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Acids & Bases

Acids & Bases. Arrhenius Definition: An acid is any substance that increases the H + ( aq ) concentration in an aqueous solution. HX( aq )  H + ( aq ) + X – ( aq ) A base is any substance that increases the OH – ( aq ) concentration in an aqueous solution.

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Acids & Bases

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  1. Acids & Bases Arrhenius Definition: • An acid is any substance that increases the H+(aq) concentration in an aqueous solution. HX(aq)  H+(aq) + X–(aq) • A base is any substance that increases the OH–(aq) concentration in an aqueous solution. MOH(aq)  M+(aq) + OH–(aq)

  2. Acids and Bases Brönsted-Lowry: • An acid is any substance that donates H+(aq) to another species in an aqueous solution. HX(aq) + H2O(l) H3O+(aq) + X–(aq) • A base is any substance that accepts an H+(aq) in an aqueous solution. H+(aq) + NH3(aq) NH4+(aq) H3O+(aq) = H+(aq)

  3. Acids

  4. Strong Acids Examples: Strong acidsare almost completely ionized in water. (strong electrolytes) HX (aq) (X = Cl, Br & I) hydro ___ ic acid HNO3 (aq) nitric acid perchloric acid HClO4 (aq) sulfuric acid H2SO4 (aq)* * Only the 1st H is strong, sulfuric acid dissociates via: H2SO4 (aq)  H+ (aq) + HSO4– (aq)

  5. Acids An acid: H3O+ in water

  6. Weak Acids Examples: Weak Acidsare incompletely ionized in water. (weak electrolytes) Weak acids are governed by dynamic equilibrium. HC2H3O2 (aq) acetic acid (vinegar) nitrous acid HNO2 (aq) H2S (aq) hydrosulfuric acid HSO4–(aq) hydrogen sulfate ion Weak acids are always written in their molecular form. See you text and home work for more examples.

  7. Strong Bases Bases:A base is a substance that produces OH– (aq)ions in water by dissociation in water: Strong bases are almost completely ionized in aqueous solution. (Strong electrolytes) Examples: Hydroxides of Group 1 (MOH(aq) where M = Li, Na, K ect…) and Ca, Sr, Ba.* *Ca(OH)2, Sr(OH)2 & Ba(OH)2 are slightly soluble, but that which dissolves is present as ions only.

  8. Bases NaOH(aq)  Na+(aq) + OH-(aq) Base: OH- in water NaOH is a strong base

  9.    Weak Bases Weak Bases: NH3 acts as a base by reacting with water: NH3(aq) + H2O(l) NH4+(aq) + OH –(aq) Ammonia can also accept H+ from an acid: NH3(aq) + H+(aq) NH4+(aq)

  10. Ammonia, NH3

  11. Reactions of Acids & Bases: Acid-Base Neutralization Salt + Water (usually) Acid + Base  HA (aq) + MOH(aq)  MA(aq) + HOH(l) Strong acid - Strong base neutralization: HBr(aq)/KOH(aq) Molecular Equation: HBr(aq) + KOH(aq)  KBr (aq) + H2O(l) Total Ionic Equation: / / H+ (aq) + Br– (aq) K+(aq) + Br– (aq) + K+(aq) + OH– (aq)  + H2O(l) Br- K+ Net Ionic equation: H+ (aq) + OH– (aq) H2O (l)

  12. Acid-Base Reactions • The “driving force” is the formation of water. NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(liq) • Net ionic equation OH-(aq) + H3O+(aq)  2 H2O(l) • This applies to ALL reactions of STRONG acids and bases.

  13. Reactions of Acids & Bases: Acid-Base Neutralization Reactions of weak acids and strong bases: Molecular Equation: HC2H3O2(aq) + NaOH(aq)  NaC2H3O2(aq) + H2O(l) Total Ionic Equation: / / HC2H3O2(aq) + Na+(aq) + OH–(aq)  Na+(aq) + C2H3O2–(aq) + H2O(l) Leave in molecular form Net Ionic: HC2H3O2(aq) + OH–(aq)  C2H3O2–(aq) + H2O(l)

  14. Non-Metal Acids Nonmetal oxides can form acids in aqueous solutions: Examples: CO2(aq) + H2O(s)  H2CO3(aq) SO3(aq) + H2O(s)  H2SO4(aq) Both gases come from the burning of fossil fuels.

