Chemical science part vii, Julia Burdge,2e (2009)
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136 Chapter iv Reactions in Aqueous Solutions Figure 4.viii Ii solutions of iodine in benzene. The solution on the left is more concentrated. The solution on the right is more dilute. Full-bodied solution: More solute particles per unit of measurement volume Dilute solution: Fewer solute particles per unit volume Concentration of Solutions The qualitative terms concentrated and dilute are relative terms, like expensive and cheap. r "' , _....: Multimedia One of the factors that can influence reactions in aqueous solution is concentration. The concentration 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 conthan the one on the right-that is, it contains a higher ratio of solute to solvent. By con. . . . . . .centrated . , . . . . . . . . . . . . . . . . . . . . . . . . . . . , . . . . . .. . . . . . . . . . . . . . . trast, the solution on the right is more dilute. The color is more than intense in the more than concentrated solution. Often the concentrations of reactants determine how fast a chemical reaction occurs. For case, the reaction of magnesium metallic and acid [ ~~ Section 4.4] happens faster if the concentration of acrid is greater. As we will see in Chapter 13, in that location are several dissimilar ways to express the concentration of a solution. In this chapter, nosotros introduce only molarity, which is one of the most commonly used units of concentration. Solutions-training of solutions. Molarity • • • • • Molarity tin can as well exist defined as millimoles per milliliter (mmol/mL), which tin can simplify some calculations. •• Molarity, or molar concentration, symbolized Grand, is defined as the number of moles of solute per liter of solution. Thus, 1 L of a 1.5 molar solution of glucose (C 6H I2 0 6), written as 1.5 M C 6H 120 6 , contains 1.5 moles of dissolved glucose. Half a liter of the same solution would contain 0.75 mole of dissolved 3 glucose, a milliliter of the solution would contain 1.5 X 10- mole of dissolved glucose, and so on. moles solute mo 50 an'ty = - --liters solution Equation four.i In order to calculate the molarity of a solution, nosotros dissever the number of moles of solute past the volume of the solution in liters. Equation iv.one can exist rearranged in three means to solve for any of the three variables: molarity (M), moles of solute (mol), or volume of solution in liters (L) . ... .. . .. .. . ..... . . .. . . ..... . . . .. . . ... , ...... .. ·mol· ....... .... ..... .mel· .. ........ . Students sometimes have difficulty seeing how units cancel in these equations. It may help to write Thou as mollL until yous become completely comfortable with these equations. (i) M = L (ii) L = One thousand (3) mol = Yard XL Sample Problem 4.9 illustrates how to use these equations to solve for molarity, volume of solution, and moles of solute. Sample Trouble 4.9 For an aqueous solution of glucose (C 6H I2 0 six), determine (a) the molarity of ii.00 Fifty of a solution that contains 50.0 yard of glucose, (b) the volume of this solution that would contain 0.250 mole of glucose, and (c) the number of moles of glucose in 0.500 L of this solution. Concentration of Solutions Section 4.v Strategy Convert the mass of glucose given to moles, and use the equations for interconversions of M, liters, and moles to calculate the answers. ... ... : : · Setup The molar mass of glucose is 180.2 grand. • • • • • • • • = A common way to land the concentration of this solution is to say, "This solution is 0. 139 M in glucose." • • fifty.0 g moles of glucose = 80 I = 0.277 mol one .two glmo Solution (a) molarity 137 0.277 mol C6H'206 ........................................................... : 200 L l ' = 0.139 1000 . so utlOn 0.250 mol C6H p 0 6 (b) volume = 0 9 = i.80 L .13 Thousand (c) moles of C6H 120 vi in 0.500 L = 0.500 L X 0.139 M = 0.0695 mol Practise Problem A For an aqueous solution of sucrose (C 12 H22 0 Retrieve Near It Check to see that the magnitudes of your answers are logical. For example, the mass given in the trouble corresponds to 0.277 mole of solute. If you lot are asked, equally 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. decide (a) the molarity of 5.00 Fifty of a solution that contains 235 m of sucrose, (b) the book of this solution that would contain ane.26 mole of sucrose, and (c) the number of moles of sucrose in i.89 L of this solution. 