Chemistry part 7, Julia Burdge,2e (2009) ppt

136 CHAPTER 4 Reactions in Aqueous Solutions Figure 4.8 Two so lutions of iodine in ben ze ne. The solution on the left is more concentrated. The so lution on the right is more dilute. The qualitative terms concentrated and dilute are relati ve terms, like expensive and cheap. r "' , _ : Multimedia Solutions-preparation of so lutions. Molarity can equally well be defined as millimoles per milliliter (mmol/mL ), which ca n simplify some calculations. • • Students sometimes have difficulty seeing how units cancel in these equations. It may help to write M as mollL unt il you become completely comfortable with these equations. Concentrated solution: Dilute solution: More solute particles per unit volume Fewer so lute particles per unit volume Concentration of Solutions One of the factors that can influence reactions in aqueous solution is concentration. The concen- tration of a solution is the amount of solute dissolved in a given quantity of solvent or solution. Consider the two solutions of iodine pictured in Figure 4.8. The solution on the left is more con- centrated than the one on the right-that is, it contains a higher ratio of solute to solvent. By con- . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . • • • • • trast, the solution on the right is more dilute. The color is more intense in the more concentrated solution. Often the concentrations of reactants determine how fast a chemical reaction occurs. For example, the reaction of magnesium metal and acid [ ~~ Section 4.4] happens faster if the concentration of acid is greater. As we will see in Chapter 13, there are several different ways to express the concentration of a solution. In this chapter, we introduce only molarity, which is one of the mo st commonly u se d units of concentration. Molarity Molarity, or molar concentration, symbolized M, is defined as the number of moles of solute per liter of solution. Thu s, 1 L of a 1.5 molar solution of glucose (C 6 H I2 0 6 ), written as 1.5 M C 6 H 12 0 6 , contains 1.5 moles of dissolved glucose. Half a liter of the same solution would contain 0.75 mole of dissolved glucose, a milliliter of the solution would contain 1.5 X 10- 3 mole of dissolved glucose, and so on. Equation 4.1 l 't moles solute mo an y = liters solution In order to calculate the molarity of a solution, we divide the number of moles of solute by the volume of the solution in liters. Equation 4.1 can be rearranged in three ways to solve for any of the three variables: molarity (M), moles of solute (mol), or volume of solution in liters (L) . . . . . . . . . . . . , ·mol· mel· . (1) M = L (2) L = M (3) mol = M XL Sample Problem 4.9 illustrates how to use these equations to solve for molarity, volume of solu- tion, and moles of solute. Sample Problem 4.9 For an aqueous solution of glucose (C 6 H I2 0 6 ), determine (a) the molarity of 2.00 L of a so lution that contains 50.0 g of glucose, (b) the volume of this so lution that would co ntain 0.250 mole of glucose, and (c) the number of mole s of glucose in 0.500 L of this solution. SECTION 4.5 Concentration of Solutions 137 Strategy Convert the mass of glucose given to mole s, and use the equations for interconversions of M, liter s, and moles to calculate the answers. Setup The molar mass of glucose is 180.2 g. 50.0 g moles of glucose = 80 I = 0.277 mol 1 .2 glmo . : A common way to state the concentration of : this solution is to say, "This solution is 0. 139 M · in glucose." • • • • • • • • • • Solution 0.277 mol C6H'206 : (a) molarity = 200 L l ' = 0.139 M Think About It Check to see that the magnitudes of your answers are logical. For example, the mass given in the problem corresponds to 0.277 mole of solute. If you are asked, . so utlOn 0.250 mol C 6 H p 0 6 (b) volume = 0 9 - = 1. 