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Lecture 2

# Lecture 2 - Water Ionization and the pH scale K eq =[H2 O =...

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1 H 2 O H + + OH - [H 2 O] = 55.5 M [H + ] = [OH - ] = 1 × 10 -7 M Water Ionization and the pH scale pH scale K W = [H + ][OH - ] ; water ion product 14 = -log [H + ] + -log[OH - ] ; Let p = -log -log K W = -log [H + ][OH - ] ; take -log and expand. 14 = pH + pOH The pH scale Since [H + ] = [OH - ] and both are << [H 2 O], we can treat [H 2 O] as a constant. Thus, (water ion product) K eq [ H 2 O ] = [ H + ][ OH " ] K W = [ H + ][ OH " ] = 1.0 # 10 " 14 M 2 K eq = [ H + ][ OH " ] [ H 2 O ] # 1.8 \$ 10 " 16 M Water You should know how to determine the pH of 0.1 M HCl based on first principles. 14 10 -14 1 14 = pH + pOH 1) 2) pH of pure water is 7 H 2 O H + + OH - Strong Acids Dissociate completely 3) HCl H + + Cl - So, what is the pH of 0.1M NaOH? HA H + + A - Henderson-Hasselbalch Equation Weak Acid The all important p K a Acid dissociation constant Take log of both sides Expand Solve for [H + ] Midpoint (p K a =4.8 ) OH - 2H 2 O Acid Base Conjugate Base Conjugate Acid CH 3 COOH + H 2 O CH 3 COO - + H 3 O + Titrating Acetic Acid p K a determines buffering range (p K a ± 1 pH unit) pK a + log 10 pK a + log 0.1 pH Equivalents of OH - 1) Given a pKa, sketch the titration curve. 2) Circle pKa on graph 3) To the left or right of the pKa, write the dominant species (HA or A - ) 5) Report the buffering range. What is the pK a of water? pK a = pH - log( [OH - ]/[H 2 O] ) ; Rearranged the H-H Eq. Values needed to solve the equation: [Water] is ~55.5 M pH of water is 7, so [OH - ] is 1 X 10 -7 M pK a of water is 15.7! What does the “pK a of water is 15.7” really mean? Water rarely dissociates to donate a proton. In pure water only 1 in 555,000,000 water molecules is ionized to H + and OH - (55.5 M / 1 X 10 -7 M).

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2 pK a tells how strongly the acid/base wants the proton Methyl Chloride pK a predicts the leaving group.
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