11 ), , Practice Trouble B For an aqueous solution of sodium chloride (NaCl), determine (a) the molarity • of 3.75 L of a solution that contains 155 grand of sodium chloride, (b) the book of this solution that would contain 4.58 moles of sodium chloride, and (c) the number of moles of sodium chloride in 22.75 Fifty of this solution. --------------------------------------------------------_.. ~. The procedure for preparing a solution of known molarity is shown in Effigy iv.9. Outset, Media Player/MPEG Blitheness: Figure iv.9, Preparing a Solution from a Solid, the solute is weighed accurately and transferred , ofttimes with a funnel, to a volumetric flask pp. 138-139. of the desired volume. Next, water is added to the flask, which is then swirled to dissolve the solid. After all the solid has dissolved, more water is added slowly to bring the level of olution exactly to the volume marker. Knowing the volume of the solution in the flask and the . . . . . . . . . . . . . . . . . . . . . . . . . . .. Information technology is important to recollect that molarity is quantity of compound dissolved, we tin can decide the molarity of the solution using Equation divers in terms of the volume of solution, not four. ane. Note that this procedure does non crave that we know the exact amount of water added. the volume of solvent. In many cases, these 2 Considering of the way molarity is divers , it is important only that nosotros know the final volume of are not the same. the solution. Dilution Concentrated "stock" solutions of commonly used substances typically are kept in the laboratory stockroom. Often we demand to dilute these stock solutions before u sing them. Dilution is the procedure of preparing a less concentrated solution from a more full-bodied 1. Suppose that nosotros desire to prepare 1.00 L of a 0.400 M KMn04 solution from a solution of 1.00 Chiliad KMn04' For thi south purpose we demand 0.400 mole of KMn04' Considering at that place is ane.00 mole of KMn04 in 1.00 L of a 1.00 M KMn04 solution , there is 0.400 mole of KMn04 in 0.400 L of the aforementioned -olution: 1.00 mol KMn04 1.00 L of solution , Multimedia Solutions-dilution. 0.400 mol KMn04 OAOO 50 of solution Therefore, we must withdraw 400 mL from the one.00 M KMn04 solution and dilute it to 1.00 L by adding h2o (in a 1.00-50 volumetric flask) . This method gives the states ane.00 L of the desired 0.400 M KMn04' r In carrying out a dilution process, it is useful to remember thilt adding more solvent to a given amount of the stock solution changes (decreases) the concentration of the solution without hanging the number of moles of solute nowadays in the solution (Effigy 4.ten, p. 140). moles of solute before dilution = moles of solute after dilution • Equation four.2 L ing arrangement (iii) of Equation 4.one , nosotros can calculate the number of moles of solute: moles of solute = moles of sol u t east . . X hters of solutlon liters of solution ! Effigy iv.9 • Transfer the weighed KMn04 to the volumetric flask The mass likely will not be exactly the calculated number Weigh out the solid KMn04 (The tare function on a digital residual automatically subtracts the mass of the weighing paper.) \ Calculate the mass of KMn04 necessary for the target concentration of 0.1 Grand. 0.ane tal 10 0.2500 Fifty = 0.02500 mol 158.04 thou 0.02500 mol X I = 3.951 1000 KMn04 rna • 138 Add water sufficient to dissolve the KMn04 Swirl the flask to dissolve the solid Add more water Fill exactly to the calibration marker using a launder bottle Summate the actual concentration of the solution. ane mol 3.896 g KMn04 Ten 158.04 chiliad = 0.024652 mol 0.024652 mol = 009861 M 0.2500 L . Whatls the betoken? The goal is to prepare a solution of precisely known concentration, with that concentration existence very close to the target concentration of 0.1 Chiliad. Note that considering 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.x Dilutio n changes the concentration of a solution; it does non change the number of moles of solute in the solution. Add together solvent • • After dilution: Fewer