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Acids and Bases Chapter 15

Acids and Bases Chapter 15. Notes One Unit Twelve. Properties of Acids Properties of Bases Structure of Bases Neutralization Reactions. Properties of Acids. Sour taste. warhead React with “active” metals produce H 2 . Al, Zn, and Fe React with carbonates, producing CO 2 and H 2 O.

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Acids and Bases Chapter 15

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  1. Acids and Bases Chapter 15

  2. Notes One Unit Twelve Properties of Acids Properties of Bases Structure of Bases Neutralization Reactions

  3. Properties of Acids Sour taste. warhead React with “active” metals produce H2.Al, Zn, and Fe React with carbonates, producing CO2 and H2O. Marble, baking soda, chalk, limestone. Change color of vegetable dyes. cabbage juice / On the top cabbage juice / baking soda (left) cabbage juice / vinegar (right).

  4. Acids acid - proton donor (H+1) strong vs. weak acids strong - lots of (H+ ) weak - very little (H+ )

  5. HCl is a strong acid Muriatic acid is hydrochloric acid. Making Bleach. Making PVC pipe. Making Table Salt. Humanstomach acid. Cleaning steel. Neutralize bases in chemical plants. Chrome tanning leather.

  6. HNO3 is a strong acid Explosives Fertilizers etc. Nitrate salts To make H2SO4 Etching copper, brass, bronze Dyes, perfumes Purification of Ag, Au, Pt

  7. Properties of Bases Also known as alkali.( Li, Na, K, Rb, Cs, Fr) Taste bitter. caffeine often poisonous. Solutions feel slippery. Change color of vegetable dyes. Different color than acid. Red litmus turns blue. React with acids to form salt and water(Neutralization). Acid + base salt +water

  8. Common Base Features Most ionic bases contain OH-1 ions. NaOH (drain cleaner) Some contain CO32-ions. CaCO3(in Tums) NaHCO3 (baking soda) Molecular bases contain structures that react with H+. Mostly amine groups(-NH2).

  9. Neutralization Reactions acid + base  salt + water Double-displacement CO2 and H2O Some make

  10. Acid Reactions Acid+carbonate Salt +Water +Carbon Dioxide Acid+Base Salt +Water Acid+Metal Salt +Hydrogen

  11. Acid and Carbonate Carbonate Water+ Carbon Dioxide Acid+ Salt+ 2 HNO3+ 1 Na2CO3  2 1 H2O + 1 CO2 NaNO3 + H+1 NO3-1 Na+1 CO3-2 Salt ? =Sodium Nitrate ( )_( )_ 1 1 Na+1 NO3-1 NaNO3

  12. Acid and Base Base  Water Acid+ Salt + 1 H2SO4+ 2 KOH  1 2 H2O K2SO4 + H+1 SO4-2 K+1 OH-1 Salt ? =Potassium Sulfate ( )_( )_ 2 1 K+1 SO4-2 K2SO4

  13. Acid and Metal Metal  Hydrogen Salt + Acid+ 1 H2SO4 + 1 Mg(s) 1 1 H2(g) MgSO4 + H+1 SO4-2 Mg(s) = Magnesium Sulfate Salt ? ( )_( )_ 2 2 Mg+2 SO4-2 MgSO4

  14. Titration Example One What is the molarity of a NaOH solution, if 12.01 mL is required to titrate 5.99 mL of 0.100 M HCl? 1. Balance the equation. 1 NaOH(aq)+ 1 HCl(aq) 1 1 NaCl(aq) + H2O(l) 2. Find moles used of known solution. MxV=n 0.100M x 0.00599L = 0.000599m HCl 3. Calculate moles used of unknown (titrant). 1 m NaOH = 0.000599m NaOH 0.000599 m HCl x 1 m HCl 4. Calculate the molar concentration of the titrant. n/V=M = 0.0499M NaOH ÷ 0.01201L 0.000599m NaOH

  15. Notes Two Unit Twelve-Text Pages 550-558 • Self Ionization of Water • Brönsted-Lowrey Acid-Base Theory • Arrhenius Theory

  16. Self Ionization of Water • Water is and acid and a base at the same time…amphoteric • H2O(l) + H2O(l)  H3O+ + OH- • Mass Action Expression • Kw = [H3O+][OH-] • Or Kw = [H+][OH-] • Kw = 1.0 x 10-14

  17. Brönsted-Lowrey Acid-Base Theory • Acid - proton donor • Base - proton acceptor • Acid-Conjugate Base / Base-Conjugate Acid • HCl(aq) + H2O(l)  Cl−1(aq) + H3O+1(aq) • NaF(aq) + H2O(l)  HF(aq) + NaOH(aq) • OH−1(aq) + H2O(l)  H2O(l)+ OH−1(aq) • NH3(aq) + H2O(l)  NH4+1(aq) + OH−1(aq) H+1 A B CA CB CB B A CA B A CA CB A CA CB B

  18. Arrhenius Theory • Bases form OH- ions in water. • Acids form H+ions in water.

  19. Arrhenius theory • HCl(aq)  H+1(aq) + Cl−1(aq) • HF(aq)  H+1(aq) + F−1(aq) • NaOH(aq)  Na+1(aq) + OH−1(aq) • NH3(aq) +H2O(aq) NH4+1(aq) +OH−1(aq)

  20. Polyprotic Acids • More than one proton to donate. • H2CO3(aq)   H+1(aq) + HCO3-1(aq) • HCO3-1(aq)       H+1(aq)    +    CO3-2(aq) • H2SO4 H+1(aq) + HSO4−1(aq) • HSO4−1 H+1(aq) + SO4−2(aq)

  21. Notes Three Unit Twelve • Titration

  22. Titration • Titration is a technique to determine the concentration of an unknown solution. • Titrant (unknown solution) • Phenolphthalein identifiesthe Endpoint.

