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Ionization of water

Ionization of water. HOH( l) H + (aq) + OH - (aq) The reverse reaction is highly favored. Water at 25 ◦ C : [H + ] = 10 -7 moles/L [OH - ] = 10 -7 moles/L Since [H + ] = [OH - ], water is neutral [H + ] ∙ [OH - ] = equilibrium constant

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Ionization of water

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  1. Ionization of water HOH(l) H+(aq) + OH-(aq) The reverse reaction is highly favored.

  2. Water at 25◦C : [H+] = 10-7moles/L [OH-] = 10-7moles/L Since [H+] = [OH-], water is neutral [H+] ∙ [OH-] = equilibrium constant = 10-14 M2 If one concentration goes up, the other must go down.

  3. [H+] = 10-pH [OH-] = 10-pOH pH + pOH = 14.0 Sample problem: A 0.01 M sample of NaOH completely ionizes. What is its pH, pOH, [H+], and [OH-] ? [OH-] = 0.01 mole/L = 10-2 pOH = 2 pH = 14 – pOH = 12 [H+] = 10-12

  4. pH = -log[H3O+] Find the pH of a solution with [H3O+] of 6.59 x 10-10M. pH = -log(6.59 x 10-10) = 9.18

  5. pH  [H+ ] antilog (-pH) • [H+ ]  pH -log [H+ ]

  6. pKa • pKa = -log Ka

  7. Indicators • large organic molecules that behave as weak acids. • HIn(aq) + H2O D In-(aq) + H3O+ (color A) (color B) • The acid (HIn) and conjugate base (In-) are different colors.

  8. Titration • Standard solution- one whose concentration is known • Endpoint- the point at which equivalent amounts of reactants are present. • M∙V = moles • MaVa=MbVb

  9. The endpoint of a titration is reached when 22.3ml of 0.240M NaOH reacts with 50.0ml of HC2H3O2. What is the concentration of HC2H3O2? HC2H3O2 + NaOH g H2O + NaC2H3O2 .0223L x 0.240mole= 0.00535 mol NaOH L 0.00535mol= 0.107 M HC2H3O2 0.050L

  10. Buffers • Can absorb acids and bases without changing its pH • A weak acid (HA) with its salt (NaA) • HA + OH-D H2O + A- • A- + H3O+D HA + H2O • A weak base (MOH) with its salt (MCl) • MOH + H3O+D M+ + 2H2O • M+ + OH-D MOH

  11. HF(aq) D H+(aq) + F-(aq) Buffered solution is mostly HF and F-. If H+ is added: H+ + F- g HF If OH- is added: OH- + HF g F- + H2O

  12. Calculating pH of a buffer • What is the pH of a buffer that is 0.12M lactic acid (HC3H5O3) and 0.10M sodium lactate? • Ka = 1.4 x 10-4  -x +x +x .12-x x .10+x

  13. (like “common ion” problem) • Ka= [H+][C3H5O3-]=x(.10+x) [HC3H5O3] (.12-x) 1.4 x 10-4= x(.10) .12 X = 1.7 x 10-4M pH = -log(1.7 x 10-4) = 3.77

  14. Henderson-Hasselbalch Conjugate base of the acid • pH = pKa + log [base] [acid] = -log(1.4x10-4) + log .10 .12 = 3.85 + (-.079) = 3.77

  15. Weak acid + strong base also creates a buffer • CH3COOH + NaOH CH3COO- + H2O + Na+ • Moles of acid anion equals the moles of strong base added. • Moles of weak acid left is original amount minus moles of strong base

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