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Chapter 20 Section 2

Chapter 20 Section 2. self-ionization - the reaction in which two water molecules produce ions H 2 O + H 2 O  H 3 0 + + OH - hydroxide ion (OH - ) – what a water molecule that loses a hydrogen ion becomes

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Chapter 20 Section 2

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  1. Chapter 20 Section 2 self-ionization- the reaction in which two water molecules produce ions H2O + H2O  H30+ + OH- hydroxide ion (OH-) – what a water molecule that loses a hydrogen ion becomes hydronium ion (H30+) – what a water molecule that gains a hydrogen ion becomes

  2. -in pure water at 25°C the concentration of [H+] and [OH-] are 1.00 x 10-7mol/L * [H+] = [OH-] in pure water neutral solution- [H+] and [OH-] are equal -in any aqueous solution, as [H+] inc. [OH-] dec. and vice versa -in aqueous solutions: [H+] x [OH-] = 1.00 x 10-14M2

  3. Kw = ion product constant for water Kw = [H+] x [OH-] = 1.0 x 10-14 M2 [H+] = Kw/ [OH-] [OH-] = Kw/ [H+]

  4. -not all solutions are neutral some release H+ when dissolved in water ex- HCℓ  H+ + Cℓ- -such solutions have a greater [H+] than [OH-] acidic solution- [H+] is greater than [OH-] -so [H+] > 1.0 x 10-7 M -some solutions release [OH-] ex- NaOH  Na+ + OH- basic solution- [H+] is less than [OH-] -so [H+] < 1.0 x 10-7 M **looking at negative exponent (ex -7 < -2)

  5. -[H+] can also be expressed using pH scale pH = potential hydrogen -pH scale ranges from 0-14 -pH < 7 = acids 0= strongly acidic -pH > 7 = bases 14= strongly basic -pH = 7 = neutral solution **based on scales of 10 -page 584 Table 20.2

  6. pH = -log [H+] pOH = -log [OH-] pH + pOH = 14 Examples: [H+]= 1.0 x 10-7 pH = -log 1.0 x 10-7 pH = 7.00 neutral solution [H+]= 4.3 x 10-3 pH = -log 4.3 x 10-3 pH = 2.37 acidic solution

  7. To find [H+] or [OH-] from a pH or pOH use: [H+] = antilog - pH [OH-] = antilog –pOH antilog = 10x button Ex: pH = 3.40 Find [H+] [H+] = antilog -3.40 = 3.98 x 10-4M

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