solute particles per unit book Before dilution: More solute particles per unit of measurement volume Considering the number of moles of solute earlier the dilution is the same as that subsequently dilution, we can write Equation iv.iii where the subscripts c and d stand for concentrated and dilute, respectively. Thus, by knowing the molarity of the full-bodied stock solution (Me) and the desired final molarity (Physician) and volume (Ld) of the dilute solution , we can calculate the volume of stock solution required for the dilution (Le). Because well-nigh volumes measured in the laboratory are in milliliters rather than liters, information technology is worth pointing out that Equation 4.three can be written with volumes of the concentrated and dilute solutions in milliliters. Students sometimes resist using the unit millimole. Yet, using the Me Ten mLc = M d X m'-d form of Equation 4.three often reduces the number of steps in a problem thereby reducing the number of opportunities to make adding errors. . .. ... . . . . . . . . . . . . .. ...... ... ...... . . . . . . . . .. .................. ...... . In this form of the equation, the product of each side is in millimoles (mmol) rather than moles. We use Equation 4.3 in Sample Problem 4.ten. Sample Problem 4.x What vo lume of 12.0 M Hel , a common laboratory stock solution, must be used to prepare 250.0 mL of 0.125 One thousand He]? • • • • It is very of import to notation that, for condom, when diluting a concentrated acrid, the acid must be added to the water, and non the other mode around. ·• • • • • • ·• • • Strategy Use Equation 4.three to determine the volume of 12.0 M Hel required for the dilution. Considering the desired last volume is given in milliliters, information technology volition be convenient to use the course of Equation 4.3 that includes milliliters . Setup Me = 12.0 M, Md = 0.125 M, mLd = 250.0 mL. • • • • Call back About It Plug the answer into Equation 4.3, and brand sure that the product of concentration and volume on both sides of the equation give the aforementioned result. • • • • • • • • Solution 12.0 M Ten mLe = 0.l25 M X 250.0 mL .. . . . . . . . . . . . . . . .. ........... . .. ............... ~ = 0.l25 M X 250.0 mL = two.60 mL e 12.0 Thousand Prac.tic.east Problem A What volume of six.0 K H 2S0 4 is needed to prepare 500.0 mL of a solution that is 0.25 M in H 2S0 4 ? Exercise ·Problem B What book of 0.20 M H 2S04 can be prepared past diluting 125 mL of half dozen.0 Yard H2 S0 4 ? Concentration of Sol utions Section 4.five 141 Solution Stoichiometry Soluble ionic compounds such equally KMn04 are strong electrolytes, so they undergo complete dissociation upon dissolution and exist in solution entirely as ions. KMn04 dissociates, for instance, to give i mole of potassium ion and 1 mole of permanganate ion for every mole of potassium permanganate. Thus, a 0.400 Chiliad solution of KMn04 volition be 0.400 M in K+ and0.400 M in MnO four . 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 chemic formula to determine the concentration of each ion in solution. Sodium sulfate (Na1 S04) dissociates, for case, to give twice as many sodium ions as sulfate ions. Na2S0is) H 20, 2Na+(aq) + And then ~-(aq) Therefore, a solution that is 0.35 Min Na2S04 is actually 0.70 M in Na + and 0.35 1000 in And so ~- . Frequently, molar concentrations of dissolved species are expressed using square brackets. Thus, of species in a 0.35 M solution of Na2S04 .can be expressed as follows: [Na +] ......... . the concentrations 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . " 0.70 M and [SO 4- ] = 0.35 K. Sample Problem 4.xi lets y'all practice relating concentrations of : compounds and concentrations of individual ions using solution stoichiometry. : ~ • Square brackets around a chemical species tin can exist read as "the concentration of" that species. For example, [Na +] is read as "the concentration of sodium ion." • • • ' Using square-subclass notation, limited the concentration of (a) chloride ion in a solution that is i.02 M in AICI three, (b) nitrate ion in a solution that is 0.451 Min Ca(N0 three)two, and (c) Na2C03 in a solution in which [Na +] = 0.124 M. .... If we only need to express the concentration of the compound, rather than the concentrations