80 L .13 M (c) moles of C 6 H 12 0 6 in 0.500 L = 0.500 L X 0.139 M = 0.0695 mol Practice Problem A For an aqueous so lution of sucrose (C 12 H 22 0 11 ), determine (a) the molarity of 5.00 L of a solution that contains 235 g of sucrose, ( b) the vo lume of this so lution that would contain 1.26 mole of sucrose, and (c) the number of moles of sucrose in 1.89 L of this so lution. Practice Problem B For an aqueous solution of sodium chloride (NaCl), determine ( a) the molarity of 3.75 L of a solution that contains 155 g of so dium chloride, (b) the vo lume of this solution that would contain 4.58 moles of so dium chloride, and (c) the number of moles of sodium chloride in 22.75 L of this so lution. ~. _ The procedure for preparing a so lution of known molarity is s hown in Figure 4.9. First, the solute is weighed accurately and tran s ferred , often with a funnel, to a volumetric flask of the desired volume. Next , water is added to the flask, which is then sw irled to dissolve t he so lid. After all the solid ha s dis so lved , more water is added slowly to bring the level of o lution exactly to the volume mark. Knowing the volume of the so lution in the fla sk and the . . . . . . . . . . . . . . . . . . . . . . . . . . q uantity of compound dissolved , we can determine the molarity of the so lution us ing Equation 4. 1. Note that this procedure doe s not require that we know the exact amount of water added. Beca u se of the way molarity is defined , it is important only that we know the final volume of the solution. Dilution Co ncentrated "stock" solutions of commonly u se d substances typicall y are kept in the labora- to ry stockroom. Often we need to dilute these s tock s olution s before us ing them. Dilution is the p rocess of preparing a less concentrated sol ution from a more co ncentr ated one. Suppose th at we want to prepare 1.00 L of a 0.400 M KMn04 so lution from a so lution of 1.00 M KMn0 4' For this purpo se we need 0.400 mole of KMn0 4' Becau se ther e is 1.00 mole of KMn0 4 in 1. 00 L of a 1.00 M KMn0 4 s olution , there is 0.400 mole of KMn0 4 in 0.400 L of the same -o lution: 1.00 mol KMn04 0.400 mol KMn0 4 1.00 L of solution OAOO L of solution The refore, we must withdraw 400 mL from the 1.00 M KMn0 4 so lution and dilute it to 1.00 L by adding water (in a 1.00-L volumetric fla sk) . This method gives us 1.00 L of the desired 0.400 M KM n04' r In carrying out a dilution proce ss, it is u se ful to remember thilt adding more solvent to a giv en amount of the stock solution changes ( decrea ses) the concentration of the solution without ha nging the number of moles of so lute pre se nt in the solution (F igure 4.10, p. 140). moles of solute befor e dilution = mole s of solute after dilution Equation 4.2 L ing arrangement (3) of Equation 4.1 , we can calculate the number of mole s of solute: mol es of so lute. . moles of solute = X hter s of solutlon liters of so lution as in part (b), for the volume that contains a number of moles smaller than 0.277, make sure your answer is smaller than the original volume. , • - Media Player/MP EG Animation: Figure 4. 9, Prepar i ng a Solution from a So l id , pp . 138-139. , It is important to remember that molarity is defined in terms of the volume of solution, not the volume of solven t. In many cases, these two are not the same. Multimedia Solutions-dilution. • ! Figure 4.9 • 138 Weigh out the solid KMn04 (The tare function on a digital balance automatically subtracts the mass of the weighing paper.) The mass likely will not be exactly the calculated number Transfer the weighed KMn04 to the volumetric flask Calculate the mass of KMn04 necessary for the target concentration of 0.1 M. 0.1 tal X 0.2500 L = 0.02500 mol 158.04 g 0.02500 mol X I = 3.951 g KMn04 rna \ • Add water sufficient to dissolve the KMn0 4 Fill exactly to the calibration mark using a wash bottle Calculate the actual concentration of the solution. 1 mol 3.896 g KMn04 X 158.04 g = 0.024652 mol 0.024652 mol = 009861 M 0.2500 L . Swirl the flask to dissolve the solid Add more water Whatls the point? The goal is to prepare a solution of precisely known concentration, with that concentration being very close to the target concentration of 0.1 M. Note that because 0.1 is a specified number, it does not limit the number of significant figures in our calculations. 