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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.

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Presentation on theme: "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."— Presentation transcript:

1 Unit 3 Acids and Bases

2 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 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 A pH scale from 0 to 14 is used. (It has no units) If pH 7: solution is basic If pOH 7 : solution is acidic pH and pOH

5 pH of Common Substances Timberlake, Chemistry 7 th Edition, page 335 1.0 M HCl 0 gastric juice 1.6 vinegar 2.8 carbonated beverage 3.0 orange 3.5 apple juice 3.8 tomato 4.2 lemon juice 2.2 coffee 5.0 bread 5.5 soil 5.5 potato 5.8 urine 6.0 milk 6.4 water (pure) 7.0 drinking water 7.2 blood 7.4 detergents 8.0 - 9.0 bile 8.0 seawater 8.5 milk of magnesia 10.5 ammonia 11.0 bleach 12.0 1.0 M NaOH (lye) 14.0 8 910 111214 13 34 5 6 2 1 70 acidic neutral basic [H + ] = [OH - ]

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 What is the [H + ] and [OH - ] of a solution if pOH=4.40 Given: pOH= 4.40 Unknown: [H + ] and [OH - ] [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 -10 M [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, HNO 3, H 2 SO 4, KOH, NaOH Weak acids and bases: ionize slightly in aqueous solution. Examples: HClO, H 3 PO 4, NH 3

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 HOH HCl + NaOH → NaCl + H 2 O 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. M 1 V 1 = M 2 V 2

14 Ex. 1 How many milliliters of 0.45M HCl will neutralize 25.0 mL of 1.00M KOH? mL HCl= 25.0mL x 1.00M KOH 0.45 M HCl mL HCl= 55.6mL HCl M 1 V 1 = M 2 V 2

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? M NaOH= 1.00 M HCl x 28.0 mL 20.0 mL M NaOH = 1.40 M Classwork: p 989 #22-25 M 1 V 1 = M 2 V 2


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