of the individual ions, we could limited the concentration of this solution every bit [Na,S04] = 0.35 M Strategy Apply the concentration given in each instance and the stoichiometry indicated in the conesponding chemical fOlTllula to make up one's mind 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 iii times the concentration of AlCl 3. • (b) There are 2 moles of nitrate ion for every 1 mole of Ca(N0 3)ii, Ca(N0 3Ms) H20, Ca2+ (aq) + 2NO](aq) so [NO ] ] will be twice [Ca(N0 3h ]. (c) There is i mole of Na2C03 for every 2 moles of sodium ion, Na2C03(S) H xx, 2Na +(aq) + CO ~ - (aq) so [Na2C0 3] will be half of [Na +]. (Assume that Na2C03 is the only source of solution.) a + ions in this Solution (a) [CI-] = [AICH X 3 mol Cl o 1 mol AICl 3 ~1~.0~ii~~~i 3 X 3 mol Cl Fifty one 3 iii.06 mol ClL = three.06 • One thousand _ (b) [N0 3 ] = [Ca(N0 three)2] = 0.451 mol X 2 mol NO] 1 mol Ca(N0 )ii Ca(H0 iii) ; X L three 2 mol NO ] _ 1 _mol Ca(N0 3h 0.902 mol NO] L = 0.902 M (Connected) 142 Chapter iv React ions in Aqueous So lutions Call back About It Make sure that units cancel properly. Remember that the concentration of an ion can never exist less than the concentration of its di ssolved parent compound. It will always be the concentration of the parent chemical compound times its stoichiometric subscript in the chemic formula. _ 0.124 ~ X 1 mol Na2C03 two~ Fifty 0.0620 mol Na2C03 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(S04)3]. Practise Problem B Using the foursquare-bracket notation, limited the concentration of chloride ions in a solution that is 0.250 Yard in sodium chloride (NaCl) and 0.25 M in magnesium chloride (MgCI 2). Checkpoint four.5 four.5.i Concentration of Solutions Calculate the molar concentration of a solution prepared by di ssolving 58.5 k NaOH in plenty water to yield one.25 L of solution. four.5.3 What book in mL of a 1.20 Chiliad HCI solution must be diluted in order to prepare 1.00 Fifty of 0.0150 Grand HCI? a) 15.0 mL a) 1.46M b) 12.5 mL b) 46.8 G c) 12.0 mL c) 2.14 Ten ten- 2 M d) 85.0 mL d) ane.l7M e) ll5 mL east) 0.855 M four.5.iv four.5.2 What mass of glucose (C 6H 120 6) in grams must be used in society to fix 500 mL of a solution that is 2.50 Min glucose? A solution that is 0.18 Min Na2C03 is too . (Cull all that utilise.) a) 0.18MinCO ~b) 0.18 Min Na+ . a) 225 thousand c) 0 .09 G in Na+ b) 125 chiliad d) 0.09 Grand in CO ~- c) 200 one thousand east) 0.36 Min Na + d) 1.25 g e) 625 g Aqueous Reactions and Chemical Assay . Experiments that measure the amount of a substance present are called quantitative analysis. .. . . . . . . . . . .. ................. ....... . . . . . . . . . .. . Sure aqueous reactions are useful for determining how much of a particular substance is nowadays in a sample. For example, if nosotros want to know the concentration of lead in a sample of water, or if nosotros need to know the concentration of an acrid, knowledge of precipitation reactions, acidbase reactions, and solution stoichiometry will be useful. 2 common types of such quantitative analyses are gravimetric analysis and acid-base titration. Gravimetric Assay According to the information in Tabular array 4.two, Agel is an insoluble exception to the chlorides, which typica lly are soluble. Gravimetric assay 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(due south): AgN03(aq) + NaCI(aq) - -+. NaN0 3(aq) + AgCI(due south) SECTION 4.6 Aqueous Reactions and Chemical Analysis This reaction is often used in gravimetric analysis considering the reactants can be obtained in pure grade. The net ionic equation is Ag+(aq) + Cl- (aq) ---. AgCI(southward) Suppose, for example, that we wanted to test the purity of a sample of NaCI by determining the percent past mass of Cl. Commencement, nosotros would accurately counterbalance 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 Clions nowadays in solution as AgCl. (In this process NaCI is the limiting reagent and AgN0 3 is the excess reagent.) We would and so separate, dry, and counterbalance the AgCl precipitate. From the measured mass of AgCl, we would exist able to summate 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 corporeality of Cl that nosotros summate is the amount that was present in the original NaCI sample. Nosotros