139 140 CHAPTER 4 Reactions in Aqueous Solutions Figure 4.10 Dilution changes the concentration of a solution; it does not change the number of moles of solute in the solution. Students sometimes resist using the unit millimole. However, using the Me X mLc = Md X m'-d form of Equation 4.3 often reduces the number of steps in a problem thereby reducing the number of opportunities to make calculation errors. It is very important to note that, for safety, when di luting a concentrated acid, the acid must be added to the wate r, and not the other way around. Think About It Plug the answer into Equation 4.3, and make s ure that the product of concentration and vo lum e on both sides of the equation give the same result. • Before dilution: More solute particles per unit vo lume Add solvent • After dilution: Fewer so lute particles per unit volume Because the number of moles of solute before the dilution is the same as that after dilution, we can write Equation 4.3 where the subscripts c and d stand for concentrated and dilute, respectively. Thus, by knowing the molarity of the concentrated stock solution (Me) and the desired final molarity (Md) and volume (Ld) of the dilute solution, we can calculate the vo lume of stock solution required for the dilution (Le). Because most vo lumes measured in the laboratory are in milliliters rather than liters, it is worth pointing out that Equation 4.3 can be written with volumes of the concentrated and dilute solutions in milliliters. . . . . . . . . . . . . . . . . . . . . . . • • • • · • • • • • • · • • • • • • • • • • • • • • • In this form of the equation, the product of each side is in millimoles (mmol) rather than moles. We apply Equation 4.3 in Sample Problem 4.10. Sample Problem 4.10 What volume of 12.0 M Hel , a co mmon laboratory stock solution, must be u sed to prepare 250.0 mL of 0.125 M He]? Strategy Use Equation 4.3 to determine the vo lume of 12.0 M Hel required for the dilution. Because the desired final vo lum e is given in milliliters, it will be convenient to use the form of Equation 4.3 that include s milliliters. Setup Me = 12.0 M, Md = 0.125 M, mLd = 250.0 mL. Solution 12.0 M X mL e = 0.l25 M X 250.0 mL . . . . . . . . . . . . . . . ~ = 0.l25 M X 250.0 mL = 2.60 mL e 12.0 M Prac.tic.e Problem A What vo lume of 6.0 M H 2 S0 4 is n ee ded to prepare 500.0 mL of a solution that is 0.25 M in H 2 S0 4 ? Practice ·Problem B What vo lum e of 0.20 M H 2 S0 4 ca n be prepared by diluting 125 mL of 6.0 M H 2 S0 4 ? SECTION 4.5 Concentration of Sol utions 141 Solution Stoichiometry Soluble ionic compounds such as KMn04 are strong electrolytes, so they undergo complete dissociation upon dissolution and exist in solution entirely as ions. KMn04 dissociates, for example, to give 1 mole of potassium ion and 1 mole of permanganate ion for every mole of potassium permanganate. Thus, a 0.400 M solution of KMn04 will be 0.400 M in K+ and- 0.400 M in MnO 4 . In the case of a soluble ionic compound with other than a 1: 1 combination of constituent ions, we must use the subscripts in the chemical formula to determine the concentration of each ion in solution. Sodium sulfate (Na 1 S04) dissociates, for example, to give twice as many sodium ions as sulfate ions. Na2S0is) H 2 0, 2Na+(aq) + SO ~-( aq) Therefore, a solution that is 0.35 Min Na2S04 is actually 0.70 M in Na + and 0.35 M in SO ~- . Frequently, molar concentrations of dissolved species are expressed using square brackets. Thus, the concentrations of species in a 0.35 M solution of Na2S04 can be expressed as follows: [Na +] . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ~ . . . . . . . . . . . " 0.70 M and [SO 4 - ] = 0.35 M. Sample Problem 4.11 lets you practice relating concentrations of : • compounds and concentrations of individual ions using solution stoichiometry. : Using square-bracket notation, express the concentration of (a) chloride ion in a solution that is 1.02 M in AICI 3 , (b) nitrate ion in a solution that is 0.451 Min Ca(N0 3 )2, and (c) Na 2 C0 3 in a solution in which [Na +] = 0.124 M. Strategy Use the concentration given in each case and the stoichiometry indicated in the conesponding chemical fOlTllula to determine the concentration of the specified ion or compound. Setup (a) There are 3 moles of Cl- ion for every 1 mole of AlCl 3 , so the concentration of Cl- will be three times the concentration of AlCl 3 . • (b) There are 2 moles of nitrate ion for every 1 mole of Ca(N0 3 )2, Ca(N0 3 Ms) H2 0, Ca 2+ (aq) + 2NO](aq) so [NO] ] will be twice [Ca(N0 3 h ]. (c) There is 1 mole of Na 2 C0 3 for every 2 moles of sodium ion, Na 2C03(S) H 2 0, 2Na +(aq) + CO ~ - (aq) so [Na 2C 0 3] will be half of [ Na +]. (Ass ume that Na2C0 3 is the only source of a + ion s in this solution.) Solution (a) [CI-] = [AICH X 3 mol Cl- o 1 mol AICl 3 ~1~.0~2~~~i 3 X 3 mol Cl- L 1 3.06 mol Cl- L = 3.06 M 3 _ 2 mol NO] (b) [N0 3 ] = [Ca(N0 3 )2] X 1 mol Ca(N0 3 )2 = 0.451 mol Ca(H0 3 ); X 2 mol NO ] _ L 1 _mol Ca(N0 3 h 0.902 mol NO] L = 0.902 M (Continued) • • • ' Square