1 IONIC COMPOUNDS Many reactions involve ionic compounds, especially reactions in water...

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Transcript of 1 IONIC COMPOUNDS Many reactions involve ionic compounds, especially reactions in water...

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  • 1 IONIC COMPOUNDS Many reactions involve ionic compounds, especially reactions in water aqueous solutions. KMnO 4 in water K + (aq) + MnO 4 - (aq)
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  • 2 An Ionic Compound, CuCl 2, in Water CCR, page 149
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  • 3 How do we know ions are present in aqueous solutions? The solutions conduct electricity! They are called ELECTROLYTES HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions
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  • 4 HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions
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  • 5 Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH 3 CO 2 H(aq) ---> CH 3 CO 2 - (aq) + H + (aq)
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  • 6 Aqueous Solutions Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH 3 CO 2 H(aq) ---> CH 3 CO 2 - (aq) + H + (aq)
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  • 7 Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugarethanol ethylene glycol Examples include: sugarethanol ethylene glycol
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  • 8 If one ion from the Soluble Compd. list is present in a compound, the compound is water soluble. Water Solubility of Ionic Compounds Screen 5.4 & Figure 5.1 Guidelines to predict the solubility of ionic compounds
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  • 9 Water Solubility of Ionic Compounds Common minerals are often formed with anions that lead to insolubility: sulfidefluoride sulfidefluoride carbonateoxide carbonateoxide Azurite, a copper carbonate Iron pyrite, a sulfide Orpiment, arsenic sulfide
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  • 10 An acid -------> H + in water ACIDSACIDS Some strong acids are HClhydrochloric H 2 SO 4 sulfuric HClO 4 perchloric HNO 3 nitric HNO 3
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  • 11 An acid -------> H + in water ACIDSACIDS HCl(aq) ---> H + (aq) + Cl - (aq)
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  • 12 The Nature of Acids HCl H 2 O Cl - H 3 O + hydronium ion
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  • 13 Weak Acids WEAK ACIDS = weak electrolytes CH 3 CO 2 H acetic acid H 2 CO 3 carbonic acid H 3 PO 4 phosphoric acid HF hydrofluoric acid Acetic acid
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  • 14 ACIDSACIDS Nonmetal oxides can be acids CO 2 (aq) + H 2 O(liq) ---> H 2 CO 3 (aq) SO 3 (aq) + H 2 O(liq) ---> H 2 SO 4 (aq) and can come from burning coal and oil.
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  • 15 Base ---> OH - in water BASES see Screen 5.9 and Table 5.2 NaOH(aq) ---> Na + (aq) + OH - (aq) NaOH is a strong base
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  • 16 Ammonia, NH 3 Ammonia, NH 3 An Important Base
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  • 17 BASESBASES Metal oxides are bases CaO(s) + H 2 O(liq) --> Ca(OH) 2 (aq) --> Ca(OH) 2 (aq) CaO in water. Indicator shows solution is basic.
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  • 18 Know the strong acids & bases!
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  • 19 Net Ionic Equations Mg(s) + 2 HCl(aq) --> H 2 (g) + MgCl 2 (aq) We really should write Mg(s) + 2 H + (aq) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) The two Cl - ions are SPECTATOR IONS they do not participate. Could have used NO 3 -.
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  • 20 Net Ionic Equations Mg(s) + 2 HCl(aq) --> H 2 (g) + MgCl 2 (aq) Mg(s) + 2 H + (aq) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) We leave the spectator ions out Mg(s) + 2 H + (aq) ---> H 2 (g) + Mg 2+ (aq) NET IONIC EQUATION to give the NET IONIC EQUATION
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  • 21 Chemical Reactions in Water Sections 5.2 & 5.4-5.6CD-ROM Ch. 5 We will look at EXCHANGE REACTIONS Pb(NO 3 ) 2 (aq) + 2 KI(aq) ----> PbI 2 (s) + 2 KNO 3 (aq) The anions exchange places between cations.
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  • 22 Precipitation Reactions The driving force is the formation of an insoluble compound a precipitate. Pb(NO 3 ) 2 (aq) + 2 KI(aq) -----> 2 KNO 3 (aq) + PbI 2 (s) Net ionic equation Pb 2+ (aq) + 2 I - (aq) ---> PbI 2 (s)
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  • 23 Acid-Base Reactions The driving force is the formation of water.The driving force is the formation of water. NaOH(aq) + HCl(aq) ---> NaCl(aq) + H 2 O(liq) Net ionic equationNet ionic equation OH - (aq) + H + (aq) ---> H 2 O(liq) OH - (aq) + H + (aq) ---> H 2 O(liq) This applies to ALL reactions of STRONG acids and bases.This applies to ALL reactions of STRONG acids and bases.
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  • 24 Acid-Base Reactions CCR, page 162
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  • 25 Acid-Base Reactions A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end.A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end. The other product of the A-B reaction is a SALT, MX.The other product of the A-B reaction is a SALT, MX. HX + MOH ---> MX + H 2 O HX + MOH ---> MX + H 2 O M n+ comes from base & X n- comes from acid M n+ comes from base & X n- comes from acid This is one way to make ionic compounds! This is one way to make ionic compounds!
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  • 26 Gas-Forming Reactions This is primarily the chemistry of metal carbonates. CO 2 and water ---> H 2 CO 3 H 2 CO 3 (aq) + Ca 2+ ---> H 2 CO 3 (aq) + Ca 2+ ---> 2 H + (aq) + CaCO 3 (s) (limestone) 2 H + (aq) + CaCO 3 (s) (limestone) Adding acid reverses this reaction. MCO 3 + acid ---> CO 2 + salt
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  • 27 Gas-Forming Reactions CaCO 3 (s) + H 2 SO 4 (aq) ---> 2 CaSO 4 (s) + H 2 CO 3 (aq) 2 CaSO 4 (s) + H 2 CO 3 (aq) Carbonic acid is unstable and forms CO 2 & H 2 O H 2 CO 3 (aq) ---> CO 2 (g) + water (Antacid tablet has citric acid + NaHCO 3 )
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  • 28 See also: Gas Forming Reactions in Biological Systems Three of the pioneers in working out the roles of NO forming reactions shared a Nobel Prize in 1988 for their discoveries. See also: Gas Forming Reactions in Biological Systems Three of the pioneers in working out the roles of NO forming reactions shared a Nobel Prize in 1988 for their discoveries.
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  • 29 Quantitative Aspects of Reactions in Solution Sections 5.8-5.10
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  • 30 TerminologyTerminology In solution we need to define the - SOLVENTSOLVENT the component whose physical state is preserved when solution forms SOLUTESOLUTE the other solution component
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  • 31 Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution Concentration (M) = [ ]
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  • 32 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 177
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  • 33 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M
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  • 34 The Nature of a CuCl 2 Solution Ion Concentrations CuCl 2 (aq) --> Cu 2+ (aq) + 2 Cl - (aq) If [CuCl 2 ] = 0.30 M, then [Cu 2+ ] = 0.30 M [Cl - ] = 2 x 0.30 M
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  • 35 USING MOLARITY What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/V this means that moles = MV
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  • 36 Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?
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  • 37 Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent.Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.Dilute a concentrated solution to give one that is less concentrated.
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  • 38 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!
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  • 39 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?
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  • 40 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution
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  • 41 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M V = M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH 0.15/Volume of final solution = 0.5 M/ 1 L Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL
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  • 42 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaO