Chapter 15 Acids and Bases Week 2

Chapter 15 Acids and Bases Week 2 - Calculations involving...

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Calculations involving Weak Bases or Salts containing Weak Bases This will involve molecules containing a N atom with a lone pair: NH 3 , RNH 2 , etc, or salts of weak acids e.g. Na(CH 3 COO) (sodium acetate) Remember : B (aq) + H 2 O (l) BH + (aq) + OH - (aq) “base-dissociation reaction” or “base-ionization reaction” ( NOTE: Base doesn’t really dissociate). K b = base dissociation (ionization) constant K b = [B] ] ][OH [BH - + ([H 2 O] is a constant and contained within K b ) e.g. NH 3 (aq) + H 2 O (aq) NH 4 + (aq) + OH - (aq) K b = ] [NH ] ][OH [NH 3 - 4 + = 1.8 x 10 -5 **Solve calculations similarly to weak acids!** Example 1 . What is K b for quinine (anti-malarial drug) if the pH of a 1.5 x 10 -3 M solution is 9.84? ( Note: pH > 7.00 quinine is a base) Qui (aq) + H 2 O (aq) HQui + (aq) + OH - (aq) [init] 1.5 x 10 -3 M O O [change] -x +x +x [equil] (1.5 x 10 -3 )-x x x K b = [Qui] ] ][OH [HQui - + = x ) 10 x 5 . 1 ( x 3 2 - - 3 2 10 x 5 . 1 x - pH = 9.84 pOH = 4.16 [OH - ] = x = 6.9 x 10 -5 M ( Note : x = [OH - ] << 1.5 x10 -3 (4.6%) assumption okay) 7
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K b = 3 - 2 -5 10 x 1.5 ) 10 x (6.9 = 3.1 x 10 -6 Example 2 . Calculate the pH of a 15.0 M solution of NH 3 (K b = 1.8 x 10 -5 )? NH 3 (aq) + H 2 O (aq) NH 4 + (aq) + OH - (aq) [init] 15.0 M O O [change] -x +x +x [equil] 15.0-x x x K b = ] NH [ ] OH ][ NH [ 3 4 - + = ) x 0 . 15 ( x 2 - 0 . 15 x 2 = 1.8 x 10 -5 x 2 = 2.7 x 10 -4 x = 1.6 x 10 -2 (<< 15.0 assumption okay) [OH - ] = 1.6 x 10 -2 M pOH = 1.80 pH = 14.00-1.80 = 12.20 NOTE: 1) Remember: pK b = -log K b 2) Just like pK a , strength of a base increases with decreasing pK b The bigger is K b , the more OH - is generated. The smaller is pK b , the more OH - is generated. 8
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Example 3. What is the pH of a 2.5 M solution of sodium acetate (NaAc)? K a for acetic acid (HAc) is 1.8 x 10 -5 . acetate (Ac - = CH 3 COO - ) is the conjugate base of acetic acid (HAc = CH 3 COOH). Ac - (aq) + H 2 O(l) HAc(aq) + OH - (aq) [init] 2.5 M O O [change] -x +x +x [equil] (2.5-x) x x K b = ] Ac [ ] OH ][ HAc [ - - = x) - (2.5 x 2 2.5 x 2 (assume x << 2.5) We need K b . We were given K a . K b = a w K K = 5 - -14 10 x 1.8 10 x 1.0 = 5.55 x 10 -10 5 . 2 x 2 = 5.55 x 10 -10 x = 3.72 x 10 -5 = 3.7 x 10 -5 M (assumption okay!) K b pK b Diethylamine NHEt 2 8.6 x 10 -4 3.07 Methylamine NH 2 Me 4.4 x 10 -4 3.36 Ammonia NH 3 1.8 x 10 -5 4.74 Pyridine NC 6 H
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This note was uploaded on 05/30/2010 for the course CHM 2046 taught by Professor Veige/martin during the Spring '07 term at University of Florida.

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Chapter 15 Acids and Bases Week 2 - Calculations involving...

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