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Tiga Jenis Utama Reaksi Kimia

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Tiga Jenis Utama Reaksi Kimia

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  1. TigaJenis Utama Reaksi Kimia

  2. Introduction • Artists did not have an affordable, stable blue pigment until 1704, when a paint maker accidentally produced Prussian blue in his laboratory. • The pigment quickly became extremely popular due to its deep color and because it faded in light much less than other blue pigments; it has since been used by countless artists, including van Gogh and Picasso. • This first modern synthetic pigment is produced by a simple reaction occurring in water and is just one of a myriad of products formed by aqueous reactions. • In nature, aqueous reactions occur unceasingly in the gigantic containers we know as oceans. And, in every cell of your body, thousands of reactions taking place right now enable you to function. With millions of reactions occurring in and around you, it would be impossible to describe them all. • Fortunately, it isn’t necessary because when we survey even a small percentage of reactions, especially those in aqueous solution, a few major classes emerge.

  3. SOLUTION CONCENTRATION AND THE ROLE OF WATER AS A SOLVENT • A solution consists of a smaller quantity of one substance, the solute, dissolved in a larger quantity of another, the solvent; in an aqueous solution, water serves as the solvent. • For any reaction in solution, the solvent plays a key role that depends on its chemical nature. • Some solvents passively disperse the substances into individual molecules. • But water is much more active, interacting strongly with the substances and even reacting with them in some cases. Let’s focus on how the water molecule interacts with both ionic and covalent solutes.

  4. The Polar Nature of Water • On the atomic scale, water’s great solvent power arises from the uneven distribution of electron charge and its bent molecular shape, which create a polar molecule: • Uneven charge distribution. The electrons in a covalent bond are shared between the atoms. In a bond between identical atoms—as in H2, Cl2, O2—the sharing is equal and electron charge is distributed evenly between the two nuclei • In covalent bonds between different atoms, the sharing is uneven because one atom attracts the electron pair more strongly than the other atom does. • For example, in each OH bond of water, the shared electrons are closer to the O atom because an O atom attracts electrons more strongly than an H atom does. • This uneven charge distribution creates a polar bond, one with partially charged “poles.”

  5. In Figure 4.1B, the asymmetrical shading shows this distribution, and the  symbol indicates a partial charge. • The O end is partially negative, represented by red shading and -, and the H end is partially positive, represented by blue shading and +. In the ball-and-stick model of Figure 4.1C, the polar arrow points to the negative pole, and the tail, shaped like a plus sign, marks the positive pole. • 2. Bent molecular shape. The sequence of the HOH atoms in water is not linear: the water molecule is bent with a bond angle of 104.5 (Figure 4.1C). • 3. Molecular polarity. The combination of polar bonds and bent shape makes water a polar molecule: the region near the O atom is partially negative, and the region between the H atoms is partially positive (Figure 4.1D)

  6. IonicCompounds in Water • In anionic solid, oppositely charged ions are held together by electrostatic attractions. • Water separates the ions by replacing these attractions with others between several water molecules and each ion. • Picture a granule of a soluble ionic compound in water: the negative ends of some water molecules are attracted to the cations, and the positive ends of other water molecules are attracted to the anions (Figure 4.2). • Dissolution occurs because the attractions between each type of ion and several water molecules outweigh the attractions between the ions. • Gradually, all the ions separate (dissociate), become solvated—surrounded closely by solvent molecules—and then move randomly in the solution.

  7. For an ionic compound that doesn’t dissolve in water, the attraction between ions is greater than the attraction between the ions and water. • Actually, these so-called insoluble substances do dissolve to a very small extent, usually several orders of magnitude less than so-called soluble substances. For example, NaCl (a “soluble” compound) is over 4104 times more soluble than AgCl (an “insoluble” compound): • Solubility of NaCl in H2O at 20 C = 365 g/L • Solubility of AgCl in H2O at 20 C = 0.009 g/L

  8. How Ionic Solutions Behave: Electrolytes and Electrical Conductivity

  9. Calculating the Number of Moles of Ions in Solution • From the formula of the soluble ionic compound, we know the number of moles of each ion in solution. • For example, the equations for dissolving KBr and CaBr2 in water to form solvated ions are • KBr(s) K+(aq) + Br-(aq) • CaBr2(s) Ca2+(aq) + 2Br-(aq) • (“H2O” above the arrows means that water is the solvent, not a reactant.) Note that 1 mol of KBr dissociates into 2 mol of ions—1 mol of K+ and 1 mol of Br-—and 1 mol of CaBr2 dissociates into 3 mol of ions—1 mol of Ca2+ and 2 mol of Br-.

