1 / 31

Example 4.1

Example 4.1. Calculate the molarity of each ion in an aqueous solution that is 0.00384 M Na 2 SO 4 and 0.00202 M NaCl. In addition, calculate the total ion concentration of the solution. Strategy.

bly
Télécharger la présentation

Example 4.1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Example 4.1 Calculate the molarity of each ion in an aqueous solution that is 0.00384 M Na2SO4 and 0.00202 M NaCl. In addition, calculate the total ion concentration of the solution. Strategy First, note that there are three types of ions in the solution—Na+, SO42–, and Cl–. Of these three types, SO42– comes only from the dissolved Na2SO4, Cl– comes only from the NaCl, and Na+ comes from both solutes. Thus, we can establish [SO42–] just from the molarity of Na2SO4(aq) and [Cl–] just from the molarity of NaCl(aq). For [Na+], we will have to work with both molarities. Finally, we sum the molarities of the individual ions to obtain the total ion concentration. Solution

  2. Example 4.1 continued Assessment As you become more familiar with calculating ion concentrations, you should be able to work a problem of this sort just by inspecting the formulas of the solutes, thereby arriving directly at the conclusions; [SO42–] = 0.00384 M, [Cl–] = 0.00202 M, and [Na+] = (2 x 0.00384) + 0.00202 = 0.00970 M. Exercise 4.1A Seawater is essentially 0.438 M NaCl and 0.0512 M MgCl2, together with several other minor solutes. What are the molarities of Na+, Mg2+, and Cl– in seawater? Exercise 4.1B Each year, oral rehydration therapy (ORT)—feeding a person an electrolyte solution—saves the lives of a million children worldwide who become dehydrated from diarrhea. Each liter of the electrolyte solution contains 3.5 g sodium chloride, 1.5 g potassium chloride, 2.9 g sodium citrate (Na3C6H5O7), and 20.0 g glucose (C6H12O6). Calculate the molarity of each species present in the solution. (Hint: Sodium citrate is a strong electrolyte, and glucose is a nonelectrolyte.)

  3. Example 4.2 Barium nitrate, used to produce a green color in fireworks, can be made by the reaction of nitric acid with barium hydroxide. Write (a) a complete-formula equation, (b) an ionic equation, and (c) a net ionic equation for this neutralization reaction. Solution

  4. Example 4.2 continued Exercise 4.2A Calcium hydroxide is used to neutralize a waste stream of hydrochloric acid. Write (a) a complete-formula equation, (b) an ionic equation, and (c) a net ionic equation for this neutralization reaction. Exercise 4.2B Write the (a) complete-formula equation, (b) ionic equation, and (c) net ionic equation for the following neutralization reaction. KHSO4(aq) + NaOH(aq) ?

  5. Example 4.3 A Conceptual Example Explain the observations illustrated in Figure 4.6. Analysis and Conclusions The bulb in Figure 4.6(a) is only dimly lit because acetic acid is a weak acid and therefore a weak electrolyte [recall Figure 4.3(c)]. The situation in (b) is similar because ammonia is a weak base and therefore also ionizes only slightly. When the two solutions are mixed, which is what has been done in (c), the H+ ions from the CH3COOH readily combine with NH3 molecules to form NH4+ ions: H+(aq) + NH3(aq)  NH4+(aq) By removing H+ ions from the solution, this reaction causes more CH3COOH molecules to ionize, producing more H+ to react with more NH3 and so on. Soon all the CH3COOH molecules ionize, and the neutralization goes to completion: The original weak acid and weak base are replaced by an aqueous solution of a salt—an ionic compound and strong electrolyte. The solution is now a good electrical conductor, as seen in Figure 4.6(c).

  6. Example 4.3 continued Exercise 4.3A In a situation similar to that in Figure 4.6, describe the observations you would expect to make if the original solutions were CH3NH2(aq) and HNO3(aq). Exercise 4.3B Describe the observations you would expect in a situation similar to that in Figure 4.6 if you start with a slurry of Mg(OH)2(s) in water (milk of magnesia) and add vinegar [essentially 1 M CH3COOH(aq)] until all the Mg(OH)2(s) dissolves and some vinegar remains in excess.

