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Unit 3

Unit 3. Acids and Bases. Hydrogen ions and pH. Ion product constant of water (K w ) H 2 O  H + + OH - In pure water : [H + ] = [OH - ] * [ ] are used to indicate concentration in mol/L K w = [H + ] [OH - ] = 1.0x10 -14

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Unit 3

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  1. Unit 3 Acids and Bases

  2. Hydrogen ions and pH • Ion product constant of water (Kw) • H2O  H+ + OH- • In pure water : [H+ ] = [OH- ] * [ ] are used to indicate concentration in mol/L • Kw = [H+ ] [OH- ] = 1.0x10 -14 • If [H+ ] > 1.0 x10-7 the solution is ACIDIC • If [OH- ] > 1.0 x10-7 the solution is BASIC • If [H+ ] or [OH- ] = 1.0 x10-7 the solution is NEUTRAL

  3. pH and pOH • Values for concentrations of H+ and OH- are small so scientist use a scale based on common logarithms. • pH= - log [H+] * [H+] : concentration hydrogen ions (related to acids) [H+] = 10-pH • pOH= -log [OH-] * [OH-] : concentration hydroxide ions (related to bases) [OH-] = 10-pOH • pH +pOH=14

  4. pH and pOH • A pH scale from 0 to 14 is used. (It has no units) • If pH< 7: solution is acidic ; if pH=7: solution is neutral; if pH>7: solution is basic • If pOH<7: solution is basic; if pOH=7: solution is neutral; if pOH>7 : solution is acidic

  5. pH of Common Substances water (pure) 7.0 vinegar 2.8 soil 5.5 gastric juice 1.6 carbonated beverage 3.0 drinking water 7.2 bread 5.5 1.0 M NaOH (lye) 14.0 blood 7.4 potato 5.8 orange 3.5 1.0 M HCl 0 milk of magnesia 10.5 urine 6.0 apple juice 3.8 detergents 8.0 - 9.0 bile 8.0 lemon juice 2.2 milk 6.4 tomato 4.2 ammonia 11.0 seawater 8.5 bleach 12.0 coffee 5.0 13 11 12 14 1 6 9 10 0 2 3 4 7 5 8 basic neutral acidic [H+] = [OH-] Timberlake, Chemistry 7th Edition, page 335

  6. Example 1 What is the pH of a solution with [H+]=1.00x10-4 M? Is it acidic, basic, or neutral? Given : [H+]=1.00x10-4 M Unknown: pH pH = -log[H+] pH = -log[1.00x10-4] pH = 4.00 (acidic)

  7. Example 2 What is the pH and pOH of a solution with [ H+]= 0.00350mol /L? Given : [ H+]= 0.00350mol /L? Unknown: pH and pOH pH = -log[H+] pH = -log[0.00350] pH = 2.46 pH + pOH = 14 pOH = 14 – pH= 14 - 2.46 = 11.54

  8. Example 3 Given: pOH= 4.40 Unknown: [H+ ] and [OH- ] What is the [H+ ] and [OH- ] of a solution if pOH=4.40 [OH-] = 10-pOH [OH-] = 10-4.40 [OH-] = 3.98x10-5 M [H+] [OH-] = 1 x10-14 [H+] = 1 x10-14 = 1 x10-14 = 2.51x10-10M [OH-] 3.98x10-5

  9. classwork • Read p650-655 • Do problems p651#22 p653 #24-26 p655 #30-31

  10. Strengths of acids and bases • Strong acids and bases : ionize completely in aqueous solution. Examples: HCl, HNO3, H2SO4, KOH, NaOH • Weak acids and bases: ionize slightly in aqueous solution. • Examples: HClO, H3PO4 , NH3

  11. Reactions between acids and bases When and acid and a base react with each other it is called neutralization reaction. ACID + BASE→ SALT + WATER HCl + NaOH→ NaCl+ H2O H-OH Salt: ionic compound made up of cation of base and anion from acid.

  12. Ex. Write the neutralization reaction between hydroiodic acid and potassium hydroxide.

  13. Titration • Titration: adding a known amount of a solution of known concentration to determine the concentration of another solution. • Equivalence point: when number of moles of hydrogen ions equals the number of moles of hydroxide ions. • Standard solution: solution of known concentration • End point: point at which the indicator changes color • Point of neutralization is the end point of the titration. • M1V1= M2V2

  14. Ex. 1 How many milliliters of 0.45M HCl will neutralize 25.0 mL of 1.00M KOH? M1V1= M2V2 • mL HCl= 25.0mL x 1.00M KOH 0.45 M HCl • mL HCl= 55.6mL HCl

  15. Ex. 2 What is the molarity of a NaOH solution if 20.0 mL of the solution is neutralized by 28.0mL of 1.00 M HCl? M1V1= M2V2 • M NaOH= 1.00 M HCl x 28.0 mL 20.0 mL • M NaOH = 1.40 M • Classwork: p 989 #22-25

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