  10. Sample Problems • What amount (mol) of each ion is in each solution? • 5.0 mol of ammonium sulfate dissolved in water • 78.5 g of cesium bromide dissolved in water • 7.421022 formula units of copper(II) nitrate dissolved in water

  11. CovalentCompounds in Water • Water dissolves many covalent (molecular) compounds also. Table sugar (sucrose, C12H22O11), beverage (grain) alcohol (ethanol, CH3CH2OH), and automobile antifreeze (ethylene glycol, HOCH2CH2OH) are some familiar examples. All contain their own polar bonds, which interact with the bonds of water. • However, most soluble covalent substances do not separate into ions, but remain intact molecules. For example, • HOCH2CH2OH(l) HOCH2CH2OH(aq) • As a result, their aqueous solutions do not conduct an electric current, and these substances are nonelectrolytes. • Many other covalent substances, such as benzene (C6H6) and octane (C8H18), do not contain polar bonds, and these substances do not dissolve appreciably in water

  12. Expressing Concentration in Terms of Molarity • When working quantitatively with any solution, it is essential to know the concentration— the quantity of solute dissolved in a given quantity of solution (or of solvent). • Concentration is an intensive property (like density or temperature) and thus is independent of the solution volume: a 50-L tank of a solution has the same concentration (solute quantity/solution quantity) as a 50-mL beaker of the solution. • Molarity (M) is the most common unit of concentration. • It expresses the concentration in units of moles of solute per liter of solution:

  13. Sample Problems • Glycine (C2H5NO2) has the simplest structure of the 20 amino acids that make up proteins. What is the molarity of a solution that contains 53.7 g of glycine dissolved in 495 mL of solution? • Calculate the molarity of the solution made when 6.97 g of KI is dissolved in enough water to give a total volume of 100. mL • Calculate the molarity of a solution that contains 175 mg of sodium nitrate in a total volume of 15.0 mL.

  14. Amount-Mass-Number Conversions Involving Solutions

  15. Sample Problems • Biochemists often study reactions in solutions containing phosphate ion, which is commonly found in cells. How many grams of solute are in 1.75 L of 0.460 M sodium hydrogenphosphate? • In biochemistry laboratories, solutions of sucrose (table sugar, C12H22O11) are used in high-speed centrifuges to separate the parts of a biological cell. How many liters of 3.30 M sucrose contain 135 g of solute? • What amount (mol) of each ion is in 35 mL of 0.84 M zinc chloride?

  16. Preparing and Diluting Molar Solutions • Notice that the volume term in the denominator of the molarity expression in Equation 4.1 is the solution volume, not the solvent volume. • This means that you cannot dissolve 1 mol of solute in 1 L of solvent to make a 1 M solution. • Because the solute volume adds to the solvent volume, the total volume (solute 1 solvent) would be more than 1 L, so the concentration would be less than 1 M.

  17. Preparing a Solutions • Correctly preparing a solution of a solid solute requires four steps. Let’s prepare 0.500 L of 0.350 M nickel(II) nitrate hexahydrate [Ni(NO3)26H2O]: • Weighthesolid. Calculate the mass of solid needed by converting from volume (L) to amount (mol) and then to mass (g): • Transferthesolid. We need 0.500 L of solution, so we choose a 500-mL volumetric flask (a flask with a fixed volume indicated by a mark on the neck), add enough distilled water to fully dissolve the solute (usually about half the final volume, or 250 mL of distilled water in this case), and transfer the solute. Wash down any solid clinging to the neck with some solvent

  18. Dissolvethesolid. Swirl the flask until all the solute is dissolved. If necessary, wait until the solution is at room temperature. (As we discuss in Chapter 13, the solution process may be accompanied by heating or cooling.) • Addsolventtothefinalvolume. Add distilled water to bring the solution volume to the line on the flask neck; cover and mix thoroughly again.