brackets around a chemical species can be read as "the concentration of" that species . For example, [Na +] is read as "the concentration of sodium ion." If we only need to express the concentration of the compound, rather than the concentrations of the individual ions, we could express the concentration of this solution as [Na,S04] = 0.35 M • 142 CHAPTER 4 Reac t ions in Aqueous So l utions Think About It Make sure that units cancel properly. Remember that the concentration of an ion can never be less than the concentration of its dissolved parent compound. It will always be the concentration of the parent compound time s it s stoichiometric subscript in the chemical formula. Ex per i ments t ha t me as ur e the amount of a s ubstan ce pres en t ar e ca ll ed qu a ntitative a na l ys i s. Ac co rdin g to the in formation in T ab le 4 .2, Ag el is an inso l ub le exce ption to the chlo r ides, w hi ch ty pica lly are so lubl e. _ 0.124 ~ X 1 mol Na 2 C0 3 L 2 ~ 0.0620 mol Na 2 C0 3 L = 0.0620M Practice Problem A Using the square-bracket notation, express the concentrations of ions in a solution that is 0. 750 M in aluminum sulfate [AI 2 ( S0 4)3 ]. Practice Problem B Using the square-bracket notation, expr ess the concentration of chloride ions in a solution that is 0.250 M in s odium chloride (NaCl) and 0.25 M in magnesium chloride (MgCI 2 ). Checkpoint 4.5 Concentration of Solutions 4. 5.1 Calculate the molar concentration of a solution prepared by dissolving 58 .5 g NaOH in enough water to yield 1.25 L of solution. a) 1.46M b) 46.8 M c) 2.14 X 10- 2 M d) 1.l7M e) 0.855 M 4.5.2 What ma ss of gl uco se (C 6 H 12 0 6 ) in grams must be used in order to prepare 500 mL of a solution that is 2.50 Min glucose? a) 225 g b) 125 g c) 200 g d) 1.25 g e) 625 g 4.5.3 4.5.4 What volume in mL of a 1.20 M HC I solution must be diluted in order to prepare 1.00 L of 0.0150 M HCI? a) 15.0 mL b) 12.5 mL c) 12.0 mL d) 85.0 mL e) ll5 mL A solution that is 0.18 Min Na 2 C0 3 is also . (Choose all that apply.) a) 0.18MinCO ~- b) 0.18 Min Na + . c) 0.09 M in Na+ d) 0.09 M in CO ~- e) 0.36 Min Na + Aqueous Reactions and Chemical Analysis . . . . . . . . . . . . . . . . . . . . Certain aqueous reactions are useful for determining how much of a particular substance is pres- ent in a sample. For example, if we want to know the concentration of lead in a sample of water, or if we need to know the concentration of an acid, knowledge of precipitation reactions, acid- base reactions, and solution stoichiometry will be useful. Two common types of such quantitative analyses are gravimetric analysis and acid-base titration. Gravimetric Analysis Gravimetric analysis is an analytical technique based on the measurement of mass. One type of gravimetric analysis experiment involves the formation and isolation of a precipitate, such as AgCI(s): AgN0 3 (aq) + NaCI(aq) - -+. NaN0 3 (aq) + AgCI(s) SECTION 4.6 Aqueous Reactions and Chemical Analysis 143 This reaction is often used in gravimetric analysis because the reactants can be obtained in pure form. The net ionic equation is Ag+(aq) + Cl - (aq) AgCI(s) Suppose, for example, that we wanted to test the purity of a sample of NaCI by determining the percent by mass of Cl. First, we would accurately weigh out some NaCl and dissolve it in water. To this mixture, we would add enough AgN0 3 solution to cause the precipitation of all the Cl- ions present in solution as AgCl. (In this procedure NaCI is the limiting reagent and AgN0 3 is the excess reagent.) We would then separate, dry, and weigh the AgCl precipitate. From the measured mass of AgCl, we would be able to calculate the mass of CI using the percent by mass of CI in AgCl. Because all the CI in the precipitate came from the dissolved NaCl, the amount of Cl that we calculate is the amount that was present in the original NaCI sample. We could then calculate the percent by mass of CI in the NaCI and compare it to the known composition of NaCl to deter- mine its purity. Gravimetric analysis is a highly accurate technique, because the mass of a sample can be measured accurately. However, this procedure is applicable only to reactions that go to comple- tion or have nearly 100 percent yield. In addition, if AgCI were soluble to any significant degree, it would not be possible to remove all the CI- ions from the original solution, and the subsequent calculation