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Section 18-3 Section 18.3 Hydrogen Ions and pH Explain pH and pOH. Relate pH and pOH to the ion product constant for water. Calculate the pH and pOH of.

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Presentation on theme: "Section 18-3 Section 18.3 Hydrogen Ions and pH Explain pH and pOH. Relate pH and pOH to the ion product constant for water. Calculate the pH and pOH of."— Presentation transcript:

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2 Section 18-3 Section 18.3 Hydrogen Ions and pH Explain pH and pOH. Relate pH and pOH to the ion product constant for water. Calculate the pH and pOH of aqueous solutions. pH and pOH are logarithmic scales that express the concentrations of hydrogen ions and hydroxide ions in aqueous solutions.

3 Acids Have a pH less than 7 Bases have a pH greater than 7 The pH Scale

4 Section 18-3 pH and pOH (cont.) Litmus paper and a pH meter with electrodes can determine the pH of a solution.

5 Self-Ionization of Water H 2 O + H 2 O  H 3 O + + OH - At 25 , [H 3 O + ] = [OH - ] = 1 x 10 -7 K w is a constant at 25  C: K w = [H 3 O + ][OH - ] K w = (1 x 10 -7 )(1 x 10 -7 ) = 1 x 10 -14

6 Section 18-3 pH and pOH Concentrations of H + ions are often small numbers expressed in exponential notation. pH is the negative logarithm of the hydrogen ion concentration of a solution. pH = –log [H + ]pH

7 Section 18-3 pH and pOH (cont.) pOH of a solution is the negative logarithm of the hydroxide ion concentration.pOH pOH = –log [OH – ] The sum of pH and pOH equals 14.

8 Calculating pH, pOH pH = -log 10 [H 3 O + ] pOH = -log 10 [OH - ] Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH

9 Calculating pH from solution concentration…

10 Section 18-3 pH and pOH (cont.) For all strong monoprotic acids, the concentration of the acid is the concentration of H + ions. For all strong bases, the concentration of the OH – ions available is the concentration of OH –. Weak acids and weak bases only partially ionize and K a and K b values must be used.

11 Dissociation Constants: Strong Acids

12 Dissociation Constants: Weak Acids

13 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #1: Write the dissociation equation HC 2 H 3 O 2  C 2 H 3 O 2 - + H +

14 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #2: ICE it! HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 0 0 - x +x 0.50 - x xx

15 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #3: Set up the law of mass action HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

16 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #4: Solve for x, which is also [H + ] HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E [H + ] = 3.0 x 10 -3 M

17 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #5: Convert [H + ] to pH HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

18 Dissociation of Strong Bases  Strong bases are metallic hydroxides  Group I hydroxides (NaOH, KOH) are very soluble  Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble  pH of strong bases is calculated directly from the concentration of the base in solution MOH(s)  M + (aq) + OH - (aq)

19 Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH - K b = 4.38 x 10 -4

20 K b for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

21 Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O  BH + + OH - (All weak bases do this.)

22 A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #1: Write the equation for the reaction NH 3 + H 2 O  NH 4 + + OH -

23 A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #2: ICE it! 0.500 0 - x +x 0.50 - xxx NH 3 + H 2 O  NH 4 + + OH -

24 A Weak Base Equilibrium Problem Step #3: Set up the law of mass action 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

25 A Weak Base Equilibrium Problem Step #4: Solve for x, which is also [OH - ] 0.50 - xxx E [OH - ] = 3.0 x 10 -3 M NH 3 + H 2 O  NH 4 + + OH - What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ?

26 A Weak Base Equilibrium Problem Step #5: Convert [OH - ] to pH 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

27 A.A B.B C.C D.D Section 18-3 Section 18.3 Assessment In dilute aqueous solution, as [H + ] increases: A.pH decreases B.pOH increases C.[OH – ] decreases D.all of the above

28 A.A B.B C.C D.D Section 18-3 Section 18.3 Assessment What is the pH of a neutral solution such as pure water? A.0 B.7 C.14 D.1.0 × 10 –14


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