- tion orders are 0, 1, or 2. However, we also occasionally encounter rate laws in which the reaction order is fractional (as is the case with Equation 14.11) or even negative.

Table 14.3 Elementary Reactions and Their Rate Laws Rate - Rate = Rate = Rate = Molecularity Unimolecular Bimolecular Bimolecular Termolecular Termolecular Termolecular Elementary Reaction A products A + A products A + B products A + A + A products A + A + B products A + B + C products Rate Law Rate = Rate =

Question 3 (d)

Mechanisms with a Slow Initial Step We can most easily see the relationship between the slow step in a mechanism and the rate law for the overall reaction by considering an example in which the first step in a multistep mechanism is the rate-determining step. Consider the reaction of N02 and CO to produce NO and CO, (Equation 14.23). Below 225 oc, it is found experi- mentally that the rate law for this reaction is second order in NO, and zero order in CO: Rate = k N0212. Can we propose a reaction mechanism consistent with this rate law? Consider the two-step mechanism:\* N03(g) + NO(g) (slow) step 1: N02(g) + N02(g) Step 2: N03(g) + CO(g) N02(g) + C02(g) (fast) Overall: N02(g) + CO(g) NO(g) + C02(g) Step 2 is much faster than step 1; that is, k2 >> kl, telling us that the intermediate N03(g) is slowly produced in step 1 and immediately consumed in step 2. Because step 1 is slow and step 2 is fast, step 1 is the rate-determining step. Thus, the rate of the overall reaction depends on the rate of step 1, and the rate law of the

overall reaction equals the rate law of step 1. Step 1 is a bimolecular process that has the rate law Rate = Thus, the rate law predicted by this mechanism agrees with the one observed experi- mentally. The reactant CO is absent from the rate law because it reacts in a step that follows the rate-determining step. A scientist would not, at this point, say that we have "proved" that this mecha- nism is correct. All we can say is that the rate law predicted by the mechanism is consistent with experiment. We can often envision a different sequence of steps that leads to the same rate law. If, however, the predicted rate law of the proposed mecha- nism disagrees with experiment, we know for certain that the mechanism cannot be correct.

Question 5 (d)

I X 100

Question 6 (b)

(b) (i)The reaction rate depends on the reaction kinetics which is determined by the value of the activation energy, Eact. If the activation energy is large, a reaction that is thermodynamically spontaneous may proceed very slowly (if at all). One point eamed for linking the rate of the reaction to the activation energy, which may be explained verbally or shown on a reaction profile diagram (ii) The catalyst has no effect on the value of AGO. The catalyst reduces the value of Eact, increasing the rate of reaction, but has no effect on AHO and ASO, so it cannot affect the thermodynamics of the reaction. One point eamed for indicating no change in the value of AGO One point eamed for indicating (verbally, or with a reaction-profile diagram) that the catalyst affects the activation energy 1 pt 1 pt the values of 1 pt

Question 7 (b)

Expected and Obsewed Values of the van't Hoff Factor for 0.05 m Solutions of Several Electrolytes Electrolyte NaCl MgC12 MgS04 HCI Glucose\* i (expected) 2.0 3.0 2.0 4.0 2.0 1.0 i (observed) 1.9 2.7 1.3 34 1.9 1.0 WA nonelectrolyte shown for comparison.

Question 7 (c)

The lowering of vapor pressure of water is directly proportional to the concentration of solute particles in solution. C2H50H is the only nonelectrolyte, so it will have the fewest solute particles in solution.

Vapor pressure (Pa)

vapor pressure of water (torr)

Question 7 (d)

These salts have a neutral cation and a basic anion. For example, NaF dissociates into Na+ and F— ions. The Na+ ion has no acidic or basic properties, but the fluoride ion is a conjugate base of a weak acid and therefore is a basic anion. NaF a salt of the NaOH, a strong base, and HF, a weak acid. hydrolysis and acid-base properties of salts (15.10) www.mhhe.com/physsci/chemistry/chang7/ssg/chap15\_10sg.html

Question 8 (b)

(ii) CF4 has a tetrahedral geometry, so the bond dipoles cancel, leading to a nonpolar molecule. With five pairs of electrons around the central S atom, SF4 exhibits a trigonal bipyramidal electronic geometry, with the lone pair of electrons. In this configuration, the bond dipoles do not cancel, and the molecule is polar.

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