  7. Example 4.4 Predict whether a precipitation reaction will occur in each of the following cases. If so, write a net ionic equation for the reaction. (a)   Na2SO4(aq) + MgCl2 (aq)  ? (c)   K2CO3 (aq) + ZnCl2(aq)  ? (b)  (NH4)2S(aq) + Cu(NO3)2(aq)  ? Strategy We must consider the simple compounds that could be formed from the combinations of ions present. Then we can use the solubility guidelines to determine whether any potential products are insoluble compounds. If one or more of these potential products are insoluble, a reaction will occur. If all potential products are soluble, there will be no reaction.

  8. Example 4.4 continued Solution

  9. Example 4.4 continued Exercise 4.4A Predict whether a reaction will occur in each of the following cases. If so, write a net ionic equation for the reaction. (a)   MgSO4(aq) + KOH(aq)  ? (c)   Sr(NO3)2(aq) + Na2SO4(aq)  ? (b)  FeCl3(aq) + Na2S(aq)  ? Exercise 4.4B In each of the following cases, predict whether a reaction will occur, and, if so, write the net ionic equation for the reaction. (a)   ZnSO4(aq) + BaS(aq)  ? (c)   NaHCO3(aq) + Ca(OH)2(aq)  ? (b)   Mg(OH)2(s) + NaOH(aq)  ?

  10. Example 4.5 A Conceptual Example Figure 4.8 shows that the dropwise addition of NH3(aq) to FeCl3(aq) produces a precipitate. What is the precipitate? Analysis and Conclusions The reactants, NH3 and FeCl3, are both soluble in water. That all ammonium compounds are soluble means the precipitate is not likely to contain NH4+. It must therefore contain Fe3+. But what is the anion? Recall that NH3 is a weak base and that it produces OH– ions in aqueous solution:

  11. Example 4.5 continued Exercise 4.5A Suppose that after all the Fe(OH)3(s) is precipitated, a large quantity of HCl(aq) is added to the beaker in Figure 4.8. Describe what you would expect to see, and write a net ionic equation for this change. Exercise 4.5B Potassium palmitate is a typical water-soluble soap. It is formed by the neutralization of palmitic acid, CH3(CH2)14COOH, with potassium hydroxide. When calcium chloride is added to an aqueous solution of potassium palmitate, a gray precipitate is observed. What is the likely precipitate? Write ionic and net ionic equations for its formation.

  12. Example 4.6 One cup (about 240 g) of a certain clear chicken broth yields 4.302 g AgCl when excess AgNO3(aq) is added to it. Assuming that all the Cl– is derived from NaCl, what is the mass of NaCl in the sample of broth? Strategy In this problem, chloride ions derived from NaCl are precipitated as AgCl. From the given mass of AgCl, we can determine the number of moles of chloride ions initially present. From the moles of Cl–, we can calculate first the moles of NaCl and then the mass of NaCl.

  13. Example 4.6 continued Solution

  14. Example 4.6 continued Assessment Note that the mass of the broth (about 240 g) does not enter into this calculation. We are interested only in the mass of NaCl present. On the other hand, if we had been required to find the percentage of NaCl in the broth, the mass of the broth would have been required [% NaCl  (1.754 g NaCl/240 g broth) x 100%  0.7% NaCl]. Exercise 4.6A What is the mass percent NaCl in a mixture of sodium chloride and sodium nitrate if a 0.9056-g sample of the mixture yields 0.9372 g AgCl(s) when allowed to react with excess AgNO3(aq)? Exercise 4.6B Consider the seawater sample described in Exercise 4.1A. How many grams of precipitate would you expect to get by adding (a) an excess of AgNO3(aq) to 225 mL of the seawater, (b) an excess of NaOH(aq) to 5.00 L of the seawater? What are the precipitates?