could so calculate the percent by mass of CI in the NaCI and compare information technology to the known composition of NaCl to determine its purity. Gravimetric analysis is a highly accurate technique, because the mass of a sample can be measured accurately. However, this procedure is applicable merely to reactions that become to completion or have well-nigh 100 percent yield. In addition, if AgCI were soluble to whatsoever meaning caste, it would not be possible to remove all the CI- ions from the original solution, and the subsequent calculation would exist in error. Sample Problem iv.12 shows the calculations involved in a gravimetric experiment. A 0.8633-m sample of an ionic.compound containing chloride ions and an unknown metal cation is dissolved in h2o and treated with an backlog of AgN0 three . If i.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 per centum 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, decide the percent Cl in the chemical compound. Setup To determine the percent Cl in AgCl, divide the molar mass of CI by the tooth mass of AgCI: 35.45 g -:-::-:::-:-:::---c-=:-:::-::---:- X (35.45 m + 107.9 g) 100 % = 24.72 % . The mass of Cl in the precipitate is 0.2472 10 ane.5615 g = 0.3860 chiliad. 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:-::-=-::-- 0.8633 one thousand 10 100% = 44.71 % Cl 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? . Exercise 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 1, then information technology 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 possibilities. Because no two compounds containing the aforementioned anion (or cation) have the same percent ;:omposition by mass, comparison of the percent by mass obtained from gravimetric analysis ":\ith that calculated from a series of known compounds could reveal the identity of the unknown ~ompounds. Recollect About Information technology Pay shut attention to which numbers correspond to which quantities. It is like shooting fish in a barrel in this type of problem to lose track of which mass is the precipitate and which is the original sample. Dividing past the wrong mass at the end will event in an incorrect answer. 143 144 Chapter four Reactions in Aqueous Solutions Acid-Base Titrations Multimedia Chemical reactions- titrations. Standardization in this context is the meticulous conclusion of concentration. Note that KHP is a monoprotic acid, so it reacts in a 1: 1 ratio with hydroxide ion . 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 betwixt the ii solutions is complete, as shown in Effigy iv.eleven. 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 of operations sodium hydroxide tin can be used equally the standard solution in a titration, only it must outset exist standardized, because sodium hydroxide in solution reacts with car.... .. hon' dIOXIde'In' the' aiT: iiiabng 'Its 'con"centratloiiliiistable 'o ver' dill·eastward. We 'can' standardize the sodium hydroxide solution past titrating it against an acrid solution of accurately known concentration. The acid often called for this task is a monoprotic acid called potassium hydrogen phthalate (KHP), for which the molecular formula is KHC sH 40 4 . KHP is a white, soluble solid that is commercially available in highly pure form. The reaction betwixt KHP and sodium hydroxide is . . <.... .. ................... . .............. .. ...... . ............ . . . . .. . ... . ... . ... . ...... .. .. . ........ .. ..... . KHC sH40iaq) NaOH(aq) • KNaC sH404(aq) H 20(fifty) + + • and the net ionic equation is . .. The endpoint in a titration is used to guess the equivalence bespeak. A careful option of indicators, which we will hash out in Chapter 16, helps make this approximation reasonable. Phenolphthalein, although very common, is not appropriate for every acid-base of operations titration. Figure 4.11 To standardize a solution of NaOH with KHP, a known amount of KHP is transferred to an Erlenmeyer flask and some distilled h2o is added to make up a solution. Adjacent, NaOH solution is carefully added to the KHP solution from a buret until all the acid has reacted with base. This betoken in ..the titration, where the ,.... ....... .... .. ........,.... ... ..acid has been completely neutralized, is chosen the equivalence