  19. Diluting a Solutions • A concentrated solution (higher molarity) is converted to a dilute solution (lower molarity) by adding solvent, which means the solution volume increases but the amount (mol) of solute stays the same. • As a result, the dilute solution contains fewer solute particles per unit volume and, thus, has a lower concentration than the concentrated solution (Figure 4.6). • Chemists often prepare a concentrated solution (stock solution) and store it and dilute it as needed

  20. Solving Dilution Problems • To solve dilution problems, we use the fact that the amount (mol) of solute does not change during the dilution process. • Therefore, once we calculate the amount of solute needed to make a particular volume of a dilute solution, we can then calculate the volume of concentrated solution that contains that amount of solute. • This two-part calculation can be combined into one step in the following relationship: • where M and V are the molarity and volume of the dilute (subscript “dil”) and concentrated (subscript “conc”) solutions

  21. Sample Problems • Isotonic saline is 0.15 M aqueous NaCl. It simulates the total con centration of ions in many cellular fluids, and its uses range from cleaning contact lenses to washing red blood cells. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? • A chemist dilutes 60.0 mL of 4.50 M potassium permanganate to make a 1.25 M solution. What is the final volume of the diluted solution?

  22. WRITING EQUATIONS FOR AQUEOUSIONIC REACTIONS • Chemists use three types of equations to represent aqueous ionic reactions. • Let’s examine a reaction to see what each type shows. When solutions of silver nitrate and sodium chromate are mixed, brick-red, solid silver chromate (Ag2CrO4) forms • Figure 4.7 depicts the reaction at the macroscopic level (photos), the atomic level (blow-up circles), and the symbolic level with the three types of equations (reacting ions are in red type):

  23. The molecular equation (top) reveals the least about the species that are actually in solution because it shows all the reactants and products as if they were intact, undissociated compounds. Only the designation for solid, (s), tells us that a change has occurred: • 2AgNO3(aq) + Na2CrO4(aq)  • Ag2CrO4(s) + 2NaNO3(aq) • • The total ionic equation (middle) is much more accurate because it shows all the soluble ionic substances as they actually exist in solution, where they are dissociated into ions. The Ag2CrO4(s) stands out as the only undissociated substance: • 2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO42-(aq)  • Ag2CrO4(s) + 2Na+(aq) + 2NO3-(aq)

  24. The charges also balance: four positive and four negative for a net zero charge on the left side, and two positive and two negative for a net zero charge on the right. • Notice that Na+(aq) and NO3-(aq) appear unchanged on both sides of the equation. These are called spectator ions (shown with pale colors in the atomic level scenes). • They are not involved in the actual chemical change but are present only as part of the reactants; that is, we can’t add an Ag+ ion without also adding an anion, in this case, the NO3- ion.

  25. The net ionic equation (bottom) is very useful because it eliminates the spectator ions and shows only the actual chemical change: • 2Ag+(aq) + CrO42-(aq)  Ag2CrO4(s) • The formation of solid silver chromate from silver ions and chromate ions is the only change. • To make that point clearly, suppose we had mixed solutions of potassium chromate, K2CrO4(aq), and silver acetate, AgC2H3O2(aq), instead of sodium chromate and silver nitrate. The three ionic equations would be • Thus, the same change would have occurred, and only the spectator ions would differ—K+(aq) and C2H3O2-(aq) instead of Na+(aq) and NO3-(aq).

  26. PRECIPITATION REACTIONS • In a precipitation reaction, two soluble ionic compounds react to form an insoluble product, a precipitate. The reaction you saw between silver nitrate and sodium chromate is one example. • Precipitates form for the same reason that some ionic compounds don’t dissolve: the electrostatic attraction between the ions outweighs the tendency of the ions to remain solvated and move throughout the solution. • When the two solutions are mixed, the ions collide and stay together, and a solid product “comes out of solution.” Thus, the key event in a precipitation reaction is the formation of an insoluble product through the net removal of ions from solution. • Figure 4.8 (on the next page) shows the process for calcium fluoride.