would be in error. Sample Problem 4.12 shows the calculations involved in a gravimet- ric experiment. A 0.8633-g sample of an ionic. compound containing chloride ions and an unknown metal cation is dissolved in water and treated with an excess of AgN0 3 . If 1.5615 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? Strategy Using the mass of AgCI precipitate and the percent composition of AgCI, determine what mass of chloride the precipitate contains. The chloride in the precipitate was originally in the unknown compound. Using the mass of chloride and the mass of the original sample, determine the percent Cl in the compound. Setup To determine the percent Cl in AgCl, divide the molar mass of CI by the molar mass of AgCI: 35.45 g -:-::-:::-:-: :: -c-=:-:::-:: :- X 100 % = 24.72 % (35.45 g + 107.9 g) . The mass of Cl in the precipitate is 0.2472 X 1.5615 g = 0.3860 g. Solution The percent Cl in the unknown compound is the mass of Cl in the precipitate divided by the mass of the original sample: 0.3860 g c:-::-=-::- - X 100% = 44.71 % Cl 0.8633 g Practice Problem A A 0.5620-g sample of an ionic compound containing the bromide ion (Br- ) is dissolved in water and treated with an excess of AgN0 3 . If the mass of the AgBr precipitate that forms is 0.8868 g, what is the percent by mass of Br in the original compound? . Practice Problem B A sample that is 63.9 percent chloride by mass is dissolved in water and treated with an excess of AgN0 3 . If the mass of the AgCI precipitate that forms is 1.085 g, what was the mass of the original sample? ~, _ Gravimetric analysis is a quantitative method, not a qualitative one, so it does not establish :he identity of the unknown substance. Thus, the results in Sample Problem 4.12 do not identify :he cation. However, knowing the percent by mass of CI greatly helps us narrow the possibili- ties. Because no two compounds containing the same anion (or cation) have the same percent ;:o mposition by mass, comparison of the percent by mass obtained from gravimetric analysis ":\i th that calculated from a series of known compounds could reveal the identity of the unknown ~ o mpounds. Think About It Pay close attention to which numbers correspond to which quantities. It is easy in this type of problem to lose track of which mass is the precipitate and which is the original sample. Dividing by the wrong mass at the end will result in an incorrect answer. 144 CHAPTER 4 Reactions in Aqueous Solutions • Multimedia Chemical reactions- titrations. Standardization in this context is the meticulous determination of concentration. Note that KHP is a monoprotic acid, so it reacts in a 1: 1 ratio with hydroxide ion . The endpoint in a titration is used to approximate the equivalence point. A careful choice of indicators, which we will discuss in Chapter 16, helps make this approximation reasonable. Phenolphthalein, although very common, is not appropriate for every acid-base titration. Figure 4.11 Apparatus for titration. Acid-Base Titrations Quantitative studies of acid-base neutralization reactions are most conveniently carried out using a technique known as titration. In titration, a solution of accurately known concentration, called a standard solution, is added gradually to another solution of unknown concentration, until the chemical reaction between the two solutions is complete, as shown in Figure 4.11. If we know the volumes of the standard and unknown solutions used in the titration, along with the concentration of the standard solution, we can calculate the concentration of the unknown solution. A solution of the strong base sodium hydroxide can be used as the standard solution in a titration, but it must first be standardized, because sodium hydroxide in solution reacts with car- hon' dIOXIde ' In ' the' aiT: iiiabng ' Its 'con"centratloiiliiistable 'over' dill · e. We 'can' standardize the sodium hydroxide solution by titrating it against an acid solution of accurately known concentration. The acid often chosen for this task is a monoprotic acid called potassium hydrogen phthalate (KHP), for which the molecular formula is KHC s H 4 0 4 . KHP is a white, soluble solid that is commercially available in highly pure form. The reaction between KHP and sodium hydroxide is < . . . . . . . . . . KHC s H 4 0iaq) + NaOH(aq) • KNaC s H 4 04(aq) + H 2 0(l) and the net ionic equation is To standardize a solution