betoken. It is ordinarily signaled past the endpoint, where an indicator causes a abrupt change in the color of the solution. In acid-base titrations, indicators are substances that have distinctly dissimilar colors in acidic and basic media. One commonly used indicator is phenolphthalein, which is colorless in acidic and neutral solutions but reddish pinkish in basic solutions. At the equivalence point, all the KHP present has been neutralized by the added NaOH and the solution is still colorless. All the same, if nosotros add together just one more drop of NaOH solution from the buret, the solution will be basic and volition immediately plow pinkish. Sample Trouble 4.13 illustrates just such a titration. Apparatus for titration. « "r -~.. 'Y " '. .. .0 ,t \ I'ItENOlPHTHAWH INDICATOR I 1. ~ • i ,,_J ,. I · « L - .;' ~ ._ I Department iv.half dozen Aqueous Reactions and Chemical Analysis 145 In a titration experiment, a student finds that 25.49 mL of an NaOH solution is needed to neutralize 0.7137 g of KHP. What is the concentration (in One thousand) of the NaOH solution? Strategy Using the mass given and the molar mass of KHP, determine the number of moles of KHP. Recognize that the number of moles of N aOH in the volume given is equal to the number of moles of KHP. Carve up moles of NaOH by volume (in liters) to become molarity. Setup The molar mass of KHP (KHC 8H iv 0 4) = [39.ane thousand 204.2 thousand/mol. + 5(fifty.008 g) + 8( 12.01 g) + 4 (16.00 g)] = Call up Nigh Information technology Remember that molarity tin can too be defined as mrnolimL. Try solving the problem again using millimoles and make sure you get the aforementioned respond. Solution 0.seven 137 g moles of KHP = 204.two m/mol = 0.003495 mol 0.003495 mol = iii.495 X 10- 3 mol = 3.495 mrnol Because moles of KHP = moles of NaOH, then moles of NaOH = 0.003495 mol. and 't f N OH 0.003495 mol 01371 M 1 mo an y 0 a = 0.02549 Fifty = . three.495 mrnol = 0 1371 M 25.49 mL . Practise Problem A How many grams of KHP are needed to neutralize 22.36 mL of a 0.1205 M NaOH solution? Practice Trouble B What book (in mL) of a 0.2550 K NaOH solution tin be neutralized past 10.75 thousand of KHP? III LI_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ The reaction between NaOH and KHP is a relatively unproblematic acid-base neutralization. Suppose, though, that instead of KHP, nosotros wanted to use a diprotic acid such equally H 2S0 4 for the titration. The reaction is represented by Because 2 mol NaOH "'" 1 mol H2S04, we need twice as much NaOH to react completely with an H,S04 solution of the aforementioned molar concentration and book every bit a monoprotic acid such as HCl. On the other paw, we would need twice the corporeality of HCI to neutrali ze a Ba(OH)2 solution compared to an NaOH solution having the same concentration and volume because i mole of Ba(OH)2 yields 2 moles of OH- ions: 2HCI(aq) + Ba(OHhCaq) ---+. BaCI2 (aq) + 2H, O(fifty) In any acid-base of operations titration, regardless of what acrid and base of operations are reacting, the full number of moles of H+ ions that have reacted at the equivalence betoken must exist equal to the total number of moles of OH- ions that accept reacted. Sample Trouble 4.14 explores the titration of an NaOH solution with a diprotic acid. ---~,,'" *' ·T§!lmp~~J.!!50,,~two~ What book (in mL) of a 0.203 M NaOH solution is needed to neutralize 25 .0 mL of a 0.ane 88 Grand H 2S0 4 solution? Strategy Get-go, write and balance the chemical equation that corresponds to the neutrali zation reaction: The base and the diprotic acid combine in a ii: ane ratio: 2N.aOH . . . . . . ."". . .H. .2.S0 . . . .iv.,. .Utilize . . . . .the . . . . .m . .olarity . . . . . . . . .and . . . . . . . . . . . .. . . . . . . . . Remember: molarity ten mL = millimoles. This the book given to determine the number of millimoles of H 2 S0 4 , Use the number of millimoles saves steps in titration problems. of H 2 S0 4 to determine the number of millimoles of NaOH. Us ing millimoles of North aOH and the concentration given, determine the book of NaOH that will contain the correct number of millimoles. ( Continued) This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service utilize.
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