  15. Example 4.7 What are the oxidation numbers assigned to the atoms of each element in (a) KClO4 (b) Cr2O72–(c) CaH2 (d) Na2O2 (e) Fe3O4 Strategy Solution In applying the rules for determining oxidation numbers, we must adhere to the priority order in which they are listed above. Note, however, that except in the case of uncombined elements, we cannot apply Rule 1 first. (a) The oxidation number of K is +1 (Rule 3). The oxidation number of O is –2 (Rule 6), and the total for four O atoms is –8. For these two elements, the total is +1 – 8 = –7. The oxidation number of the Cl atom in this ternary compound must be +7, to give a total of zero (+1 – 8 + 7 = 0) for all atoms in the formula unit (Rule 1). (b) The oxidation number of O is –2 (Rule 6), and the total for seven O atoms is –14. The total of the oxidation numbers in this ion must be –2 (Rule 2). Therefore the total of the oxidation numbers of two Cr atoms is +12, and that of one Cr atom is +6. (c) Keeping in mind that the total for the formula unit must be 0 (Rule 1) and that the oxidation number of Ca is +2 (Rule 3), the oxidation number of H must be –1 rather than its usual +1 (Rule 5). Thus, for CaH2 the sum of the oxidation numbers is +2 + (2 x –1) = 0.

  16. Example 4.7 continued (d) The oxidation number of Na is +1 (Rule 3), and for the two Na atoms, +2. The total for the formula unit must be 0 (Rule 1). Even though the oxidation number of O is usually –2 (Rule 6), here it must be –1 so that the total for the two O atoms is –2. Rule 3 takes priority over Rule 6. (e) The oxidation number of O is –2 (Rule 6). For four O atoms, the total is –8. The total for the formula unit must be 0 (Rule 1). The total for three Fe atoms must be +8, and for each Fe atom, +8/3. Assessment Remember that the sum of the oxidation numbers of all the atoms present must add up to equal the total charge on the molecular or ionic formula. Notice how Rules 1 and 2 were used at one point or another in each part of this example. Usually, fractional oxidation numbers, as seen in part (e) signify an average. The compound Fe3O4 is actually Fe2O3 · FeO. Two of the Fe atoms have oxidation numbers of +3, and one has an oxidation number of +2. The average is (3 + 3 + 2)/3 = +8/3. Exercise 4.7A Assign an oxidation number to each atom in (a) Al2O3, (b) P4, (c) CH3F, (d) HAsO42–, (e) NaMnO4, (f) ClO2–, (g) CsO2 Exercise 4.7B Assign an oxidation number to each underlined atom: (a) HSbF6, (b)CHCl3, (c)P3O105–, (d)S4O62–, (e)C3O2, (f)NO2+, (g)C2O42–

  17. Example 4.8 Explain the difference in what happens when a copper-clad penny is immersed in (a) hydrochloric acid and (b) nitric acid, as shown in Figure 4.14.

  18. Example 4.8 continued Analysis and Conclusions (a) Because copper lies below hydrogen in the activity series of the metals, Cu(s) cannot reduce H+(aq) to H2(g) and be oxidized to Cu2+(aq). Looking at it the other way, H+is not a strong enough oxidizing agent to oxidize Cu(s) to Cu2+(aq). Chloride ion in HCl(aq) can only be a reducing agent. As neither H+ nor Cl– can oxidize Cu, we expect no reaction between Cu(s) and HCl(aq). (b) In contrast, we clearly observe a reaction between copper and nitric acid. The Cu(s) is oxidized to Cu2+(aq) (blue color). From the HCl analysis, we know that H+(aq) is not strong enough as an oxidizing agent to oxidize Cu(s) to Cu2+(aq), and we know that NO3–(aq) is an oxidizing agent here because the N atom has its highest possible oxidation number, +5. Therefore, copper is being oxidized to Cu2+(aq) by NO3–(aq). Figure 4.12 suggests that NO3–(aq) might be reduced to any one of several products. We have no way at this point of predicting what that product might be, but the red-brown gas proves to be nitrogen dioxide, NO2. The oxidation number of nitrogen in NO2 is the same as in N2O4 that is, +4. Using this information, we can write a net ionic equation: (not balanced) Cu(s) + H+(aq) + NO3–(aq)  Cu2+(aq) + NO2(g) + H2O(1) (not balanced) Finally, the equation is balanced by inspection. Cu(s) + 4 H+(aq) + NO3–(aq)  Cu2+(aq) + 2 NO2(g) + 2 H2O(1)

  19. Example 4.8 continued Exercise 4.8A The oxidizing agent potassium dichromate, K2Cr2O7(s) reacts one way when heated with hydrochloric acid and another way when heated with nitric acid. With one of the acids, a gas is evolved and the solution color changes from red-orange to green. With the other acid, the original red-orange color remains unchanged; that is, no reaction occurs. Explain this difference in behavior. Exercise 4.8B Since 1982, U.S. pennies have been made of zinc with a thin copper coating. If the edge of a new cent is notched with a knife and dropped in hydrochloric acid overnight, only a hollow shell of copper remains the next morning. How would the resulting solution differ from the two solutions in Figure 4.14?

