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IONIC COMPOUNDS Compounds in Aqueous Solution

K + (aq) + MnO 4 - (aq). IONIC COMPOUNDS Compounds in Aqueous Solution. Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO 4 in water. An Ionic Compound, CuCl 2 , in Water. CCR, page 149. Aqueous Solutions.

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IONIC COMPOUNDS Compounds in Aqueous Solution

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  1. K+(aq) + MnO4-(aq) IONIC COMPOUNDSCompounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO4 in water

  2. An Ionic Compound, CuCl2, in Water CCR, page 149

  3. Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions conduct electricity! They are called ELECTROLYTES HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

  4. Aqueous Solutions HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

  5. Aqueous Solutions Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq)

  6. Aqueous Solutions Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq)

  7. Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol

  8. Water Solubility of Ionic Compounds If one ion from the “Soluble Compd.” list is present in a compound, the compound is water soluble. Screen 5.4 & Figure 5.1

  9. Iron pyrite, a sulfide Orpiment, arsenic sulfide Azurite, a copper carbonate Water Solubility of Ionic Compounds Common minerals are often formed with anions that lead to insolubility: sulfide fluoride carbonate oxide

  10. HNO3 ACIDS An acid -------> H+ in water Some strongacids are HCl hydrochloric H2SO4 sulfuric HClO4 perchloric HNO3 nitric

  11. ACIDS An acid -------> H+ in water HCl(aq) ---> H+(aq) + Cl-(aq)

  12. HCl - Cl H O + H O 2 3 hydronium ion The Nature of Acids

  13. Acetic acid Weak Acids WEAK ACIDS = weak electrolytes CH3CO2H acetic acid H2CO3 carbonic acid H3PO4 phosphoric acid HF hydrofluoric acid

  14. ACIDS Nonmetal oxides can be acids CO2(aq) + H2O(liq) ---> H2CO3(aq) SO3(aq) + H2O(liq) ---> H2SO4(aq) and can come from burning coal and oil.

  15. BASESsee Screen 5.9 and Table 5.2 Base ---> OH- in water NaOH(aq) ---> Na+(aq) + OH-(aq) NaOH is a strong base

  16. Ammonia, NH3An Important Base

  17. BASES Metal oxides are bases CaO(s) + H2O(liq) --> Ca(OH)2(aq) CaO in water. Indicator shows solution is basic.

  18. Know the strong acids & bases!

  19. Net Ionic Equations Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq) We really should write Mg(s) + 2 H+(aq) + 2 Cl-(aq) ---> H2(g) + Mg2+(aq) + 2 Cl-(aq) The two Cl- ions are SPECTATOR IONS — they do not participate. Could have used NO3-.

  20. Net Ionic Equations Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq) Mg(s) + 2 H+(aq) + 2 Cl-(aq) ---> H2(g) + Mg2+(aq) + 2 Cl-(aq) We leave the spectator ions out — Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq) to give the NET IONIC EQUATION

  21. Chemical Reactions in WaterSections 5.2 & 5.4-5.6—CD-ROM Ch. 5 Pb(NO3) 2(aq) + 2 KI(aq) ----> PbI2(s) + 2 KNO3 (aq) We will look at EXCHANGE REACTIONS The anions exchange places between cations.

  22. Precipitation Reactions The “driving force” is the formation of an insoluble compound — a precipitate. Pb(NO3)2(aq) + 2 KI(aq) -----> 2 KNO3(aq) + PbI2(s) Net ionic equation Pb2+(aq) + 2 I-(aq) ---> PbI2(s)

  23. Acid-Base Reactions • The “driving force” is the formation of water. NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(liq) • Net ionic equation OH-(aq) + H+(aq) ---> H2O(liq) • This applies to ALL reactions of STRONG acids and bases.

  24. Acid-Base Reactions CCR, page 162

  25. Acid-Base Reactions • A-B reactions are sometimes called NEUTRALIZATIONSbecause the solution is neither acidic nor basic at the end. • The other product of the A-B reaction is a SALT, MX. HX + MOH ---> MX + H2O Mn+ comes from base &Xn- comes from acid This is one way to make compounds!

  26. Gas-Forming Reactions This is primarily the chemistry of metal carbonates. CO2 and water ---> H2CO3 H2CO3(aq) + Ca2+ ---> 2 H+(aq) + CaCO3(s) (limestone) Adding acid reverses this reaction. MCO3 + acid ---> CO2 + salt

  27. Gas-Forming Reactions CaCO3(s) + H2SO4(aq) ---> 2 CaSO4(s) + H2CO3(aq) Carbonic acid is unstable and forms CO2 & H2O H2CO3(aq) ---> CO2 + water (Antacid tablet has citric acid + NaHCO3)

  28. Quantitative Aspects of Reactions in SolutionSections 5.8-5.10

  29. Terminology In solution we need to define the - • SOLVENT the component whose physical state is preserved when solution forms • SOLUTE the other solution component

  30. moles solute ( M ) = Molarity liters of solution Concentration of Solute The amount of solute in a solution is given by its concentration. Concentration (M) = [ …]

  31. 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 177

  32. PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity. Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate molarity [NiCl2•6 H2O] = 0.0841 M

  33. The Nature of a CuCl2 SolutionIon Concentrations CuCl2(aq) --> Cu2+(aq) + 2 Cl-(aq) If [CuCl2] = 0.30 M, then [Cu2+] = 0.30 M [Cl-] = 2 x 0.30 M

  34. moles = M•V USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/V this means that

  35. USING MOLARITY What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? 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 moles = M•V

  36. Preparing Solutions • Weigh out a solid solute and dissolve in a given quantity of solvent. • Dilute a concentrated solution to give one that is less concentrated.

  37. 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!

  38. 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?

  39. moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution 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 --->

  40. 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 = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL

  41. 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 NaOH to make 300 mL of 0.50 M NaOH.

  42. Preparing Solutions by Dilution A shortcut Cinitial • Vinitial = Cfinal • Vfinal

  43. pH, a Concentration Scale pH: a way to express acidity -- the concentration of H+ in solution. Low pH: high [H+] High pH: low [H+] Acidic solution pH < 7 Neutral pH = 7 Basic solution pH > 7

  44. The pH Scale pH = log (1/ [H+]) = - log [H+] In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC pH = - log [H+] = -log (1.00 x 10-7) = - (-7) = 7 See CD Screen 5.17 for a tutorial

  45. [H+] and pH If the [H+] of soda is 1.6 x 10-3 M, the pH is ____? Because pH = - log [H+] then pH= - log (1.6 x 10-3) pH = - (-2.80) pH = 2.80

  46. pH and [H+] If the pH of Coke is 3.12, it is ____________. Because pH = - log [H+] then log [H+] = - pH Take antilog and get [H+] = 10-pH [H+] = 10-3.12 = 7.6 x 10-4 M

  47. SOLUTION STOICHIOMETRYSection 5.10 • Zinc reacts with acids to produce H2 gas. • Have 10.0 g of Zn • What volume of 2.50 M HCl is needed to convert the Zn completely?

  48. Mass HCl Mass zinc Stoichiometric factor Moles zinc Moles HCl Volume HCl GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS

  49. Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g) Step 2: Calculate amount of Zn Step 3: Use the stoichiometric factor

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