of NaOH with KHP, a known amount of KHP is transferred to an Erlenmeyer flask and some distilled water is added to make up a solution. Next, NaOH solution is carefully added to the KHP solution from a buret until all the acid has reacted with base. This point in the titration, where the acid has been completely neutralized, is called the equivalence point. It . , , is usually signaled by the endpoint, where an indicator causes a sharp change in the color of the solution. In acid-base titrations, indicators are substances that have distinctly different colors in acidic and basic media. One commonly used indicator is phenolphthalein, which is colorless in acidic and neutral solutions but reddish pink in basic solutions. At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. However, if we add just one more drop of NaOH solution from the buret, the solution will be basic and will immediately turn pink. Sample Problem 4.13 illustrates just such a titration. 'Y " « " '. r - ~ L- .;' i \ I'ItENOlPHTHAWH INDICATOR ~ ._ ,, _J ,. I· 1. ~ . 0 • I I , t « SECTION 4.6 Aqueous Reactions and Chemical Analysis 145 In a titration experiment, a student finds that 25.49 mL of an NaOH solution is needed to ne utralize 0.7137 g of KHP. What is the concentration (in M) of the N aOH solution? Strategy Using the ma ss given and the molar ma ss of KHP, determine the number of moles of KHP. Recognize that the number of moles of N aOH in the vo lume giv en is equal to the numb er of mole s of KHP. Divide mole s of NaOH by v olume (in liters) to get molarity. Setup The molar mas s of KHP ( KHC 8 H 4 0 4 ) = [39.1 g + 5( l.008 g) + 8(12.01 g) + 4(16.00 g)] = 204.2 g/mol. Solution 0.7 13 7 g mole s of KHP = 204.2 g/ mol = 0.003495 mol Because moles of KHP = mole s of N aOH , then mo les of Na OH = 0.003495 mol. 1 't f N OH 0.00 3495 mol 01371 M mo an y 0 a = 0. 02549 L = . Practice Problem A How man y gram s of KHP are n ee ded to neutralize 22 .36 mL of a 0.1205 M NaOH solution? Practice Problem B What volume (in mL ) of a 0.2550 M N aOH solution can be neutraliz ed by 10.75 g of KHP? LI __________________________________________________________________________ III The reaction between NaOH and KHP is a relatively simple acid-base neutralization. Sup- pose, though, that instead of KHP, we wanted to use a diprotic acid such as H 2 S0 4 for the titration. The reaction is represented by Because 2 mol NaOH "'" 1 mol H 2 S0 4 , we need twice as much NaOH to react completely with an H, S0 4 solution of the same molar concentration and volume as a monoprotic acid such as HCl. On the other hand, we would need twice the amount of HCI to neutralize a Ba ( OH )2 solution com- pared to an NaOH solution having the same concentration and volume because 1 mole of Ba ( OH )2 yields 2 moles of OH - ions: 2HCI(aq) + Ba(OH hC aq) + . Ba CI 2 ( aq ) + 2H ,O(l) In any acid-base titration, regardless of what acid and base are reacting, the total number of moles of H+ ions that have reacted at the equivalence point must be equal to the total number of moles of OH- ions that have reacted. Sample Problem 4. 14 explores the titration of an NaOH solution with a diprotic acid. ~,,'" *' ·T §!l mp ~~J.!!l,,~ii~ What volume (in mL) of a 0.203 M N aOH solution is n ee ded to neutralize 25.0 mL of a 0.1 88 M H 2 S0 4 solution? Strategy First, write and balance the chemical equation that corres pond s to the neutralization reaction: The base and the diprotic acid combine in a 2: 1 ratio: 2N aOH "" H 2 S0 4 , Use th e mo larit y and . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . the volume given to determine the number of millimole s of H 2 S0 4 , Us e the numb er of millim oles of H 2 S0 4 to determine the number of millimole s of N aOH. Us ing millim o le s of N aOH and th e concentration given, determine the v olume of N aOH that will contain the c orr ect numb er of millimole s. ( Continued) Think About It Remember that molarit y can also be defined as mrnolimL. Try solving the problem again us ing millimoles and make s ure you g et the s ame answer. 0.003495 mol = 3.495 X 10- 3 mol = 3.495 mrnol and 3.495 mrnol = 0 1371 M 25.49 mL . Re member: molarity x mL = mil lim ol es. This saves st e ps in titrat ion problems . [...]... H 3P0 4 , (e) H 4P 20 7, (f) H SP30IQ 4.45 Give the oxidation numbers for the underlined atoms in the following molecules and ions: (a) CIF, (b) IF7, (c) CH4, (d) C 2Hz, (e) C ZH 4 , (f) K2Cr04, (g) K2Cr2 07, (h) KMn02' (i) NaHC0 3, (j) Liz, (k) NaI0 3, (I) KOz, (m) PF6", (n) KAuCI 4· Write the equation for calculating molarity Why is molarity a convenient concentration unit in chemistry? Describe the... the types and arrangements of atoms in the structural units that make up the sample Electrostatic energy is potential energy that results from the interaction of charged particles Oppositely charged particles attract each other, whereas particles of like charges repel each other The magnitude of the resulting electrostatic potential energy is proportional to the product of the two charges (Q\ and Q2)... oxidation number? How is it used to identify redox reactions? Explain why, except for ionic compounds, the oxidation number does not have any physical significance 4.37 (a) Without referring to Figure 4 .7, give the oxidation numbers of the alkali and alkaline earth metals in their compounds (b) Give the highest oxidation numbers that the Groups 3A-7 A elements can have 4.38 How is the activity series... be ionic or molecular Ionic electrolytes undergo dissociation in solution; molecular electrolytes undergo ionization Strong electrolytes dissociate (or ionize) completely Weak electrolytes ionize only partially Oxidation-reduction, or redox, reactions are those in which electrons are exchanged Oxidation and reduction always occur simultaneously You cannot have one without the other • Oxidation is the... N aCI and an apparatus like that shown in Figure 4.1 for comparison 4.114 A quantitative definition of solubility is the number of grams of a solute that will dissolve in a given volume of water at a particular temperature Describe an experiment that would enable you to determine the solubility of a soluble compound You are given two colorless solutions, one containing NaCI and the other sucrose (C... preparation of hydrogen? 4.119 Describe how you would prepare the following compounds: (a) Mg(OH)2, (b) AgI, (c) Ba3(P04)z 4.120 4.106 4.113 Someone spilled concentrated sulfuric acid on the floor of a chemistry laboratory To neutralize the acid, would it be preferable to pour concentrated sodium hydroxide solution or spray solid sodium bicarbonate over the acid? Explain your choice and the chemical... vegetables and in wine making In an experiment to test the presence of sulfite in fruit, a student first soaked several dried apricots in water overnight and then filtered the solution to remove all solid particles She then treated the solution with hydrogen peroxide (H20 2) to oxidize the sulfite ions to sulfate ions Finally, the sulfate ions were precipitated by treating the solution with a few drops... 4.134 Barium sulfate (BaS04) has important medical uses The dense salt absorbs X rays and acts as an opaque barrier Thus, Xray examination of a patient who has swallowed an aqueous suspension of BaS04 particles allows the radiologist to diagnose an ailment of the patient's digestive tract Given the following starting compounds, describe how you would prepare BaS04 by neutralization and by precipitation:... concentration (in molarity) of the final solution (b) Calculate the mass of KMn04 needed to directly prepare the final solution The following "cycle of copper" experiment is performed in some general chemistry laboratories The series of reactions starts with copper and ends with metallic copper The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated... Calorimetry Constant-Volume Calorimetry Hess's Law Standard Enthalpies of Formation Calories in Food Nutrition Abou t 24 biSCUits Facts Serving Size The Food and Drug Administration (FDA) and the U.S Department of Agriculture (USDA) require "Nutrition Facts" labels on nearly all prepared foods These labels indicate such things as serving size, grams of fat per serving, and Calories per serving Health-conscious . these equations. Concentrated solution: Dilute solution: More solute particles per unit volume Fewer so lute particles per unit volume Concentration of Solutions One of the. = X hter s of solutlon liters of so lution as in part (b), for the volume that contains a number of moles smaller than 0.2 77, make sure your answer is smaller than the original. the same result. • Before dilution: More solute particles per unit vo lume Add solvent • After dilution: Fewer so lute particles per unit volume Because the number of moles