  20. Example 4.9 What volume (mL) of 0.2010 M NaOH is required to neutralize 20.00 mL of 0.1030 M HCl in an acid–base titration? Strategy We need to do four things to solve this problem. In general, other titration problems can be solved with a similar approach. Step 1: Write an equation describing the neutralization and obtain a stoichiometric factor relating moles of NaOH and moles of HCl. Step 2: Determine how many moles of HCl are to be neutralized. Step 3: Find the number of moles of NaOH required in the neutralization. Step 4: Determine the volume of the solution containing this number of moles of NaOH. Solution

  21. Example 4.9 continued Assessment The molarity of the NaOH(aq) is just about twice that of the HCl(aq), and the combining mole ratio of the acid and base is 1:1. This suggests that the volume of NaOH(aq) required should be just about one-half that of the HCl(aq), and it is: 10.25 mL of the NaOH(aq) neutralizes 20.00 mL of the HCl(aq).

  22. Example 4.9 continued Exercise 4.9A What volume of 0.01060 M HBr(aq) is required to neutralize 25.00 mL of 0.01580 M Ba(OH)2 2 HBr(aq) + Ba(OH)2(aq)  BaBr2(aq) + 2 H2O(l) Exercise 4.9B A 2.000-g sample of a sulfuric acid solution that is 96.5% H2SO4 by mass is dissolved in a quantity of water and titrated. What volume of 0.3580 M KOH(aq) is required for the titration? Assume that at the equivalence point the solution in the flask is K2SO4(aq).

  23. Example 4.10 A 10.00-mL sample of an aqueous solution of calcium hydroxide is neutralized by 23.30 mL of 0.02000 M HNO3(aq). What is the molarity of the calcium hydroxide solution? Solution Let’s start with a balanced equation for this reaction: Strategy To determine the molarity of the Ca(OH)2(aq), we need to determine how many moles of Ca(OH)2 are consumed in the titration and divide this number by the volume of the sample (0.01000 L, or 10.00 mL). To determine the number of moles of Ca(OH)2, we can proceed through much of the problem as we did in Example 4.9.

  24. Example 4.10 continued Exercise 4.10A What volume of 0.550 M NaOH(aq) is required to neutralize a 10.00-mL sample of vinegar that is 4.12% by mass acetic acid, CH3COOH? Assume that the vinegar has a density of 1.01 g/mL. Exercise 4.10B A sample of battery acid is to be analyzed for its sulfuric acid content. A 1.00-mL sample has a mass of 1.239 g. This 1.00-mL sample is diluted to 250.0 mL with water, and 10.00 mL of the diluted acid is neutralized by 32.44 mL of 0.00986 M NaOH. What is the mass percent H2SO4 in the battery acid?

  25. Example 4.11 Suppose a 0.4096-g sample from a box of commercial table salt is dissolved in water and requires 49.57 mL of 0.1387 M AgNO3(aq) to completely precipitate the chloride ion. If the chloride ion present in solution comes only from the sodium chloride, find the mass of NaCl in the sample. Is commercial table salt pure sodium chloride? Solution Strategy As with other solution stoichiometry problems, we can determine first the number of moles of NaCl and then the NaCl mass from the volume and molarity of the AgNO3(aq), the appropriate stoichiometric factor from the balanced equation for the precipitation reaction, and the molar mass of NaCl.