  23. Titration Endpoint • Add 10mL of HCl and three drops phenolphthalein to the flask. • Add about 8mL base, swirl and add the rest of the base using increasingly faster spins of the valve.(?????)

  24. Titration-Acid Volume • Acid Burette • Initial Reading? • 1.98mL • Final Reading? • 7.97mL • Volume Used? • 5.99mL

  25. Titration-Volume of Base Used. • Base Burette • Initial Reading? • 0.00mL • Final Reading? • 12.01mL • Volume Used? • 12.01mL

  26. Titration Example Two • Lactic acid Concentration of Sauerkraut • For Joe’s final in chemistry, he was asked to find the concentration of lactic acid in homemade sauerkraut. • He did not have any home made sauerkraut. • Therefore, Joe was left to make the home made sauerkraut. • He looked on line and found the following recipe.

  27. Clean and Quarter 35lb of Fresh Cabbage

  28. Shred the Cabbage into a Crock

  29. Add 3 Tbsp salt per 5 pounds of cabbage.

  30. Mix salt and cabbage.

  31. Pack the cabbage into a crock and weight it down. It should be fermented in one month.

  32. Titration Example Two What Is the molarity of a sauerkraut juice if 10.0 mL is titrated using 89.9 mL of 1.00 M NaOH? 1. Balance the equation. 1 NaOH + 1 HC2H4OHCO2 1 1 NaC2H4OHCO2 + H2O 2. Find moles used of known solution. MxV=n 1.00M x 0.0899L = 0.0899m NaOH 3. Calculate moles used of unknown (titrant). 1m HC2H4OHCO2 0.0899m HC2H4OHCO2 = 0.0899m NaOHx 1 m NaOH 4. Calculate the Molar Concentration of theTitrant. n/V=M M = ÷ 0.01201L 0.0899m HC2H4OHCO2 M=0.0899M HC2H4OHCO2

  33. Titration Example Three What is the volume of a 0.0622M Ba(OH)2 solution, if it is titrated using 43.8 mL of 0.1057 M HCl? 1. Balance the equation. 1 Ba(OH)2(q) + 2 HCl(aq 1 2 BaCl2(aq)+ H2O(l) 2. Find moles used of known solution. MxV=n 0.00463m HCl 0.1057M x 0.0438L= 3. Calculate moles used of unknown (titrant). 1 m Ba(OH)2 = 0.00232 m Ba(OH)2 0.00463 m HCl x 2 m HCl 4. Calculate volume of titrant. n/M=V = 0.0376M Ba(OH)2 ÷ 0.0622M 0.00234m Ba(OH)2

  34. Conclusions continued • 2. Define these terms: standard solution; titration; endpoint. • Standard Solution: When the concentration of a solution is known to a high degree of accuracy and precision. • titration:  When the concentration of an acid or base is determined by neutralizing it. • endpoint:  The point where you actually stop a titration, usually because an indicator has changed color.  This is different than the "equivalence point" because the indicator might not change colors at the exact instant that the solution is neutral.

  35. Notes Three Unit Twelve Text Pages 559-567

  36. Weak Acids and Bases • A weak acid • little H+1 • A weak base • little OH-1 • [H+] or [OH-] from a Keq.

  37. pH = -log[H+1] pH is a measure of the amount of hydrogen in a solution. It is based on the water. Stomach Acid pH Scale Vinegar Blood 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Neutral Acid Base 1M NaOH 1M HCl Ammonia Cleaner Lemon Juice Milk Water pH +pOH =14

  38. pH of strong acid • Find the pH of a 0.15 M solution of Hydrochloric acid • pH = - log 0.15 • pH = - (- 0.82) • pH = 0.82

  39. pH of a strong Base • What is the pH of the 0.0010 M NaOH solution? • pOH = - log (0.0010) • pOH = 3 • pH = 14 – 3 = 11 pH +pOH =14

  40. Ionization Constants for Acids/Bases

  41. Calculating pH (pH=-log[H+1]) The Ka for nitrous acid is 4.5 x 10-4. Calculate the [H+1] and pH in 0.15 M nitrous acid solution. 1) Balanced Equation HNO2 NO2-1 + H+1 [NO2-1] [H+1] 2) Mass Action Expression Ka= [HNO2] 3) What do we know? 0.15 0 0 4.5 x 10-4 Very small +x +x -x 1.5 x 10-1 x 0.15-x x 4) Calculate the [H+1] and pH. [x] [x] pH=-log[H+1] Ka= = 4.5 x 10-4 [0.15] pH=-log[ ] 0.0082M =2.09 =[H+1] X= [0.0082M]

  42. Calculating Concentrations Using Ka The Ka for benzoic acid is 6.5 x 10-5. (a) Calculate the concentrations of C6H5COO-1 and H+ in a 0.10 M benzoic acid solution. (b) Calculate pH. 1) Balanced Equation C6H5COOH C6H5COO-1 + H+1 [C6H5COO-1] [H+1] 2) Mass Action Expression Ka= [C6H5COOH] 3) What do we know? 6.5 x 10-5 0.10 0 0 Very small 1.0 x 10-1 +x +x -x x 0.10-x x 4) Calculate the [H+1] and pH. X= [0.0025M] =[H+1] [x] [x] Ka= = 6.5 x 10-5 pH=-log[ ] =2.60 0.0025M [0.10]

  43. End

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