  15. Bases Metal oxides form bases in aqueous solution CaO(s) + H2O(l)  Ca(OH)2(aq) CaO in water. Indicator shows solution is basic.

  16. Gas-Forming Reactions

  17. Gas-Forming Reactions Metal carbonate salts react with acids to the corresponding metal salt, water and carbon dioxide gas. 2HCl(aq) + CaCO3(s)  CaCl2(aq) + H2CO3(aq) decomposes H2O(l) + CO2(g) Similarly: HCl(aq) + NaHCO3(s)  NaCl(aq) + H2O(l) + CO2(g) acid base salt water Neutralization!!!

  18. Gas-Forming Reactions Group I metals: Na, K, Cs etc.. react vigorously with water 2K(s) 2KOH(aq) + H2(g) + 2H2O(l)  Metals & acid: Some metals react vigorously with acid solutions: Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)

  19. Gas-Forming Reactions CaCO3(s) + H2SO4(aq)  2 CaSO4(s) + H2CO3(aq) Carbonic acid is unstable and forms CO2 & H2O H2CO3(aq) CO2 + water (The antacid tablet contains citric acid + NaHCO3)

  20. Measuring Concentrations of Compounds in a Solution • Most chemical studies require quantitative measurements. • Experiments involving aqueous solutions are measured in volumes of solutions rather than masses of solids, liquids, or gases. • Solution concentration, expressed as molarity, relates the volume of solution in liters to the amount of substance in moles.

  21. That which is dissolves (greater amount) That which is dissolved (lesser amount) Terminology Solution = solute + solvent

  22. Solutions and Concentration Molarity: Moles of solute per liter of solution. moles of solute {units: mol/L} Molarity of X (cX) = L of Solution cNaCl = [NaCl] = 1.00 M Molarity is a conversion factor that transforms units of volume to mole and vise–versa.

  23. Moles, Liters, & Molarity If you know Molarity and volume, you know moles! Molarity  Volume = moles  L = moles If you know mols and molarity, you know volume! moles  = Volume L mol

  24. Preparing Solution of Known Concentration

  25. 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

  26. Ion Concentrations in a Solution [CuCl2] = 0.30 M = [Cu2+] = 0.30 M [Cl-] = 2  0.30 M = 0.60 M CuCl2(aq)  Cu2+(aq) + 2Cl(aq) For every one copper (II) ion in solution, there are two chloride ions.

  27. Concentration Problem: Determine the nitrate ion concentration (in mols/L) of a solution that is made from 50.1451 g of iron (III) nitrate diluted to a final volume of 250.0 mL.

  28. Concentration Problem: Determine the nitrate ion concentration (in mols/L) of a solution that is made from 50.1451 g of iron (III) nitrate diluted to a final volume of 250.0 mL. Solution:  molar mass  mole ratio

  29. Concentration Problem: Determine the nitrate ion concentration (in mols/L) of a solution that is made from 50.1451 g of iron (III) nitrate diluted to a final volume of 250.0 mL. Solution:  molar mass  mole ratio Fe3+(aq) Fe(NO3)3 (aq)  + 3NO3– (aq)

  30. Concentration Problem: Determine the nitrate ion concentration (in mols/L) of a solution that is made from 50.1451 g of iron (III) nitrate diluted to a final volume of 250.0 mL. Solution:  molar mass  mole ratio Fe3+(aq) Fe(NO3)3 (aq)  + 3NO3– (aq) = 2.488 M NO3–

  31. from the balanced equation! Concentration Problem: Determine the nitrate ion concentration (in mols/L) of a solution that is made from 50.1451 g of iron (III) nitrate diluted to a final volume of 250.0 mL. Solution:  molar mass  mole ratio Fe3+(aq) Fe(NO3)3 (aq)  + 3NO3– (aq) = 2.488 M NO3–