  27. Predicting Whether a Precipitate Will Form • To predict whether a precipitate will form when we mix two aqueous ionic solutions, we refer to the short list of solubility rules in Table 4.1. • Let’s see how to apply these rules. When sodium iodide and potassium nitrate are each dissolved in water, their solutions consist of solvated, dispersed ions:

  28. Three steps help us predict if a precipitate forms: • Note the ions in the reactants. The reactant ions are • Na+(aq) + I-(aq) + K+(aq) + NO3-(aq)  ? • Consider all possible cation-anion combinations. In addition to NaI and KNO3, which we know are soluble, the other cation-anion combinations are NaNO3 and KI. • Decide whether any combination is insoluble. According to Table 4.1, all compounds of Group lA(l) ions and all nitrate compounds are soluble. No reaction occurs because all the cation-anion combinations—NaI, KNO3, NaNO3, and KI—are soluble: • Na+(aq) + I-(aq) + K+(aq) + NO3-(aq)  • Na+(aq) + NO3-(aq) + K+(aq) + I-(aq) • All the ions are spectator ions, so when we eliminate them, we have no net ionic equation.

  29. Stability Rules

  30. Now, what happens if we substitute a solution of lead(II) nitrate, Pb(NO3)2, for the KNO3 solution? The reactant ions are Na+, I-, Pb2+, and NO3-. • In addition to the two soluble reactants, NaI and Pb(NO3)2, the other two possible cation-anion combinations are NaNO3 and PbI2. According to Table 4.1, NaNO3 is soluble, but PbI2 is not. • The total ionic equation shows the reaction that occurs as Pb2+ and I- ions collide and form a precipitate: • 2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq)  • 2Na+(aq) + 2NO3-(aq) + PbI2(s) • And the net ionic equation confirms it: • Pb2+(aq) + 2I-(aq) PbI2(s)

  31. Sample Problem • Does a reaction occur when each of these pairs of solutions is mixed? If so, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. • Potassium fluoride(aq) + strontium nitrate(aq)  • Ammonium perchlorate(aq) + sodium bromide(aq) 

  32. StoichiometryofPrecipitationReactions • Solving stoichiometry problems for any reaction that takes place in solution, such as a precipitation reaction, requires the additional step of converting the volume of reactant or product in solution to amount (mol): • Balancetheequation. • Find the amount (mol) of one substance using its molar mass (for a pure substance) or the volume and molarity (for a substance in solution). • Relate that amount to the stoichiometrically equivalent amount of another substance. • Convert to the desired units.

  33. Summary

  34. Sample Problem • Magnesium is the second most abundant metal in seawater, after sodium. The first step in its industrial extraction involves the reaction of the magnesium ion with calcium hydroxide to precipitate magnesium hydroxide. What mass of magnesium hydroxide is formed when 0.180 L of 0.0155 M magnesium chloride reacts with excess calciumhydroxide? • Solution • Write the balanced equation: • MgCl2(aq) + Ca(OH)2(aq) Mg(OH)2(s) + CaCl2(aq)

  35. Using the molar ratio to convert amount (mol) of MgCl2 to amount (mol) of Mg(OH)2 Converting from amount (mol) of Mg(OH)2 to mass (g):

  36. TugasMandiri - 4 • It is desirable to remove calcium ion from hard water to prevent the formation of precipitates known as boiler scale that reduce heating efficiency. The calcium ion is reacted with sodium phosphate to form solid calcium phosphate, which is easier to remove than boiler scale. What volume of 0.260 M sodium phosphate is needed to react completely with 0.300 L of 0.175 M calcium chloride? • To lift fingerprints from a crime scene, a solution of silver nitrate is sprayed on a surface to react with the sodium chloride left behind by perspiration. What is the molarity of a silver nitrate solution if 45.0 mL of it reacts with excess sodium chloride to produce 0.148 g of precipitate?

  37. LimitingReactantProblemfor a PrecipitationReaction • Mercury and its compounds have uses from fillings for teeth (as a mixture with silver, copper, and tin) to the production of chlorine. Because of their toxicity, however, soluble mercury compounds, such as mercury(II) nitrate, must be removed from industrial wastewater. One removal method reacts the wastewater with sodium sulfide solution to produce solid mercury(II) sulfide and sodium nitrate solution. In a laboratory simulation, 0.050 L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. • How many grams of mercury(II) sulfide form? • Write a reaction table for this process.