  26. Example 4.11 continued Assessment The calculated mass of the NaCl is less than the mass of the sample of table salt, as it should be because the sample is not pure NaCl. Also, notice that we did not need to use the mass of table salt in our calculation, although we would need that information if we wanted to find the mass percent of NaCl in the table salt. Note that we could have combined the two steps into a single setup in which we would not have had to record the intermediate result: 6.875 x 10–3 mol AgNO3. The final answer proves to be the same by either method; 0.4018 g NaCl. Exercise 4.11A A solution containing a water-soluble salt of the radioactive element thorium can be titrated with oxalic acid solution to form insoluble thorium(IV) oxalate. If 25.00 mL of a solution of thorium(IV) ion requires 19.63 mL of 0.02500 M H2C2O4 for complete precipitation, find the molar concentration of thorium(IV) ion in the unknown solution. Exercise 4.11B A mixture of 0.1015 g of NaCl and 0.1324 g KCl is dissolved in water and titrated with 0.1500 M AgNO3(aq). What volume of the AgNO3(aq) will be needed to reach the endpoint?

  27. Example 4.12 A 0.2865-g sample of an iron ore is dissolved in acid, and the iron is converted entirely to Fe2+(aq). To titrate the resulting solution, 0.02645 L of 0.02250 M KMnO4(aq) is required. What is the mass percent of iron in the ore? Solution The purpose of each factor in the calculation involving titration data is indicated in the following outline: Strategy We can work this problem in two parts. The first requires finding the number of grams of Fe in the unknown solution from the titration data. This calculation is similar to ones we have done for acid–base and precipitation titrations. The second part is a straightforward calculation of a percentage, using the sample mass of 0.2865 g.

  28. Example 4.12 continued Assessment Because iron is only one component of iron ore, the mass of iron in the ore sample must be less than 0.2865 g and the percent iron must be less than 100%. These facts can be used to assess the plausibility of the results obtained in the calculation. Exercise 4.12A Suppose the titration in Example 4.12 was carried out with 0.02250 M K2Cr2O7(aq) rather than KMnO4(aq). What volume of the K2Cr2O7(aq) would be required? The net ionic equation is 6 Fe2+(aq) + Cr2O72–(aq) + 14 H+(aq)  6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l) Exercise 4.12B A 20.00-mL sample of KMnO4(aq) is required to titrate 0.2378 g sodium oxalate in an acidic solution. How many milliliters of this same KMnO4(aq) are required to titrate a 25.00-mL sample of 0.1010 M FeSO4 in an acidic solution? 2 MnO4–(aq) + 16 H+(aq) + 5 C2O42–(aq)  2 Mn2+(aq) + 8 H2O(l) + 10 CO2(g)

  29. Cumulative Example Sodium nitrite is used in the production of fabric dyes, as a meat preservative, as a bleach for fibers, and in photography. It is prepared by passing nitrogen monoxide gas and oxygen gas into an aqueous solution of sodium carbonate. Carbon dioxide gas is the other product of the reaction. (a) Write a balanced equation for the reaction. (b) What mass of sodium nitrite should be produced in the reaction of 748 g of Na2CO3, with the other reactants in excess? (c) In another preparation, the reactants are 225 mL of 1.50 M Na2CO3(aq), 22.1 g nitrogen monoxide, and excess O2. What mass of sodium nitrite should be produced if the reaction has a yield of 95.1%? Strategy For part (a), we can write the formulas for reactants and products (Sections 2.6 and 2.7), and then construct and balance the equation (Section 3.7). Using the balanced equation, we can solve part (b), starting with grams of Na2CO3 and ending with grams of NaNO2 (Section 3.8). Part (c) can be broken down into three steps: Convert volume of Na2CO3 to moles using molarity (Section 3.11); solve as a limiting reactant problem (Section 3.9) for the mass of produced from NO and Na2CO3; use that theoretical yield of product along with percentage yield (Section 3.10) to find the actual yield.

  30. Cumulative Example continued Solution

  31. Cumulative Example continued The balanced equation has four N atoms, twelve O atoms, four Na atoms, and two C atoms on each side. Because the molar mass of NaNO2 is roughly half that of Na2CO3 and because the mole ratio between NaNO2 and Na2CO3 is 2:1, the mass of NaNO2 we get in part (b) should be comparable to the mass of Na2CO3 used. In part (c), because the yields of product from the two reactants are so similar, it is difficult to determine whether the limiting reactant is correct except by close inspection of the setup. However, we can note that the actual yield of NaNO2 is indeed less than the theoretical yield, as it should be. Assessment

More Related