  32. Concentration Problem: How many grams of sodium phosphate are in 35.0 mL of a 1.51 M Na3PO4 solution? Solution: mols Na3PO4 g Na3PO4 mL solution  L   use M as a conversion factor  molar mass

  33. Concentration Problem: How many grams of sodium phosphate are in 35.0 mL of a 1.51 M Na3PO4 solution? Solution: mols Na3PO4 g Na3PO4 mL solution  L   use M as a conversion factor  molar mass

  34. Preparing Solutions • Weigh out the solid solute and carefully transfer it to the volumetric flask. • Dissolve in a volume of solvent less than the flask volume. • Fill the volumetric flask to the calibration mark with solvent. • Invert the flask many times to ensure homogeneous mixing.

  35. Preparing a Solution by Dilution

  36. Dilutions: Volume by Volume as solvent is added, the denominator increases in magnitude… since the moles of solute remains constant… When more solvent is added to a solution, the concentration of solute decreases: Why? The magnitude of the concentration must decrease!!!

  37. A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution to a 500.0 mL volumetric flask. • She then filled the volumetric flask to the calibration mark with water. • What is the new molarity of the HCl solution?

  38. A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution to a 500.0 mL volumetric flask. • She then filled the volumetric flask to the calibration mark with water. • What is the new molarity of the HCl solution? Solution

  39. at this point I have mols HCl • A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution to a 500.0 mL volumetric flask. • She then filled the volumetric flask to the calibration mark with water. • What is the new molarity of the HCl solution? Solution

  40. at this point I have mols HCl • A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution to a 500.0 mL volumetric flask. • She then filled the volumetric flask to the calibration mark with water. • What is the new molarity of the HCl solution? Solution

  41. at this point I have mols HCl dividing by the “new volume” results in a new M • A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution to a 500.0 mL volumetric flask. • She then filled the volumetric flask to the calibration mark with water. • What is the new molarity of the HCl solution? Solution

  42. converting to L gives: at this point I have mols HCl dividing by the “new volume” results in a new M • A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution to a 500.0 mL volumetric flask. • She then filled the volumetric flask to the calibration mark with water. • What is the new molarity of the HCl solution? Solution 1/10th the original M

  43. The pH Scale About one in 10 million water molecules will do this spontaneously 1/10,000,000 = 1 x 10-7 moles/liter is the normal H+ concentration in water = [H+] Shorthand [ ] = concentration If acid is added, [H+] increases. [H+] is a good measure of acidity, but the #s involved are impractically small. Logarithms solve this difficulty. H20  H+ + OH-

  44. The pH Scale What we call pH is actually a mathematical function, "p" p is a shorthand notation for " log10" (the negative logarithm, based 10 ) pH = -log [H+] 3

  45. pH = log[H+] Find the pH of a solution that has hydrogen ion concentration of [H+] = 1 x10-7 • [H+] = 2 x10-7 10 x10-7 1 x10-8

  46. The pH Scale What we call pH is actually a mathematical function, "p" p is a shorthand notation for " log10" (the negative logarithm, based 10 ) Quick Review of logs…(see Appendix A, p. A2 of your text) log x = n where x = 10n log 1000 = log(103) = 3 log 10 = log(101) = 1 3 log 0.001 = log(10 3) =

  47. Example: A student is given a solution that is labeled pH = 4.72, what is the molarity of H+ in this solution? plugging in 104.72 into you calculators yields: 1.90546  105 but wait… how many sig. figs. are allowed? 100.28 = 1.9 2 sig. figs.! therefore the concentration should be reported as:

  48. The pH Scale: 0 to 14 low pH = high [H+] acidic solution high pH = low [H+] basic solution The pH of a solution provides a way to express the acidity, or the concentration of H+ in solution: a pH of 7 indicates that the solution is neutral

  49. Solution Stoichiometry Prior to now, we have discussed reactions in solution from a qualitative aspect. Reactants  Products With the addition of molarity to our tools of chemistry, we now can perform quantitative calculations for reactions in aqueous solutions. volume  moles  moles  grams grams  moles  moles  volume

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