  38. Hg(NO3)2(aq) + Na2S(aq)  HgS(s) + 2NaNO3(aq) • Finding the amount (mol) of HgS formed from Hg(NO3)2: • Finding the amount (mol) of HgS from Na2S: • Hg(NO3)2 is the limiting reactant because it forms fewer moles of HgS. • Converting the amount (mol) of HgS formed from Hg(NO3)2 to mass (g):

  39. With Hg(NO3)2 as the limiting reactant, the reaction table is: • A large excess of Na2S remains after the reaction. Note that the amount of NaNO3 formed is twice the amount of Hg(NO3)2 consumed, as the balanced equation shows.

  40. Sample Problems • Despite the toxicity of lead, many of its compounds are still used to make pigments. • When 268 mL of 1.50 M lead(II) acetate reacts with 130. mL of 3.40 M sodium chloride, how many grams of solid lead(II) chloride can form? • Using the abbreviation “Ac” for the acetate ion, write a reaction table for the process.

  41. TugasMandiri - 5 • Iron(III) hydroxide has been used to adsorb arsenic and heavy metals from contaminated soil and water. Solid iron(III) hydroxide is produced by reacting aqueous solutions of iron(III) chloride and sodium hydroxide. • What mass of iron(III) hydroxide is formed when 155 mL of 0.250 M iron(III) chloride reacts with 215 mL of 0.300 M sodium hydroxide? • Write a reaction table for this process.

  42. ACID-BASE REACTIONS • Aqueous acid-base reactions occur in processes as diverse as the metabolic action of proteins and carbohydrates, the industrial production of fertilizer, and the revitalization of lakes damaged by acid rain • These reactions involve water as reactant or product, in addition to its common role as solvent. Of course, an acid-base reaction (also called a neutralization reaction) occurs when an acid reacts with a base, but the definitions of these terms and the scope of this reaction class have changed over the years. For our purposes at this point, we’ll use definitions that apply to substances found commonly in the lab:

  43. Acids and the Solvated Proton • Acidic solutionsarisewhencertaincovalent H-containing molecules dissociate into ions in water. In every case, these molecules contain a polar bond to H in which the other atom pulls much more strongly on the electron pair. • A good example is HBr. The Br end of the HBr bond is partially negative, and the H end is partially positive. When hydrogen bromide gas dissolves in water, the poles of H2O molecules are attracted to the oppositely charged poles of the HBr. The bond breaks, with H becoming the solvated cation H+(aq) and Br becoming the solvated anion Br-(aq): • HBr(g) H+(aq) + Br-(aq)

  44. The solvated H+ ion is a very unusual species. The H atom is a proton surrounded by an electron, so H+ is just a proton. With a full positive charge concentrated in such a tiny volume, H+ attracts the negative pole of water molecules so strongly that it forms a covalent bond to one of them. We can show this interaction by writing the solvated H+ ion as an H3O+ ion (hydronium ion) that also is solvated: • The hydronium ion, which we write as H3O+ [or (H2O)H+], associates with other water molecules to give species such as H5O2+ [or (H2O)2H+], H7O3+ [or (H2O)3H+],H9O4+ [or (H2O)4H+], and so forth. Figure 4.11 shows H7O3+, an H3O+ ion associated (dotted lines) with two H2O molecules. As a general notation for these various species, we write H+(aq), but later in this chapter and in much of the text, we’ll use H3O+(aq).

  45. AcidsandBases as Electrolytes • Strong acids and strong bases dissociate completely into ions. Therefore, like soluble ionic compounds, they are strong electrolytes and conduct a large current, as shown by the brightly lit bulb (Figure). • Weak acids and weak bases dissociate very little into ions. Most of their molecules remain intact. Therefore, they are weak electrolytes, which means they conduct a small current. • Because a strong acid (or strong base) dissociates completely, we can find the molarity of H+ (or OH-) and the amount (mol) or number of each ion in solution.

  46. Sample Problem • Nitric acid is a major chemical in the fertilizer and explosives industries. How many H+(aq) ions are in 25.3 mL of 1.4 M nitric acid?